WBCHSE Class 12 Physics Notes For Refraction Of Light

Refraction Of Light

Refraction Of Light Definition:

When a ray of light travelling in one medium enters another medium obliquely, the ray changes its direction at the interface. This phenomenon is known as the refraction of light. CD be the plane of separation of the media—air and glass

It is called a refracting surface. The ray AO incident obliquely at 0, changes direction after refraction and goes along the line OB. NON’ is drawn perpendicular to the surface of separation. The perpendicular to CD, NON’ is called normal. AO is the incident ray and OB is the refracted ray. The angle between the incident ray and the normal to the surface of separation at the point of incidence is called the angle of incidence

The angle between the refracted ray and the normal to the surface of separation is called the angle of refraction (r).

Refraction from rarer to denser medium:

If the ray travels from an optically rarer medium to a denser medium, say from air to glass, the refracted ray bends towards the normal. Here i>r.

Refraction from denser to the rarer medium:

If the ray travels from denser to rarer medium, say from glass to air, the refracted ray bends away from the normal Here r>i.

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Refraction Of Light Laws Of Refraction

Refraction of light is governed by the following two laws, called
laws of refraction.

1. The incident ray, the refracted ray and the normal at the point of incidence lie on the same plane.
2. The ratio of the sine of the angle of Incidence to the sine of the angle of refraction is constant.
3. This constant depends on the nature of the two concerned media and the colour of the ray used.

This second law of refraction is known as Snell’s law, named after Willebrord Snellius (1580-1626)

Refractive Index

If i is the die angle of incidence and r is the angle of refraction, then according to the second law,

⇒ \(\frac{\sin i}{\sin r}=\mu\) (pronounced as ‘mu’) = constant.

This constant is called the refractive index of the second medium concerning the first medium.

The value of the refractive index depends on:

1. Nature of the two concerned media and
2. Colour of the Incident light.

Whatever may be the value of the angle of incidence, the value of the refractive index will remain constant if the colour of the incident light (i.e., frequency) and the two media remain unchanged

Μ has no uni

Normal Incidence:

If a ray of light is incident perpendicularly on a refracting surface, then / = 0. According to Snell’s law

π sinr = sin 0 = 0 or, r = 0

So, the ray suffers no deviation

Relative refractive index:

When light passes from medium a into another medium b, the ratio of the sine of the angle of incidence to the sine of the angle of refraction is called the refractive index of medium b concerning medium a. It is denoted by and i.e.,

_aμ_b=  \(\frac{\sin i}{\sin r}\)

[i = angle of incidence, r = angle of refraction]

This refractive index is called the relative refractive index. According to the principle of reversibility of light—a ray of light will follow the same path if its direction of travel is reversed.

Following this principle, we can say that the ray BO in medium b when incidents at an angle r on the interface of the second medium a, refracts at an angle i along the path OA. Comparing this figure with it is the opposite phenomenon.

Thus, \(b^{\mu_a}=\frac{\sin r}{\sin i}\)

Here _bμ_a is the refractive index of medium a concerning medium b.

So, \(a_a \mu_b \times{ }_b \mu_a=\frac{\sin i}{\sin r} \times \frac{\sin r}{\sin i}\)

= 1

Or, _aμ_b = \(\frac{1}{b^\mu}\)

For example, if the refractive index of water concerning air is \(\frac{4}{3}\), then the refractive index of air concerning water is\(\frac{3}{4}\)

Absolute refractive index:

When light is refracted from a vacuum to another medium, the ratio of, the sine of the angle of incidence to the sine of the angle of refraction is called the absolute refractive index of the medium

If the angle of incidence is i and the angle of refraction is r, then the absolute refractive index of the medium,

⇒ \(\mu=\frac{\sin i}{\sin r}\)

Therefore, the relative refractive index of a medium concerning a vacuum is the absolute refractive index of that medium. The refractive index of a vacuum is 1.

In general, the refractive index of a medium relative to an air medium is considered as the refractive index of that medium. But it is not absolute, 1 refractive index of the medium

It is an experimental fact that the difference in the values of the refractive index of a medium concerning air and its absolute refractive index is very small. For example, at STP, the absolute refractive index of air is 1.0002918. So the refractive index of any medium concerning air may be considered as its absolute refractive index

For example, the refractive index of glass is 1.5 which means that the refractive index of glass concerning air is 1.5. At STP, the absolute refractive index of air is 1.0002918 and the refractive index of glass concerning air is 1.5. Thus, the absolute refractive index of glass = \(\frac{1.5}{1.0002918}\) = 49956 ≈1.5 The Absolute refractive index of a medium is denoted by μ. If there is more than one medium μ₁, μ₂, μ₃ then is used.

Optical Density of a Medium

If the absolute refractive index (μ₁) of any medium is greater than that of another medium (μ₂), then the first medium is called optically denser and the second medium is called optically rarer. So, μ₁>μ₂ if then medium 1 is optically denser concerning medium 2 i.e., medium 2, is optically rarer concerning medium 1.

The optical density of a medium has no relation with its physical density or specific gravity. For example, the specific gravity of turpentine oil is 0.87 and that of water is 1. But the refractive index of turpentine oil is 1.47 and that of water is 1.33.

The refractive index of a medium depends on the colour of the incident light. It is greater for blue or violet than for red. The refracted ray in the case of violet light bends more than in the case of red light. The refractive index for yellow light is midway between these two.

So unless otherwise stated, the refractive index of a medium refers to yellow light.

Refractive indices of a few substances:

Refractive Index and Related Terms

Relation between the velocity of light and refractive index:

According to the wave theory of light, the velocity of light is different in different media. If pt is the absolute refractive index of a medium, then

μ = \(\frac{\text { velocity of light in vacuum }(c)}{\text { velocity of light in that medium }(v)}\)

∴ Speed of light in free space = ft x speed of light in that medium

For any medium μ > 1, so speed of light in a vacuum is greater than that in any medium. Thus, the speed of light is maximum in a vacuum.

Accordingly, if and is the refractive index of the medium b concerning medium a, then

⇒ \(a^{\mu_b}=\frac{\text { velocity of light in medium } a}{\text { velocity of light in medium } b}\)

= \(\frac{v_a}{v_b}\)

Medium b is denser than medium a then _aμ_b > 1 and in this case v_a>v_b.

The velocity of light in a denser medium is less than the velocity of light in a rarer medium. The velocity of light in a medium decreases with the increase of its refractive index.

Thus if the velocity of light in medium b is lesser, i.e., medium b is optically denser, and _aμ_b > 1 i.e., sin i>sinr or i< r. Thus, the angle of refraction is less than the angle of incidence, so the refracted ray bends towards the normal.

From the above discussion, it is clear that refraction variation in the speed of light in different media

Relation between relative refractive index and absolute refractive index:

If μ _a and μ_b are absolute refractive indices of media a and b respectively

⇒ \(\mu_a=\frac{c}{v_a} \text { and } \mu_b=\frac{c}{v_b}\)

So relative refractive index of medium b concerning medium a,

⇒ \({ }_a=\frac{v_a}{v_b}=\frac{\frac{c}{v_b}}{\frac{c}{v_a}}=\frac{\mu_b}{\mu_a}\)

Relation of the wavelength of light with refractive index:

The relative refractive index between two media depends on the wavelength of light. Cauchy’s equation for the dependence of refractive index on the wavelength of light is

⇒ \(\mu=A+\frac{B}{\lambda^2}\)

Here A and B are two constants; their values are different in different media. The refractive index of a medium decreases if the wavelength of light increases. The graph shows the variation of n with μ  with λ of BK7 glass.

In a medium apart from free space different coloured light travels at different speeds. In a particular medium red travels the fastest and violet travels the slowest. Thus refractive index of any medium for the red colour is the lowest and that for the violet colour is the highest. This is the reason why white light is dispersed

Relation of temperature with refractive index:

Generally, the refractive index of a medium decreases if the temperature of the medium increases. For a solid medium this change is small, for a liquid it is moderate and for a gas it is remarkable

It is to be remembered that, velocity, intensity and wave¬ length of light change due to refraction but its frequency and phase remain unchanged

Generalised Form of Snell’s Law

Let AB be the surface of the separation of two media 1 and 2. Medium 2 is denser and medium 1 is rarer. PO is the incident ray at the point O on the surface of separation and OQ is A the refracted ray at the point O . Let angle of incidence = i₁ angle of refraction = i₂

By Snells law \(\frac{\sin l_1}{\sin i_2}\)

We known \({ }_1 \mu_2=\frac{\mu_2}{\mu_1}\)

∴ \(\begin{equation}\frac{\sin i_1}{\sin i_2}=\frac{\mu_2}{\mu_1} \quad \text { or, } \mu_1 \sin i_1=\mu_2 \sin i_2\end{equation}\) …………………………. (1)

So for n number of media, it can be written as

\(\mu_1 \sin i_1=\mu_2 \sin i_2=\cdots=\mu_n \sin i_n\) …………………. (2)

This equation is known as the generalised form ofSnell’s law

Refraction Of Light Laws Of Refraction Numerical Examples

Example 1. A ray of light is incident from water on the surface of separation of air and water at an angle of 30°. Calculate the angle of refraction in air mu \(\frac{4}{3}\)
Solution:

Let the angle of refraction of the ray of light in the air be r

Since the ray of light is refracted from water to air,

⇒ \(w^\mu{ }_a=\frac{\sin i}{\sin r}\)

Or, \(\frac{1}{a^{\mu_w}}=\frac{\sin 30^{\circ}}{\sin r}\)

Or, \(\frac{1}{\frac{4}{3}}=\frac{1}{2 \sin r}\)

Or, \(2 \sin r=\frac{4}{3}\)

Or, \(\sin r=\frac{2}{3}\)

= 0.666

= sin 41.8°

or = r= 41.8°

Example 2. A ray of light is incident on a block of glass in such a way that the angle between the refracted ray and the refracted ray is 90°. Determine the relation between the angle of incidence refractive index of
Solution:

Here angle of incidence = i

The angle of reflection = i, angle of refraction = r

The angle between the reflected ray and the refracted ray = 90°

According to the

i+90°+r= 180°

Or, r = 90°-i

The refractive index of glass,

μ = \(\mu=\frac{\sin i}{\sin r}=\frac{\sin i}{\sin \left(90^{\circ}-i\right)}\)

= \(\frac{\sin i}{\cos i}\)

= tan i

Example 3.  The refractive index of glass is 1.5 and the refractive index of water is 1.33. If the velocity of light in glass is 2 × 10⁸ m s^–1 what is the velocity of light In water?
Solution:

μ_g = \(\frac{\text { velocity of light in vacuum }}{\text { velocity of light in glass }\left(v_g\right)}\)

Or, velocity light in a vacuum

Again, \(\mu_w=\frac{\text { velocity of light in vacuum }}{\text { velocity of light in water }\left(v_w\right)}\)

Or, Velocity of light in vacuum = μ_w. v_w

μ_g v_g= μ_w v_w

Or, \(\frac{\mu_g v_g}{\mu_w}=\frac{1.5 \times 2 \times 10^8}{1.33}\)

= 2.26x × 10⁸ m s^–1

Example 4. A monochromatic -ray of light-, is refracted from the vacuum. to a medium of refractive index pi. Determine the relation of the wavelengths of light in vacuum and in glass
Solution:

μ = \(\frac{c}{v}=\frac{n \lambda_0}{n \lambda}=\frac{\lambda_0}{\lambda}\)

So, λ₀ = μ λ

[Here, c and v are the velocities of light in vacuum and the medium respectively; n = frequency of light, which remains unchanged on refraction λ₀, λ = wavelengths in vacuum and in the medium respectively.]

Example 5. If a ray of light is incident on a plate inside the water at an angle of 45°, what is the angle of refraction inside the plate? Given that the absolute refractive Indices of the plate and water are 1.88 and 1.33 respectively.
Solution:

Let the angle of refraction inside,n-rriDT.tile plate be r

Here , \(w^{\mu_g}=\frac{\sin i{ }^0}{\sin r} \text { or, } \frac{\mu_g}{\mu_w}=\frac{\sin i}{\sin r}\)

Or, \(\frac{1.88}{1.33}=\frac{\sin 45^{\circ}}{\sin r}\)

Or, in r = \(\frac{1}{\sqrt{2}} \times \frac{1.33}{1.88}\) = 0.5

= \(\frac{1}{2}\)= sin 30°

r = 30°

Example 6. How much time will sunlight take to pass through the glass window of thickness 4 mm ? μ of glass =1.5.
Solution:

Velocity of sunlight in a vacuum or air,

C = 3×10^-8m s^-1

Thus velocity in a medium of refractive index μ

u = \(v\frac{c}{\mu}\)

So, to cross a thickness d, the time taken by light,

t = \(\frac{d}{v}=\frac{d \mu}{c}\)

= \(\frac{\left(4 \times 10^{-3}\right) \times 1.5}{3 \times 10^8}\)

[here d= 4 mm = 4 ×10m]

= 2 × 10^-11 s

Example 7. Green light of wavelength 5460 A°Is incident on an airglass interface. If the refractive index of glass is 1.5 what will be the wavelength of light In glass?
Solution:

The wavelength of the light in air, A0 = 5460 A°

The refractive index of glass concerning air, μ = 1.5

If the wavelength of the light In glass is λ, then

= \(\frac{\lambda_0}{\lambda}\)

Or, λ = \(\frac{\lambda_0}{\mu}=\frac{5460}{1.5}\)

= 3640 A°

Refraction Of Light Deviation Of A Ray

During reflection or refraction, the change In the direction of the tight is called its deviation.

The angle between the refracted ray and the direction of the incident ray gives the measure of deviation.

Deviation of incident ray AO, which after refraction proceeds along OD instead of OC. So the deviation of ray, S = ∠BOC

Now, δ = ∠BOC = ∠N’OC- ∠N’OB

= ∠AON- ∠N’OB = i-r

We know. If the angle of Incidence Increases, the angle of refraction also Increases

For normal Incidence, i = 0 thus r = 0 and so δ = 0 (minimum).

For  i = 90°, δ Is maximum

For refraction to a rarer medium from a denser medium, the angle of refraction is greater than the angle of incidence, l.e., r> i.

Now, δ = ∠BOC = ∠N’OB- ∠N’OC

=∠N’OB-  ∠AON =r – i

Optics

Refraction Of Light Image Due To Refraction

Suppose, a beam of rays from a point object after refraction reaches our eyes in another medium. Now if dierefracted rays are produced backwards they are nice at a point It seems that the refracted rays are diverging from die second point. The second point Is the image of the lift’s first point. The image of any point object is formed in the same way. Thus a complete image of the object is formed.

If the object Is situated In a denser medium and Is viewed from a rarer medium, It appears (closer to the surface of separation. For example, If we look at a fish inside water in a pond it appeals nearer to the surface than the actual position.

If the object Is situated in a rarer medium and Is viewed from a denser medium, it appears to move away from the surface of separation, for example, our earth is surrounded by a thick atmospheric layer composed of different gases. Starlight comes to our eyes through this atmosphere. Hence we are the observers on Earth in a denser medium whereas the stars are in a vacuum i.e., in the rarer medium. So, the animal position of a star Is far behind its normal viewing position.

Object In denser medium end eyes In rarer medium:

Let the refractive indices of the two media a and b be μ₁ and μ₂, respectively and μ₁ >μ₁, An object P situated in a is viewed from b. The surface of separation of a and b is a plane surface. A ray of light from P incident perpendicularly at A proceeds along AB without changing its direction. Another oblique ray PC incident at C is refracted along CD.

The refracted rays A B and CP when produced backwards meet at Q . So when the two refracted rays reach the rays of the observer. It will appear as if tyre two rays are coming from Q . So Q h the virtual Image of p.  In this case, the image rises towards the surface of the separation

If the angle of incidence and the angle of refraction are l and r.

⇒ \(\frac{\mu_2}{\mu_1}=\frac{\sin r}{\sin i}=\frac{\tan r}{\tan i}=\frac{A C / A Q}{A C / A P}=\frac{A P}{A Q}\)

For near-normal viewing, points A and C are close enough. So sin0 *s tarif

Hence, \(\frac{\mu_2}{\mu_1}=\frac{\sin r}{\sin i}=\frac{\tan r}{\tan l}=\frac{A C / A Q}{A C / A P}=\frac{A P}{A Q}\)………………………….(1)

If the observer is In the air then μ₁  = 1

Putting μ₁ = 1 and μ₂ =μ  (say) In equation (1) and writing AP = u, AQ =  v

We have \(\mu=\frac{A P}{A Q}=\frac{u}{v}\)

If d is die real deeds of the object then

μ = \(\frac{d}{\text { apparent depth of the object }}\)

Or, Apparent depth of the object = \(\frac{d}{\mu}\)

Hence apparent displacement

x = PQ = AP-AQ = d- \(\frac{d}{\mu}\)

= \(d\left(1-\frac{1}{\mu}\right)\)

Therefore, the apparent displacement of an object depends on the real depth (d) of the object and the refractive index (/z) of the denser medium.

The refractive index of water concerning air is|. If an object immersed in water is observed vertically from above the water, then its apparent displacement.

x = \(d\left(1-\frac{1}{4 / 3}\right)=\frac{d}{4}\)

General case:

The apparent depth of an object when viewed from the air through successive media of thicknesses d₁ d₂, d₃ ‘ ……….. d_n having refractive indices μ₁,  μ₂, μ₃ ‘ ……….. μ_n  respectively is

= \(\frac{d_1}{\mu_1}+\frac{d_2}{\mu_2}+\frac{d_3}{\mu_3}+\cdots+\frac{d_n}{\mu_n}=\sum_{i=1}^n \frac{d_i}{\mu_i}\)

Its apparent displacement is

⇒ \(d_1\left(1-\frac{1}{\mu_1}\right)+d_2\left(1-\frac{1}{\mu_2}\right)+d_3\left(1-\frac{1}{\mu_3}\right)+\cdots+d_n\left(1-\frac{1}{\mu_n}\right)\)

= \(\sum_{i=1}^n d_i\left(1-\frac{1}{\mu_i}\right)\)

Object in rarer medium and eye in denser medium:

Let the refractive indices of the two media a and b be μ₂ and μ₂ respectively and μ₂ > μ₁ An object P is situated in the medium b and it is seen from the medium. The surface of separation of a and b is a plane surface.

A ray of light from P is incident perpendicularly at point A on the surface of separation and proceeds straight along AB through the medium a without changing its direction. Another oblique ray PC is incident at C and proceeds along CD after j refraction. The refracted rays AB and CD when produced backwards meet at Q. So when the two refracted rays reach the observer, they will.

Appears to him that the two rays are coming from Q. So Q is the virtual image of P. In this case, the image appears to move farther away from the surface of separation.

If the angle of incidence and the angle of refraction of the incident ray at C are i and r respectively, then according to Snell’s law

μ₁ sin i= μ₂sin r

∴ \(\frac{\mu_2}{\mu_1}=\frac{\sin i}{\sin r}=\frac{\sin \angle P C N_1}{\sin \angle N C D}\)

Since the two lines PAB and NjCiV are parallel

∠PCN₁ = ∠APC and ∠NCD = ∠AQC

⇒ \(\frac{\mu_2}{\mu_1}=\frac{\sin \angle A P C}{\sin \angle A Q C}=\frac{\frac{A C}{C P}}{\frac{A C}{C Q}}=\frac{C Q}{C P}\)

If the points A and C are very close to each other i.e., if the ray PC is not so oblique, then CP ≈ AP and CQ ≈ AQ

∴ \(\frac{\mu_2}{\mu_1}=\frac{A Q}{A P}\)…………. (3)

If the rarer medium in which the object is situated is air then = 1.

Putting μ₁ = 1 and μ₂= p (say) in equation (3) we get

⇒ \(\mu=\frac{A Q}{A P}\)

The refractive index of the denser medium concerning air apparent height of the object from

= \(\frac{\text { the surface of separation }}{\text { real height of the object from the surface of separation }}\)

If the real height of the object, AP = μd, then

= \(\mu\frac{\text { apparent height of the object }(A Q)}{d}\)

Or, AQ = μd ………………… (4)

So the apparent displacement of the object

=PQ = AQ – AP – μd-d = (μ-1) d

General case:

The apparent height of an object in the air when viewed from a medium of refractive index pn through successive media of thicknesses d₁ d₂, d₃ ‘ ……….. d_n having refractive indices, μ₁,  μ₂, μ₃ ‘ ……….. μ_n   respectively is

⇒ \(\mu_1 d_1+\mu_2 d_2+\cdots+\mu_n d_n=\sum_{i=1}^n \mu_i d_i\)

Its apparent displacement is

⇒ \(\left(\mu_1-1\right) d_1+\left(\mu_2-1\right) d_2+\cdots+\left(\mu_n-1\right) d_n\)

= \(\sum_{i=1}^n\left(\mu_i-1\right) d_i\)

Let an object in a medium of refractive index fly be viewed from a medium of refractive index  Then we have,

= \(\frac{\text { Apparent depth of the object }}{\text { real depth of the object }}=\frac{\mu_2}{\mu_1}\)

=  \({ }^1 \mu_2\)

If the object is situated in a comparatively denser medium, then μ₁>μ₂.

In that case, apparent depth < real depth. If the object is situated in a comparatively rarer medium then μ₁<μ₂ In that case, apparent depth > real depth

Image Formed by Oblique Incident Rays

In the last section, we talked of almost normal viewing. For more B oblique incidence, the apparent displacement is higher

A point object O in an optically denser medium (say, water) is viewed from a rarer medium (say, air), at different angles. For different positions of the observer, the locus of the different positions of the image is a curved line. This curved line is called a caustic curve. The curve has two parts. These two parts meet at a point O’, known as the cusp. The image of an object at O when viewed vertically downward, is formed, at O’.  shows how the images A’, B’, C’, etc. of different points A, B, C, etc. on the base of a vessel or tank containing water will appear to an observer located at a given position. It is evident

The image for normal incidence is at the lowest position. The other images go on rising as the oblique rays from the base produce the images. The image of the base of the vessel will be a curved surface indicated by A’B’C’. With the increase of the distance of the base of the vessel from the eye, the curved surface appears, to rise higher. So if an observer stands in a shallow pond having equal depth everywhere, it appears to him that the pond near his feet is the deepest

Image of an Object under a Parallel Slab

ABCD is a parallel glass slab. Its thickness is d and its refractive index μ,  P is a point object placed in air under the surface AB of the slab. A ray of light PX normally incident on AB goes undeviated along XY. Another ray PQ incident obliquely is refracted along QR. After that, the ray is further refracted along RS. Since the two faces AB and DC of the glass slab are parallel, the rays

PQ and RS will be parallel. So an observer from above the slab will see the image of P at P’. PP’ is the apparent displacement of the point object towards the observer. The normal NQN₁ at Q intersects P’R at M. The quadrilateral PP’MQ is a parallelogram

PP’ = QM

The Apparent Displacement of the point object

∴ \(d\left(1-\frac{1}{\mu}\right)\)

PQ and RS will be parallel. So an observer from above the slab will see the image of P at P’. PP’ is the apparent displacement of the point object towards the observer.

The normal NQN₁ at Q intersects P’R at M. The quadrilateral PP’MQ is a parallelogram.

∴ PP’ = QM

∴ The apparent displacement of the point object

= PP’ = QM

= d(1- \(\frac{1}{\mu}\))

So the apparent displacement of an object does not depend on the position of the object under the lower face of the glass slab. It only depends on the thickness of the slab (d) and the refrac¬ index of its material (μ).

 Optics

Refraction Of Light Some Examples Of Refraction

A coin immersed in water:

A coin is placed at the bottom of a pot such that the coin is just not visible. If eyes are set at the same position and the pot is now filled with water, the coin becomes visible. Because the rays from P are refracted from denser to rarer medium and are bent away from the normal, they reach our eyes. As a result, the refracted rays appear to diverge from P’, i.e., the virtual image of the coin is formed at P’, situated above the coin.

A rod partly immersed in water:

Let a straight rod be partly immersed in water. When the rod is held obliquely in water, the portion of the rod in water will appear to be bent upward [Fig. 2.20]. The reason is the same as above. The light rays coming from the portion immersed in water are refracted from denser to rarer medium and hence bent away from the normal. So point A of the rod appears to be raised at B. This happens for every point of the immersed portion of the rod

Object and medium having approximately equal refractive index:

An object becomes invisible when it is sur¬ rounded by a medium having a nearly equal refractive index. Since the refractive indices of both of them are almost the same, negligible refraction does occur from their surface of separation, and for refraction lending of light is negligible. i.e., it travels undefeated. As a result, the surface of separation is not visible. ff99Wt9rf . The refractive indices of glycerine and glass are almost equal. So when a glass rod is immersed in glycerine thiqÿrod is not visible

Multiple images in a thick glass mirror:

If an object is placed in front of a thick mirror with silvered glass at the mirror at the back if a surface object is viewed from a slanting direction, a series of images are formed.

When ray PA is incident on the front face at point A, a very small portion of the light is reflected along AK producing a faint image at P₁. The remaining larger portion of the light is refracted into the mirror along AB and is reflected along BC from the silvered surface. A large portion of this reflected ray within the glass comes out along CL producing the second image P₂ which is the brightest of all the images.

The remaining part of the ray CD is reflected from surface Y to the silvered surface where it is again reflected. The process continues and gradually fainter images are formed. The different images lie on the line drawn perpendicular to the surface X from the object P.

The ray BC, which is the first reflected ray from the silvered surface, is the brightest and it travels along CL. Consequently, image P₂ is the brightest. Hence the brightest second image is considered to be the image of the object. The more oblique the incident rays are, the more the amount of reflection from the front surface of the mirror and the brightness of the image P₁ will increase accordingly

Apparent thickness of thick glass mirror:

In the case of a water-filled bowl, the depth of the bowl appears to be less. Similarly, a thick glass mirror seems to be less thick than it is.

We have, \( \frac{\text { real thickness of mirror }}{\text { apparent thickness of mirror }}\)

= Refractive index of glass = \(\frac{3}{2}\)

Therefore, the apparent thickness of a mirror

= \(\frac{2}{3 }\) × Real thickness of the mirror

Optics

Refraction Of Light Some Examples Of Refraction Numerical Examples

Example 1. There is a mark at the bottom of the beaker. A liquid with a refractive index of 1.4 is poured into it. If the depth then determines how much the (liquid is 3.5 cm mark appears to rise when it is viewed from above.
Solution:

If the object is in a denser medium and the observer is in a rarer medium, the refractive index of the denser medium relative to the rarer medium

= \(\frac{\text { real depth of the object }}{\text { apparent depth of the object }}\)

Or, 1.4 = \(\frac{3.5}{\text { apparent depth of the object }}\)

Or, apparent depth of the object = \(\frac{3.5}{1.4}\)

= 2.5 cm

Apparent upward displacement of the mark

= 3.5 – 2.5 = 1cm

Example 2.  There is a black spot at the bottom of a rectangular glass slab of thickness d and refractive index μ. When the spot is viewed perpendicularly from above, the spot appears to be shifted through a distance \(\frac{(\mu-1) d}{\mu}\) towards the observer. Prove it.
Solution:

The real depth of the black spot from the upper surface of

Let the apparent depth of the black spot from the upper surface of the glass slab be d₂

Refractive index of glass \(\mu=\frac{d}{d_1}\)

Or, \(d_1=\frac{d}{\mu}\)

∴ The apparent displacement of the black spot towards the observed

= \(d-d_1=d-\frac{d}{u}\)

= \(d\left(1-\frac{1}{\mu}\right)=d\left(\frac{\mu-1}{\mu}\right)\)

Example 3.  In a beaker partly filled with water, the depth of water seems to be 9 cm. On pouring more water into it, the real depth of water is increased by 4cm. Now the apparent depth of water seems to be 12cm. Determine the refractive index of water and the initial depth of water in the beaker.
Solution:

Let the refractive index of water be ft and the initial| depth of water in the beaker be x.

μ = \(\frac{x}{9}\)

or, x = 9μ

When more water is poured into the beaker, the real depth of water becomes (x + 4) cm.

In the second case, μ = \(\frac{x+4}{12}\)

Or, 12 μ = x + 4 or; 12 μ = 9 μ + 4 Or, 3 μ= 4

∴ μ = \(\frac{4}{3}\)

μ = 1.33

Initial depth of water in the beaker.

x = 9 μ = 9 ×  \(\frac{4}{3}\)

= 12cm

Example 4. A small air—bubble exists inside a transparent cube of side 15 cm each. The apparent distance of the bubble observed from one face is 6 cm and from the opposite face its apparent distance becomes 4 cm. Determine the real distance of the bubble from the first face and the refractive index of the material of the cube
Solution:

Let the real distance of the bubble from the first face be x cm.

The real distance of the bubble from the opposite face = (15-x) cm

Let the refractive index of the material of the cube be ft.

We know if the object lies in a denser medium and eye in the rarer medium,

μ = \(\frac{\text { real distance }}{\text { apparent distance }}\)

In the first case , μ=  \(\frac{x}{6}\)

In the second case,  μ  = \(\frac{15-x}{4}\)

⇒ \(\frac{x}{6}=\frac{15-x}{4}\)

Or, 4x = 90- 6x

Or, 10x = 90

Or, x = 9cm and

mu = \(\frac{9}{6}\) = 1.5

Therefore, the real distance of the bubble from the first face is 9 cm and the refractive index of the material of the cube is 1.5

Example 5. A vessel is filled with two mutually immiscible liquids with refractive indices μ₁  and μ₂. The depths of the two liquids are d₁ and d₂ respectively. There is a mark at the bottom of the vessel. Show that the apparent depth of the mark when viewed normally is given by \(\left(\frac{d_1}{\mu_1}+\frac{d_2}{\mu_2}\right)\)
Solution:

The image of P is formed at Q due to refraction at the surface of separation B of the 1st and 2nd liquid. Another final image due to refraction in air from the second liquid is formed at R

For the first refraction:

⇒ \(\frac{\mu_1}{\mu_2}=\frac{B P}{B Q}\)

Or, \(B Q=\frac{\mu_2}{\mu_1} \cdot B P\)

⇒ \(\frac{\mu_2}{\mu_1} d_1\)

For the second refraction:

⇒ \(\frac{\mu_2}{1}=\frac{A Q}{A R}\)

Or, \(A R=\frac{A Q}{\mu_2}\)

= \(\frac{1}{\mu_2}(A B+B Q)\)

Or, \(A R=\frac{1}{\mu_2}\left(d_2+\frac{\mu_2}{\mu_1} d_1\right)\)

= \(\frac{d_2}{\mu_2}+\frac{d_1}{\mu_1}\)

The apparent depth of the mark P when viewed normally

= AR = \(\frac{d_1}{\mu_1}+\frac{d_2}{\mu_2}\)

Example 6. A rectangular slab of refractive index [i is placed on another slab of refractive index 3. Both the slabs are of ror. the displacement of the image? the same dimensions. There is a coin at the bottom of the lower slab. What should be the value of n such that when viewed normally from above, the coin appears to be at the surface of separation of the two slabs?
Solution:

Let the thickness of each slab be d. According to the question the apparent depth of the coin =d

d = \(\frac{d}{\mu}+\frac{d}{3}\)

Or, 1= \(\frac{1}{\mu}+\frac{1}{3}\)

Or, \(\frac{1}{\mu}=\frac{2}{3}\)

Or, \(\frac{3}{2}\)

= 1.5

Example 7. A tank contains ethyl alcohol of a refractive index of 1.35 The depth of alcohol is 308 cm. A plane mirror is placed horizontally at a depth of 154 cm in it. An; object is placed 254 mm above the mirror. Calculate the apparent depth of the image formed by the mirror
Solution:

Depth of the mirror =1.54 m; object distance from the mirror = 0.254 m

The image of the object is formed at a distance of 0.254 m behind the mirror

= \(\frac{\text { real depth }}{\mu}\)

=\(\frac{1.794}{1.35}\)

= 1.33m

Example 8. A 20mm thick layer of water \(\left(\mu=\frac{4}{3}\right)\) 35mm thick layer of another liquid \(\left(\mu=\frac{7}{5}\right)\) = ‘ in a tank. A small coin lies at the bottom of the tank. Determine the apparent depth of the coin when viewed normally from above the water
Solution:

Real depth of the coin d+d = 20+ 35 = 55m m

∴ A parent depth of the coin

= \(\frac{d_1}{\mu_1}+\frac{d_2}{\mu_2}=\frac{20}{\frac{4}{3}}+\frac{35}{\frac{7}{5}}\)

= 15+ 25

= 40 mm

Example 9. If a point source is placed at a distance of 18 cm from the pole of a concave mirror, its image is formed at a distance of 9 cm from the mirror. A glass slab of thickness 6 cm is placed between the point source and the mirror such that the parallel faces of the glass slab remain perpendicular to the principal axis of the mirror the refractive index of glass is 1.5 what will be the displacement of the image?
Solution:

Let P be the position of the object and in the absence of the glass slab, Q be the position of the image formed by the concave mirror

u = 18 cm,

OQ = v= 9 cm

According to \(\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\) we get,

= \(\frac{1}{-9}+\frac{1}{-18}=\frac{1}{f}\)

Or, f= -6 cm

If the glass slab of thickness 6 cm is placed between the point I source and the concave mirror, apparent displacement of the I Point source will take place towards the mirror. The rays coming from P appear to come from P’ after refraction.

The apparent displacement of P

PP’ = \(d\left(1-\frac{1}{\mu}\right)\)

= 6\(\left(1-\frac{1}{1.5}\right)\)

= 2cm

So in the second case, object distance u = -(18- 2) = -16 cm ; f = -6 cm; v = ?

We know, \(\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\) Or, \(\frac{1}{v}+\frac{1}{-16}=\frac{1}{-6}\)

Or, \(\frac{1}{v}=\frac{1}{-6}+\frac{1}{16}=\frac{-10}{96}\)

Or, v = \(-\frac{96}{10}\)

= – 9.6 cm

Displacement of the image

QQ’ = OQ’ OQ – OQ = 9.6- 9 = 0.6 cm

Example 10. A plane is made of glass having a thickness of 1.5 cm. Its back surface is coated with memory. A man is standing at a distance of 50 cm from the front face of the mirror. If he looks at the mirror normally, where can he find his image behind the front face of the mirror? The refractive index of glass  = 1.5 
Solution:

The real depth of the mercury-coated surface from the upper surface of the mirror = 1.5 cm

If the apparent depth of the mercury coated mercury  cm, then \(\frac{1.5}{x}\) = 1.5 or, x= 1 cm

So the mercury-coated surface appears to be at n distance of 1 cm from the front face of the mirror.

‘Therefore, the distance of the man from the appetent position of the mercury-coated surface a = 50 +1  cm

So the distance of the Image from the apparent position of the mercury-coated surface =  51 cm

The distance of the Image of the man from the front tuifnee of the mirror =51 + 1 = 52 an

Example 11.  An observer can see the topmost point of a narrow rod of height through a small hole  The rod Is placed Inside a beaker. The beaker’s height is 3h and its radius Is h. When the beaker Is filled up to 2 h of its height with a liquid the observer can see the entire rod. What Is the value of the refractive Index of the liquid?

Solution:

Let us assume that  PQ is a thin straight rod kept in a beaker. B Is a small hole in the wall of the beaker. When the beaker is filled with a liquid up to a height of 2h, then Q can be seen through hole B.

Here, QD ray propagates through the liquid and gets refracted along DB in the air

According to ABRP is a square midpoint of the diagonal PB.

Here DE= PE = h

Again, ∠BDP = 45°  [∴ ∠DPE ]

Since the object was placed in the denser medium, according to Snell’s law

⇒ \(\frac{1}{\mu}=\frac{\sin i}{\sin r}=\frac{\frac{Q G}{Q D}}{\sin 45}\)

Or, \(\frac{1}{\mu}=\frac{\frac{h}{\sqrt{5 h}}}{\frac{1}{\sqrt{2}}}\)

Since,  QD²=QG²+GD²=h²+(2 h)²= 5 h²

Or, \(\frac{1}{\mu}=\sqrt{\frac{2}{5}} \quad \text { of, } \mu=\sqrt{\frac{5}{2}}\)

Therefore, the required refractive Index of the liquid is \(\sqrt{\frac{5}{2}}\)

Example 12. A ray of light Incident at the Interface of glass and water at an angle of Incidence i. If the ray finally emerges parallel to the surface of water Then what will be the value of μ_g?

Solution:

When refraction occurs due to the propagation of light rays from glim to water, we may write from Snell’s law,

⇒ \(\mu_g \sin i=\mu_w \sin r\) …………..(1)

Again, when refraction occurs due to the propagation of light rays from water to air, we may write from Snail * law

⇒ \(\mu_g \sin i=\mu_w \sin r\) …………..(2)

Therefore, the required refractive Index of glass is \(\frac{1}{\sin i}\)

Optics

Refraction Of Light Critical Angle’s Total Internal Reflection

We know, in refraction from denser to a rarer medium light | bents away from the normal. As a result angle of refraction ’ becomes larger than the angle of incidence.

L line AB represents the surface of the separation of water and air.

Ray P₁ O travelling through water is incident at O on the surface- of separation. A part of the ray is reflected into the water along OR₁ and another part is refracted into the air along OQ₁:  The angle of refraction ∠Q₁ON is greater than the angle of incidence ∠P₁ ON₁. The greater the angle of incidence, the greater the angle of refraction and in each case, both reflection and refraction will take place.

For a particular value of the angle of incidence, the angle of refraction becomes 90°, so that the refracted ray grazes the surface of separation. This limiting angle of incidence in the denser medium is called the critical angle for the two given media. Thus, ∠P₂ON₁ = critical angle (θ_c). In this case, the angle of refraction ∠NOQ₂ = 90°. Here also a part of the incident ray is reflected to water along OR₂.

If the angle of incidence exceeds the critical angle i.e., if i  > θ_c [as in the case of the incident rayP₃O ] no part of the incident ray is refracted in the second medium. The ray is completely reflected along OR₃ into the first medium. This phenomenon is called total internal reflection. In this case, the surface of the separation of the two media behaves as a mirror.

Critical angle:

It is that particular angle of incidence of a ray of light for a given pair of media, passing from denser one to rarer one, for which the corresponding angle of refraction is equal to 90 0 and the refracted ray grazes along the surface of the interface separating the two media. The critical angle of a pair of media depends on the colour of the incident light and the nature of the two media.

For example, in the case of two particular media, the critical angle for red light is greater than that for violet light. Again, the critical angle of water to air is 49°, while that of glass to air is 42°. The statement, ‘critical angle of glass to air is 42° ‘ means that a . ray of light from glass being incident on the surface of separation of glass and water at an angle of 42°, should go along the surface of separation after refraction i.e., the refracted angle will be 90°.

Total internal reflection:

When a ray of light travelling from a denser medium to a rarer medium is incident at the surface of separation of the two media at an angle greater than the critical angle for the media, there is no refraction; rather the whole of the incident ray is reflected. This phenomenon is known as total internal reflection.

Condition of total internal reflection:

The conditions to be satisfied for total internal reflection are as follows

1. The light must travel from a denser to a rarer medium.
2. The angle of incidence must be greater than the critical angle for the two media

Reason for using the term “total’:

In ordinary reflection, a part of the incident light is reflected from the surface of separation and the rest is refracted. But in the case of internal reflection, no part of the incident light is refracted, rather the entire portion of the incident light is reflected to the first medium from the surface of separation of the two media. So this reflection is called total reflection.

Relation between critical angle and refractive index of the denser medium:

Let ∠P₂ON₁ = θ_c = critical angle between the two media, water and air, which

Impliesair concerning the angle to water of refraction is _aμ_w, which is then 90°. If the refractive index

_aμ_w = \(\frac{\sin \theta_c}{\sin 90^{\circ}}\)

Or,  sin θ_c = \(\frac{1}{a^{\mu_w}}\)

So, the value of critical depends on the refractive index of one medium concerning another

If the medium is a and b the,

⇒ \(\sin \theta_c=\frac{1}{b^{\mu_a}}=\frac{1}{\begin{array}{r}
\text { refractive index of denser medium } \\
\text {concerning rarer medium }
\end{array}}\)

= \(\frac{\mu_b}{\mu_a}=\frac{\text { absolute refractive index of medium } b}{\text { absolute refractive index of medium } a}\)

Optics

Refraction Of Light Critical Angle’s Total Internal Reflection Numerical Examples

Example 1. If the absolute refractive index of a medium is \(\sqrt{2}\), calculate the critical angle of glass to the medium. Given
Solution:

If the refractive index of glass concerning air is _aμ_g, then

_aμ_g= \(\frac{1}{\sin \theta_c}=\frac{1}{\sin 30^{\circ}}\)

= 2

If the refractive index of glass concerning the medium is _mμ_g, then

_aμ_g= \(\frac{a^{\mu_g}}{{ }_a \mu_m}\)

= \(\frac{2}{\sqrt{2}}\)

= \(\sqrt{2}\)

If the critical angle of glass to the medium is #c, then

sin = \(\sin \theta_c=\frac{1}{m^\mu}\)

= \(\frac{1}{m^\mu}=\frac{1}{\sqrt{2}}\)

= sin 45

Or, θ_c = 45

Example 2.  The refractive index of carbon disulphide for red light is 1.634 and the difference in the values of the critical angle for red and blue light at the surface of separation of carbon disulphide and air is 0°56/. What is the value of the refractive index of carbon disulphide for blue light
Solution:

Let the refractive index of carbon disulphide for red light = and circle angle =

Now, sin θ_r = \(\frac{1}{\mu_r}=\frac{1}{1.634}\)

= 0.6119 = sin 37.73

θ_r = 37.73

Let the critical angle for blue light be θ_b. The refractive index increases as the wavelength of light decreases. So the critical angle decreases.

∴ θ_b<θ_r

According to the Question,

θ_b = 37.73°-0°.56′

= 37.73°-0.93°

= 36.8°

So refractive index of carbon disulphide for blue light,

μ_b = \(\frac{1}{\sin \theta_b}=\frac{1}{\sin 36^{\circ} 48^{\prime}}\)

= \(\frac{1}{\sin 36.8^{\circ}}=\frac{1}{0.599}\)

= 1.669

Example 3. The refractive index of-diamond is 2.42,-which proves that all the beams of rays having an angle of incidence of more than 25° will be reflected, [sin 24.41° = 0.4132]
Solution:

If the critical angle is θ_c then

Sin θ_c = \(\frac{1}{\mu}=\frac{1}{2.42}\)

= 0.4312° = sin 24.41°

θ_c  = 24.41°

We know that if the angle of incidence of a ray of light is greater than the critical angle, the ray will be reflected. Here the critical angle is 24.41°. So rays of light having an angle of incidence greater than 25° will be reflected

Example 4. A ray of light will go from diamond to glass. What should be the minimum angle of incidence at the surface of separation of the two media, diamond and glass, so that the ray of light cannot be refracted in glass? μ of glass =1.51 and μ of diamond = 2.47; sin37.69° = 0.61134
Solution:

If the light ray is incident at the surface of separation of the two media diamond and glass at a critical angle, the ray is grazingly refracted in the glass. If the critical angle is QQ then

sin θ_c = \(\frac{1}{g^{\mu_d}}=\frac{1}{\frac{\mu_d}{\mu_g}}\)

= \(\frac{\mu_g}{\mu_d}=\frac{1.51}{2.47}\)

= 0.61134

= sin 37.69°

∴ θ_c  = 37.69°

So if the angle of incidence of a light ray is greater than 37.69° it | cannot be refracted in glass.

∴  Required minimum angle of incidence =37.69°

Example 5. cube has a refractive Index μ₁. There is a plate of refractive Index μ₂(μ₂<μ₁) A ray travelling through the air is incident on the side face of the cube. The refracted ray Is an Incident on the upper face of the cube at the minimum angle for total internal reflection to occur. Finally, the reflected ray emerges from the opposite face. Show that if the angle of emergence Is Φ then sin = \(\sqrt{\mu_1^2-\mu_2^2}\)
Solution:

PQ = incident ray on the side face of the cube, QR = refracted ray inside 0′ the cube, S = reflected ray from the upper face of the cube, ST = emergent ray from the opposite face.

Let the critical angle for total reflection be θ_c

According to the question θ’_c ≈θ_c

⇒ \(\sin \theta_c^{\prime} \approx \sin \theta_c=\frac{1}{2^{\mu_1}}=\frac{1}{\frac{\mu_1}{\mu_2}}=\frac{\mu_2}{\mu_1}\)

Angle of incidence of the ray RS = l = 90° – θ_c and angle of refraction -tf>

⇒ \(\sin \theta_c^{\prime} \approx \sin \theta_c=\frac{1}{2^{\mu_1}}=\frac{1}{\frac{\mu_1}{\mu_2}}=\frac{\mu_2}{\mu_1}\)  = Refractive index of air with respect to the cube

= \(\frac{1}{\text { refractive index of the cube with respect to air }}\)

Or, \(\frac{\sin i}{\sin \phi}=\frac{1}{\mu_1}\)

Or, sin Φ = μ₁ sin i

= \(\mu_1 \sin \left(90^{\circ}-\theta_c\right)=\mu_1 \cos \theta_c\)

= \(\mu_1, \sqrt{1-\sin ^2 \theta_c}\)

= \(\mu_1 \sqrt{1-\frac{\mu_2^2}{\mu_1^2}}=\sqrt{\mu_1^2-\mu_2^2}\)

Example 6. A ray of light travelling through a denser medium is incident at an angle i in a rarer medium. If the angle between the reflected ray and the refracted ray is 90° show that the critical angle of the two media, \(\theta_c=\sin ^{-1}(\tan i)\) 
Solution:

Suppose the angle of refraction in the medium =r

From  we get

i+ 90° + r = 180°

Or, r = 90° – i

According to Snell’s law.

⇒\(\frac{\sin l}{\sin r}={ }_1 \mu_2=\frac{\mu_2}{\mu_1}\)

Or, \(\frac{\sin i}{\sin \left(90^{\circ}-i\right)}=\frac{\mu_2}{\mu_1}\)

Or, \(\frac{\sin i}{\cos i}=\frac{\mu_2}{\mu_1}\)

Or, \(\tan i=\frac{\mu_2}{\mu_1}\)

If the critical angle for the two media is θ_c, then

⇒ \(\sin \theta_c=\frac{1}{{ }_2 \mu_1}\)

= \(\frac{1}{\frac{\mu_1}{\mu_2}}=\frac{\mu_2}{\mu_1}\)

= tan i

Or, \(\theta_c=\sin ^{-1}(\tan i)\)

Example 7. A nail Is fixed up perpendicularly at the centre of a circular wooden plate. Keeping the nail at the bottom, the circular plate Is made to float In water. What should be the maximum ratio of the radius of the plate and the length of the nail so that the nail will be out of vision? Refractive index of water \(\frac{4}{3}\)
Solution:

AB Is the circular wooden plate and CD is the nail. Suppose, the radius of the plate =r and the length of the nail =h Since the nail Is not seen from the air, the angle of incidence of the ray DA will be greater than 0 and the ray will be reflected.

We know, \(\sin \theta_c=\frac{1}{a^\mu{ }_w}=\frac{3}{4}\)

⇒  \(\cos \theta_c=\sqrt{1-\frac{9}{16}}=\frac{\sqrt{7}}{4}\)

⇒ \(\tan \theta_c=\frac{\frac{3}{4}}{\frac{\sqrt{7}}{4}}=\frac{3}{\sqrt{7}}\)

Or, \(\frac{r}{h}=\frac{3}{\sqrt{7}}\)

This is the required ratio

Example 8.  The Critical angle of glass relative to a liquid is 57° 20′. Calculate the velocity of light in the liquid. Given,  μ of glass = 1.58, velocity of light in vacuum = 3×10⁸ m s^-1 sin 57° 20′ = 0.8418
Solution:

Critical angle of glass relative to the liquid,

θ_c = 57°20′

If the refractive index of glass concerning the liquid is {fiR then.

sinθ_c =\(\frac{1}{\nu_g}=\frac{1}{\frac{\mu_g}{\mu_l}}=\frac{\mu_l}{\mu_g}\)

⇒ \(\frac{\mu_l}{\mu_g}=\sin 57^{\circ} 20^{\prime}\)

= 0.8418

Or, \(\mu_l=0.8418 \times \mu_g=0.8418 \times 1.58\)

Again , \(\mu_l=\frac{\text { velocity of light in vacuum }}{\text { velocity of light in the liquid }}\)

The velocity of the light in the liquid

= \(\frac{3 \times 10^8}{\mu_l}=\frac{3 \times 10^8}{0.8418 \times 1.58}\)

= \(2.255 \times 10^8 \mathrm{~m} \cdot \mathrm{s}^{-1}\)

Example 9.  A transparent solid cylindrical rod has a refractive Index of. It Is surrounded by air. A light ray Is Incident at the midpoint of one end of the rod.  Determine the incident angle 8 for which the light ray grazes along the wall of the rod

Solution:

For refraction of light at point B, we can write by applying Snell’s law

1 × sin θ = μ sin r

[where μ is the refractive index of the solid material)

or, sin θ = \(\frac{2}{\sqrt{3}} \sin r\) ……………………… (1)

The light ray BC is incident on point C making critical angle θ_c and propagates along CD

Thus, from Snell’s law,

μ = \(\frac{1}{\sin \theta_c}\)

Or, \(\theta_c=\sin ^{-1}\left(\frac{1}{\mu}\right)\)

=\(\sin ^{-1}\left(\frac{\sqrt{3}}{2}\right)\)

= 60°

r = 180°- (60° + 90°) = 30°

Hence from equation (1), we can write

θ_c = \(\frac{2}{\sqrt{3}} \sin 30^{\circ}=\frac{1}{\sqrt{3}}\)

Or, \(\sin ^{-1}\left(\frac{1}{\sqrt{3}}\right)\)

Examples of Total Internal Reflection

A test tube dipped in water: A glass test tube half filled with water is held obliquely in a beaker containing water

The empty portion of the Immersed test tube appears shining if It is seen from above. This happens due to the total internal reflection of light. For the empty portion of the tube, the light goes from a denser to a rarer medium. Rays which are incident at angles greater than the critical angle of glass and air (48.5°) arc are reflected. So this portion of the glass appears shining.

A portion of the tube filled with water does not glow because here light enters water In test tube from water in a beaker. Thus, total internal reflection does not occur here. In this discussion, we do not take into account the existence of the glass wall of the tube due to its negligible thickness.

A metal ball coated wHb lampblack Immersed In water:

If a metal ball coated will lampblack Is Immersed in water, the ball appears shining. Due to the coating of the lamp¬ black. a thin layer of air surrounds the surface of the ball. Rays incident at an angle greater than the critical angle of water and air, are reflected. The ball appears shining when the reflected rays reach the eyes of the observer.

Glass tumbler full of water:

A glass tumbler full of water Is held above eye level. If the upper surface of water In the tumbler is seen from any one side, the surface appears shining. Rays coming from the side of the tumbler are incident on the surface of the separation of water and air. Hence total internal reflection takes place at particular angles of slantness and the surface of the water appears shining.

Air bubbles:

The air bubbles rising through the water look shiny. Rays travelling through water are incident on the surface of the air bubbles. Those rays which are incident at angles greater than the critical angle are reflected. When these reflected rays reach the eyes of the observer, the hubbies appear shining. For the same reason, air bubbles existing In paper weights appear to be shining

Natural Examples Of Total Internal Reflection

Mirage: It Is an optical Illusion brought about by total internal reflection. There are two types of mirage, one observed in hot regions and the other observed In extremely cold regions.

1. Inferior mirage or mirage in the desert:

People travelling through the desert sometimes see water at a distant place which is an optical illusion, called an inferior mirage, or simply, a mirage.

During daytime, the lower regions of the atmosphere become hot¬ ter than the upper regions. So density of air in the lower regions is less than that in the higher regions. Let us consider the atmo¬ sphere to be made up of layers of air, one above the other. A ray of light starting from a distant tree (P) and travelling downward happens to be going from a denser to a rarer medium.

So its angle of incidence at consecutive layers goes on increasing gradually till it exceeds the critical value when it is reflected due to total internal reflection. The traveller sees an inverted virtual image (P’) of the tree. Secondly, due to continuous temperature changes, there exists a temperature gradient in the layers which undergo a continuous change of density and hence in the refractive index as well. So the path of the rays coming through the layers of air is also continuously changing. Hence to the traveller, the image of the tree appears to be swaying. This completes the illusion of a pond lined with trees.

 2. Superior mirage or mirage in cold countries:

In cold countries, the temperature of air In the lower regions is lower than that of the upper region. So the density of air in the lower region is greater than that of the upper region. A ray of light starting from an object (P) travelling upwards, finds itself going from denser to rarer medium

So its angle of incidence at consecutive layers of air gradually increases till it reaches the critical value. Then it is reflected due to total internal reflection. To an observer, the ray appears to come from a point above, thus giving the impression that an inverted object (P’) is floating in the air which is an optical illusion. This phenomenon is called a superior mirage

View of an observer inside water:

To the eye of an observer or a fish Inside water, all objects above water appear to exist in n cone of semi-vertical angle <19° which Is the critical angle of water and air. This happens due to total Internal reflec¬ tion of light.

If a ray of light travelling from a denser medium is Incident at the critical angle, the refracted ray grazes the surface of separation. Conversely, if a ray of light travelling from a rarer medium is incident at an angle of 90°, the angle of refraction In the denser medium becomes equal to the critical angle. The critical angle of water and air is 49°. So if a ray of light S₁A coming from the rising sun AS₁, along the surface of water reaches eye E along the direction AE, then the angle of refraction in water becomes 49°

As the eye cannot follow ray AS₁, an observer inside water will sec the rising sun along the line EAC and this line will make an angle of 49° with the line OE. Similarly, the setting sun S₂ will be seen along the line EBD and this line also will make an angle of 49° with the line OE. So all the objects above water appear to exist in a cone of angle 98° to the eye of a fish or observer underwater.

It Is to be noted that the sun describes an arc of 180° to earthbound observers but to the eyes of a fish it describes an arc of 98°.

1.  Surface of water to the eye of an observer inside water:

The diameter of the circular base of the cone AEB is AB. If an observer keeping his eye on E looks at the circular section of water, he can see any object lying above water. But If the observer looks at the rest of the portion of water other than the circular portion, then

1. He cannot see any object above water, rather
2. He can see the images of the objects inside the water.

Reason explaining 1st Incident:

Any ray of light coming from outside water can reach point E only through the circular section but cannot reach point E if it comes through die remaining portion.

Reason explaining 2nd Incident:

Suppose the ray of light emerging from the object situated In water, reaches die point E after reflection from the surface of water. This reflection will take place from the surface of the water excluding the circular portion.

This reflection will be a total reflection. For example, if a ray of light from the object P situated inside water, is incident on the surface of water, the angle of incidence exceeds 49°. So, the ray after total reflection from the surface of the water reaches the eye of the observer and he observes the dead object, at P’.

So to the observer situated inside water, the surface of the water appears as a mirror with a circular hole in it, because he sees the objects situated outside water through the circular section and sees the images of the objects inside water in die remaining action of the surface of the water. The radius of the circular hole is OA or OB.

2. Determination of the radius of the hole:

Let the radius of the hole =OA = OB = r and OE – h. If the critical angle is θ then ∠OEA = Qc

⇒ \(\tan \theta_c=\frac{O A}{O E}=\frac{r}{h}\)

Or, \(r=h \tan \theta_c=h \frac{\sin \theta_c}{\cos \theta_c}\)

= \(h \frac{\sin \theta_c}{\sqrt{1-\sin ^2 \theta_c}}\)

Since ( sin \(\theta_c=\frac{1}{\mu}\))

= \(h \frac{\frac{1}{\mu}}{\sqrt{1-\frac{1}{\mu^2}}}\)

r = \(\frac{h}{\sqrt{\mu^2-1}}\)

Sparkling of diamond:

Diamond is notable for its sparkle and shine. This characteristic of a diamond is based on total internal reflection. The refractive index of a diamond is 2.42 and its critical angle relative to air is only 24.4°. This value is quite small as compared to other pairs of media.

Therefore, there is a high probability of total internal reflection in the case of diamond. If the diamond is cut properly, it will have a large number of faces. Ray of light entering through one face undergoes total internal reflection at several faces. As the rays of light enter through many faces and are confined inside they emerge together through only a few faces, these faces appear to sparkle and shine

Transmission of Light through Optical Fibre

Optical fibre:

A beam of light can be sent from one place to another through an optical fibre made of glass, quartz or optical-grade plastic, by following successive total internal reflections. As water can be sent from one place to another through a hollow pipe, a fibre can allow light to flow through it from one place to another. Hence, an optical fibre is often loosely called a light pipe.

Construction and principle of action:

An optical fibre is a long and very thin pipe. Its diameter is about 10 × 10^-6 metres. The internal section of the die pipe is called the core. It is a die core through which light travels from one point to another. Above the core, there is a coating of a substance having a refractive index less than that of the core. This coating is called cladding.

A ray of light entering die fibre through one face undergoes successive total internal reflections at the surface of separation of core and cladding and emerges through the other face [Fig. 2.38].: As total internal reflection of light takes place inside a fibre, the intensity of the light remains almost the same.

The image of a large object cannot be sent through a single fibre. In that case, bundles of fibres or cables of fibres are used. A cable contains about a thousand fibres. The image of an object is focused on one end of the bundle. If the order of die fibres is properly maintained, the image obtained at the other end will be an exact reproduction. In, the letter ‘T’ has been focussed at one end of the die bundle and an exact image of ‘T’ has been obtained. Light rays from the different portions of ‘T’ travel through the different fibres and form a die image at the other end

Application of optical fibre:

• Optical fibres are extensively used in medical science and the field of communication.
• These are used to study the interior parts of the body which are inaccessible to the bare eye, for example, lungs, tissues, intestines etc. It can be used to transmit high-intensity laser light inside the body for medical purposes.
• These are used for sending signals from one place to another. This signal is mainly digital. It is information that the signal carries.
• This information is used in telephone, television, fax, computer etc. It is to be noted that, different digital signals may be sent through the same fibre at the same time, without any chance of overlapping.
• So many times it is needed to collect samples inside from human body to identify disease. For this purpose, optical fibre is used. Besides, optical fibre is used for operation inside the human body. Thus, in most cases, no major excising is needed outer part of the body

Advantages of optical fibre over copper wire:

• Comparatively less power Is required to send a signal,
• The loss in energy Is much less
• The capacity of carrying information is approximately times.
• There exists no influence of any external electromagnetic wave signal.
• Electrical resistance is much more.
• It is very light.
• killVelocity of the signal is very fast (approximately equal to that of light in vacuum).
• The possibilities of the illegal usage of the signal are very low.

Optics

Refraction Of Light  Transmission of Light through Optical Fibre Numerical Examples

Example 1. A point source of light is placed at a depth of h below the calm surface of the water. From the source, light rays can only be transmitted to air through a definite circular section,

1. Draw the circular section of the surface of the water by ray diagram and mark its radius r.
2. Determine the angle of incidence of a ray of light incident at any point on the circumference of the circular plane. [Given: refractive index of water,\(\frac{4}{3}\) = 48°36′ = sin^-1 0.7501 ]
3. Show that r= \(\frac{3}{\sqrt{7}} h\)

Solution:

Let MN be the open surface of water. O is the source of light at a depth h below the surface of water. Light rays incident on the surface of the water from 0 at angles less than critical angle transmit in air after refraction. At points A and B the light rays are incident at angles equal to the critical angle (θ). So the refracted rays at these two

Points graze along the surface of separation. So the light rays will transmit outside water only through the circular section of radius r =  AP = PB.  If the rays are Incident on the surface of water excluding this circular section, the»7 will be reflected from (be surface of the water and will return to water,

Let the angle of Incidence be θ.

⇒ \(\sin \theta=\frac{1}{a^{\mu_w}}=\frac{1}{\mu}=\frac{3}{4}\)

= sin 48°36′ or, = 48° 36′

From the triangle AOP,

⇒ \(\tan \theta=\frac{A P}{O P}=\frac{r}{h} \quad \text { or, } \frac{\sin \theta}{\cos \theta}=\frac{r}{h}\)

Or,\(\frac{\frac{1}{\mu}}{\sqrt{1-\frac{1}{\mu^2}}}=\frac{r}{h}\)

∴ \(\sin \theta=\frac{1}{\mu}\)

Or, \(r=\frac{h}{\sqrt{\mu^2-1}}=\frac{h}{\sqrt{\frac{16}{9}-1}}=\frac{3}{\sqrt{7}}\)

Example 2.  The water in a pond has a refractive Index| of light and is placed 4 m below the surface of the water. Calculate the minimum radius of an opaque disc that needs to be floated on water so that light does not come out.
Solution:

Minimum radius of the opaque disc,

r = \(\frac{h}{\sqrt{\mu^2-1}}=\frac{4}{\sqrt{\left(\frac{5}{3}\right)^2-1}}\)

= \(\frac{4}{\frac{4}{3}}\)

= 3m

Example 3. Shows a longitudinal cross-section of an optical fibre made of glass with a refractive index of 1.68. The pipe is coated with a material of a refractive index of 1.44. What is the range of the angles of the incident rays with the axis of the pipe for which total reflection inside the fibre can take place?

Solution:

The refractive index of the outer coating concerning the glass pipe

⇒ \(g_g \mu_c=\frac{a^{\mu_c}}{a^{\mu_g}}=\frac{1.44}{1.68}\)

If the critical angle for the total reflection is θ, then

sinθ = \(\frac{1}{c^{\mu_g}}\)

= \(\frac{1.44}{1.68}\)

= 0.857 or, 59°

Thus’ total reflection takes place when i’ > 59° or when r < r_maxwhere r_max= 90° – 59° = 31°

So if the maximum angle of incidence on the fibre is imax then,

sin i_max  = μ_gsir r _max= 1.68 sin 31° = 0.865

_Imax  = 60°

So, the range of the angles of incidence for total internal reflection inside the fibre is from 0° to 60°.

Example 4. In which direction will the sun appear to set if the observer is inside the water of a pond? Refractive index of water, μ – 1.33.
Solution:

For a setting sun, the incident rays graze along the surface of the water, i.e., angle of incidence = 90°

∴ According to Snell’s law

μ_w= \(\frac{\sin i}{\sin r} \)

Or,   1.33 \(=\frac{\sin 90^{\circ}}{\sin r}\)

sir r  = \( \frac{1}{1.33}\)

= 0.7518

= sin 48.75°

r = 48.75°

Therefore, to see the setting sun the observer in water should look at an angle of 48.75° with the normal.

Optics

Refraction Of Light Atmospheric Refraction

Apparent position Of a star: The whole atmosphere surrounding the earth may be supposed to be divided into different horizontal layers. As the height above the earth’s surface increases, the density of the air decreases. Due to this, the refractive index of air also decreases with the increase in altitude. For this reason, the ray from a star S (say) proceeding towards the earth’s surface cannot travel straight but continually bends towards the normal at the surface of separation due to refraction as it penetrates from rarer to denser layers

This ray after several refractions reaches the observer at O. But our vision cannot follow the curved path OS. A tangent OS’ is drawn on OS at O . So, the observer sees the star at S’. This phenomenon is called atmospheric refraction.

Visibility of the sun before sunrise and after sunset:

The diameter of the sun subtends an angle of 0.5° at the eye of an observer on Earth. This value is equal to the deviation of sun¬ light due to atmospheric refraction, when at the horizon. So, the sun appears to just touch the horizon during sunset and sunrise when it is actually below it. What we see therefore is the raised image of the sun, formed due to atmospheric refraction. As a result, we see the sun a few minutes after sunset or before sunrise. Now, the sun covers a distance equal to its diameter in 2 min.

So, the sun becomes visible another 2 min earlier at sunrise and also remains visible for another 2 min after the actual sunset. Consequently, 4 min are added to the length of a day. This value is valid for observations from the equatorial region. At higher latitudes, this time increases

The oval shape of the sun when It Is near the horizon:

During sunrise or sunset, the lower edge of the sun remains nearer to the horizon than its upper edge. So, the rays coming from the lower edge of the sun are incident on an atmospheric layer at a larger angle than that for the rays coming from its upper edge. As the refracting angle increases with the increase of the incident angle, the rays coming from the lower edge bend more than the others due to multiple refractions at different atmo¬ spheric layers. As a result, the vertical diameter of the Sun appears to be reduced whereas the horizontal diameter remains unaffected.

Twinkling of Stars:

Due to atmospheric refraction, we see the stars twinkle. Light rays from the stars situated far and far away from us come to our eyes passing through various layers of air. The temperature of the layers does not remain constant and changes continuously. So the density of the various layers also. changes.

Again the refractive index of the layers changes with the change of density. So, when the rays of light from a star come to our eyes, the direction of the path of the rays changes continuously. As a result, the amount of light reaching our eyes also changes continuously. It seems as if the brightness of the stars is changing. So the stars appear twinkling.

As the planets are nearer to us than the stars, more amount of light comes to us. Therefore, the change in brightness of the planets due to changes in the refractive index of various layers of air is negligible. We cannot detect it with our eyes. So it appears that the planets are emitting light steadily.

Optics

Refraction Of Light Thin Prism

Thin Prism Definition:

The prism, whose refracting angle is very small (not more than 10°), is called a thin prism.

Deviation produced by a thin prism:

ABC is a thin prism. A ray PQ is incident on the refracting face AB nearly normally. For nearly normal incidence, i₁ ≈ 0, i₂≈0. If n is the refractive index of the material of the prism, then

μ = \(\frac{\sin i_1}{\sin r_1}=\frac{i_1}{r_1} \quad \text { or, } i_1=\mu r_1\)

And \(\mu=\frac{\sin i_2}{\sin r_2}=\frac{i_2}{r_2} \quad \text { or, } i_2=\mu r_2\)

Or,

So, the deviation of the ray

⇒ \(\delta=i_1+i_2-A=\mu r_1+\mu r_2-A=\mu\left(r_1+r_2\right)-A\)

= μA – A

Since = r₁+r₂ = A

= (μ- 1)

Again, if the ray PQ is incident on the face AB normally, then

i₁= r₁ = 0. So, A = r₂
.
Therefore, the deviation of the ray,

δ = i₁ + i₂ – A = μr₂-A =μA-A = (μ -1 )A

So, for normal and nearly normal incidence, the deviation of the array in a thin prism, δ = (μ-1)A

Thus it is seen that for normal or nearly normal incidence, the deviation of a ray in a thin prism depends only on the refract¬ ing angle of the prism and the refractive index of its material but not on the angle of incidence. So if the angle of the incidence is small, the deviation of a ray in the case of a thin prism remains constant.

Optics

Refraction Of Light Thin Prism Numerical Examples

Example 1. A very thin prism deviates a ray of light through 5°. If the refractive index of the material of the prism is 1.5, what is the value of the angle of the prism?
Solution:

The angle of deviation for a thin prism,

δ = (μ-1)A

Here  δ = 5° and μ = 1.5

Therefore from equation (1) we get,

5° = (1.5 -1)A or, A = 10°

Example 2. A prism haying refracting angle 4°. Is placed in the air. Calculate the angle of deviation of a ray incident nor¬ mally or nearly normally on It. The refractive index of the material of the prism| = \(\frac{3}{2}\)
Solution:

The refracting angle of the prism, A = 4°. So it is a thin prism. We know that the deviation of a ray in a thin prism for. normal or nearly normal incidence is given by,

⇒  \(\delta=(\mu-1)\)A

⇒ \(\delta=\left(\frac{3}{2}-1\right) \times 4^{\circ}\)

= 2°

Example 3. A thin prism with a refracting angle of 5° and having refractive index of 1.6 is kept adjacent to another thin prism having a refractive index of 1.5 such that one is inverted concerning the other. An incident ray falling vertically on the first prism passes through the second prism without any deviation. Calculate the refracting angle of the second prism.
Solution:

According to the condition,

(μ₁– 1)A₂ = (μ₂– 1)A₂

(1.6-1) × 5° = (1.5-1)A₂

Or, A₂ =\(\frac{0.6 \times 5^{\circ}}{0.5}\)

So the refracting angle of the second prism = 6°

Limiting Angle of a Prism for No Emergent Ray

A ray of light incident on a refracting surface of a prism may not emerge from the second refracting surface. It depends on the refracting angle of the prism. Every prism has a limiting value of its refracting angle. Light can emerge from the prism if the angle of the prism is equal to or less than this critical value, otherwise, no light can emerge from the prism.

Let ABC be the principal section of a prism  PQRS is the path of a ray through the prism placed in the air where ray RS grazes along the second face AC.

Let the angles of incidence and refraction at the face AB be i₁ and r₁ respectively and the corresponding angles at the face AC be r₂ and i₂, where i₂ = 90°.

So, r₂ = θ_c, the critical angle between glass and air. If the refracting angle of prism A is equal to the limiting angle, then the ray incident at an angle of incidence fj to the face AB of the prism makes a grazing emergence along the second refracting surface AC

A= r₁+r₂ ……………….(1)

For refraction at Q

⇒ \(\sin i_1=\mu \sin r_1 \quad \text { or, } r_1=\sin ^{-1}\left(\frac{\sin i_1}{\mu}\right)\)

For refraction At R,

⇒ \(\sin 90^{\circ}=\mu \sin r_2 \quad \text { or, } r_2=\sin ^{-1}\left(\frac{1}{\mu}\right)\)

Or, \(\theta_c=\sin ^{-1}\left(\frac{1}{\mu}\right)\)

From equation (1) we get, A = \(A=\sin ^{-1}\left(\frac{\sin i_1}{\mu}\right)+\sin ^{-1}\left(\frac{1}{\mu}\right)\) …………. (2)

Special cases:

Limiting angle of the prism for normal incidence on the first face: When ray, PQ is incident on the face AB normally, then fj = 0. In this case, if the emergent ray grazes along the surface AC then from equation (2) we get,

A = \(\sin ^{-1}\left(\frac{\sin 0}{\mu}\right)+\sin ^{-1}\left(\frac{1}{\mu}\right)=\sin ^{-1}\left(\frac{1}{\mu}\right)=\theta_c\)

Hence, the ray can emerge from the prism through its second surface till the refracting angle of the prism remains less than its critical angle. But as the refracting angle of the prism becomes greater than its critical angle, no ray emerges from the surface AC. Then the face AC acts as a total reflecting surface.

Limiting angle of the prism for grazing incidence on the first face:

For grazing incidence on the face AB, i₁ = 90° Then from equation (2) we get

A = \(=\sin ^{-1}\left(\frac{\sin 90^{\circ}}{\mu}\right)+\sin ^{-1}\left(\frac{1}{\mu}\right)\)

= \(\sin ^{-1}\left(\frac{1}{\mu}\right)+\sin ^{-1}\left(\frac{1}{\mu}\right)=2 \sin ^{-1}\left(\frac{1}{\mu}\right)=2 \theta_c\)

So, if the refracting angle of the prism is greater than 20C and if a ray is incident on the face AB grazing the surface, then it will be reflected from the face AC. It means the ray will not emerge in the air through the face AC. Hence, no emergent ray will be obtained.

Thus, from the above discussions we conclude that for any incidence no ray can emerge from the prism If the angle of the prism is greater than twice the critical angle for the material concerning the surrounding medium.

Optics

Refraction Of Light Limiting Angle Of Incidence For No Emergent Ray From A Given Prism

Just like a prism has a limiting refracting angle for no emergent ray, a prism with a definite refracting angle also possesses a limiting angle of incidence for no emergent ray from it. If the angle of incidence i₁ becomes less than this limiting incident angle, then there will be no corresponding emergent ray.

Let ABC be the principal section of a prism. The ray of light PQ is incident at Q on the face AB. After refraction through the prism, the emergent ray RS grazes the second face AC of the prism.

Let the angles of incidence and refraction at the face AB be i₁ and r₁ the corresponding angles at the face AC be r₂ and i₂ respectively. Here, i₂ = 90° .

Now, the angle of the prism, A = r₁ + r₂ – constant.

From, A = r₁ + r₂; we get, r₂ = A – r₁ , reduces with the decrease of it. Again, r₁ increases with the decrease of r₁

Now, if r₂ is greater than θ_c the ray QR is reflected from the face AC inside the prism and it does not emerge in the air.

So when r₁ = θ_c then i₁ = limiting angle of incidence.

If μ is the refractive index of the material of the prism then

⇒ \(\sin \theta_c=\frac{1}{\mu}\)

⇒ \(\cos \theta_c=\sqrt{1-\sin ^2 \theta_c}=\sqrt{1-\frac{1}{\mu^2}}=\frac{\sqrt{\mu^2-1}}{\mu}\)

Considering the refraction of the ray at Q we have,

⇒ \(\cos \theta_c=\sqrt{1-\sin ^2 \theta_c}=\sqrt{1-\frac{1}{\mu^2}}=\frac{\sqrt{\mu^2-1}}{\mu}\)

= \(\cos \theta_c=\sqrt{1-\sin ^2 \theta_c}=\sqrt{1-\frac{1}{\mu^2}}=\frac{\sqrt{\mu^2-1}}{\mu}\)

= \(\sin i_1=\mu \sin r_1=\mu \sin \left(A-r_2\right)\)

Since A = r₁+r₂

= \(\mu \sin \left(A-\theta_c\right)=\mu\left[\sin A \cos \theta_c-\cos A \sin \theta_c\right]\)

= \(\mu\left[\sin A \cdot \frac{\sqrt{\mu^2-1}}{\mu}-\cos A \cdot \frac{1}{\mu}\right]\)

= \(\mu\left[\sin A \cdot \frac{\sqrt{\mu^2-1}}{\mu}-\cos A \cdot \frac{1}{\mu}\right]\)

= \(\sin A \sqrt{\mu^2-1}-\cos A\)

Or, i₁ = \(i_1=\sin ^{-1}\left[\sin A \sqrt{\mu^2-1}-\cos A\right]\)

This is the limiting angle of incidence. If the angle of incidence is less than this limiting angle, no ray will emerge from the second face of the prism

Optics

Refraction Of Light Limiting Angle Of Incidence For No Emergent Ray From A Given Prism Numerical Examples

Example 1. To get an emergent ray from a right-angled prism its refractive index should not exceed \(\sqrt{2}\) —prove it.
Solution:

The condition of getting an emergent ray from a prism is that the refracting angle of the prism should be equal to or less than twice the value of the critical angle

A≤ 2θ_c Or, 90°≤ 2θ_c 

Or, θ_c ≥ 45°

∴ sin θ_c ≥ sin 45° ,Or, sin θ_c ≥ \(\frac{1}{\sqrt{2}}\)

∴  \(\sin \theta_c=\frac{1}{\mu}\)

∴ \(\frac{1}{\mu}\frac{1}{\sqrt{2}}\)

Or, \(\sqrt{2}\)

Example 2. The refractive index of a prism having a refracting angle of 75° is \(\sqrt{2}\). What should be the minimum angle of incidence on a refracting surface so that the ray will emerge from the other refracting surface of the prism?

Solution: According to the question, the emergent angle is i₂ = 90°.

So for refraction at the second face of the prism

μ = \(\frac{\sin i_2}{\sin r_2}=\frac{\sin 90^{\circ}}{\sin r_2}\)

Or, \(\sin r_2=\frac{1}{\mu}=\frac{1}{\sqrt{2}}\)

= sin 45°

Or, r₂ = 45°

We, know A= r₁ +r₂

75= r₁ + 45°

Or, r₁ = 30°

For refraction at the first face

μ = \(\frac{\sin i_1}{\sin r_1}\)

Or, \(\sqrt{2}=\frac{\sin i_1}{\sin 30^{\circ}}\)

Or, \(\sin i_1=\sqrt{2} \times \frac{1}{2}=\frac{1}{\sqrt{2}}\)

= sin 45°

Or, i₁ = 45°

The required angle of incidence = 45°

Example 3. Find the value of the limiting angle of incidence if the refractive index of the material of the prism is 1.333 and the angle of the prism is 60°
Solution:

Here, the refractive index of the material of the prism, mu = 1.333; the angle of the prism, A = 60°

⇒ \(i_L=\sin ^{-1}\left[\sin A \sqrt{\mu^2-1}-\cos A\right]\)

= \(\sin ^{-1}\left[\frac{\sqrt{3}}{2} \sqrt{(1.333)^2-1}-\frac{1}{2}\right]\)

= \(\sin ^{-1}(0.2633)\)

= 15.27°

Example 4. The refracting angle of the prism is 60° and its refractive J index is Jl. What should be the minimum angle of | incidence on the first refracting surface so that the ray | can emerge somehow from the second refracting our face?
Solution:

Let i be the limiting angle of incidence, then

sin i = \(\sqrt{\mu^2-1} \cdot \sin A-\cos A\)

= \(\sqrt{\frac{7}{3}-1} \cdot \sin 60^{\circ}-\cos 60^{\circ}\)

= \(\frac{2}{\sqrt{3}} \cdot \frac{\sqrt{3}}{2}-\frac{1}{2}\)

= \(1-\frac{1}{2}=\frac{1}{2}\)

i = 30

Example 5. The refractive index of the material of a prism is \(\)  and the refracting angle is 90°. Calculate the angle of minimum deviation and the corresponding angle of incidence. Show that the limiting angle of incidence for getting emergent ray is 45°
Solution:

Here, the refractive index of the material of the prism,

M = \(\sqrt{\frac{3}{2}}\)

The angle of prism A =  90°

μ = \(\frac{\sin \frac{A+\delta_m}{2}}{\sin \frac{A}{2}}\)

Or, \(\sqrt{\frac{3}{2}}=\frac{\sin \frac{A+\delta_m}{2}}{\sin 45^{\circ}}\)

⇒ \(\sin \frac{A+\delta_m}{2}=\sqrt{\frac{3}{2}} \times \frac{1}{\sqrt{2}}=\frac{\sqrt{3}}{2}\)  = sin 60°

⇒  \(\frac{A+\delta_m}{2}\)  = 60°

Or, A + δ_m = 120°

δ_m = 120° – 90° = 30°

For minimum deviation, i₁= i₂

δ_m = i₁+ i₂ -A

30° = 2i₁ – 90

Or, 2i₁ = 120°

Or, i₂= 60°

Or minimum deviation angle of indecency = 60

To obtain the emergent ray I be the limiting angle of incidence. Then

sini = \(\sqrt{x^2-1} \sin x-\cos 4\)

= \(\sqrt{\frac{3}{2}-1} \cdot \sin 90^{\circ}-\cos 90^{\circ}=\frac{1}{\sqrt{2}}\)

= sin 45°

= 45°

Example 6. The refractive index of a prism is \(\sqrt{2}\). A ray of light is incident on the prism grazing along one of its refracting surfaces. What should be the limiting angle of the prism for no emergent ray from the other face?
Solution:

If The limiting angle of the prism for no emergent ray is A, then

A = \(2 \sin ^{-1} \frac{1}{\mu}\)

= \(2 \sin ^{-1} \frac{1}{\sqrt{2}}=2 \times 45^{\circ}\)

= 90°

Optics

Refraction Of Light Conclusion

1. When a ray of light enters a medium from another medium through the interface of the two media, then the path of the rav changes its direction and this phenomenon is known as refraction of light.

2. Laws of refraction:

• The incident ray, the refracted ray and the normal to the refracting surface at the point of incidence lie on the same plane.
• The sine of the angle of incidence bears a constant ratio to the sine of the angle of refraction. The value of this constant depends on the nature of the pair of media concerned and the colour of the incident light.

3. Relative refractive index:

When a ray of light is refracted from a medium a to another medium b then the ratio of the sine of the angle of Incidence to the sine of the angle of refraction is called the refractive index of the medium b concerning the medium a. This refractive index is called the relative refractive index.

4. Absolute refractive index:

When a ray of light is refracted from the vacuum to any other medium then the ratio of the sine of the angle of incidence to the sine of the angle of refraction is called the absolute refractive index of that medium.

5. Critical angle:

When a ray of light is refracted from a denser medium to a rarer medium than for a particular angle of incidence in the denser medium, the angle of refraction in the rarer medium is 90° i.e., the refracted ray grazes along the interface of the .two media. Then that particular angle of incidence is called the critical angle for the pair of media.

6. Total internal reflection:

While passing from a denser medium to a rarer medium, if a ray of light is incident on the surface of separation between the angle greater than the critical angle for the two media involved, the ray of light is reflected to the denser medium without undergoing refraction in the rarer medium. This phenomenon is known as the total internal reflection of light.

Through optical fibres tight rays can be transmitted from one place to another in the straight or curved path by successive total internal reflections. f If a ray of light from a rarer medium is incident on a prism of denser medium then after refraction through the prism, the ray of tight bends towards the base of tyre prism. -f In the case of reflection or refraction, the change of direction of tight is called deviation.

4- Minimum deviation:

For any prism, there is a certain angle of incidence for which the angle of deviation becomes minimum or the least. This angle of deviation is called the angle of minimum deviation for the prism.

At minimum deviation, the angle of incidence becomes equal to the angle of emergence.

5. Thin prism:

If the refracting angle of a prism is very small (not more than 10° ) then it is called a thin prism.

6. Total reflecting prism:

If the principal section of a prior made of crown glass is a right-angled isosceles triangle then the prism makes a total Internal reflection of light for any incidence on it hence this type of prism is called a total reflecting prism.

7. During the refraction of light, the velocity, intensity and wavelength of light change but its frequency and phase remain constant.

8. If there is a parallel glass slab in the path of light then Ugh rays remain undefeated after refraction through it but lateral displacement of light rays will occur.

9.  \(\frac{\sin i}{\sin r}={ }_a \mu_b=\frac{\mu_b}{\mu_a}\)

μ_a = absolute refractive index of medium a, nb = absolute refractive index of medium b, afib = relative refractive index of medium b concerning the medium a.

10.  \(\frac{\sin i}{\sin r}={ }_a \mu_b=\frac{\mu_b}{\mu_a}\)

[ va = velocity of light in the medium a, vb= velocity of tight in the medium b ]

⇒  \(\mu=\frac{c}{v}\)

[n = absolute refractive index of a medium, v = velocity of light in that medium, c = velocity of tight in vacuum]

[v= – frequency of the wave, A = wavelength]

11. \(a^{\mu_b}=\frac{v_a}{v_b}\)

The relation between (i and A given by scientist Cauchy

12. \(\mu=\frac{c}{v}\)

Deviation of a ray of tight due to refraction,

= i~r [i = angle of incidence, r = angle of refraction]

13. Lateral displacement of a ray of tight after suffering refraction through a parallel plate glass slab

= \(t \sin i_1\left[1-\frac{\cos i_1}{\sqrt{\mu^2-\sin ^2 i_1}}\right]\)

[where, i- the angle of incidence on the front surface of the tin slab, t = thickness of the slab, (J- = refractive index of the material of the slab]

If the angle of incidence ij is very small then lateral displacement = \(t i_1\left(1-\frac{1}{\mu}\right)\)

14. General formula ofSnell’s law:

μ₁ sin₁  = μ₂sin₂ =  μ₃ sin₃ = ………………. = μ₃ sin i₃

15. The apparent position of an object due to refraction:

The refractive index of rarer medium =; refractive index c denser medium = μ₁

1. If the object is in a denser medium and the observer is in: a rarer medium:

The refractive index of the denser medium with respect to the rarer medium real depth of the object from
= \(=\frac{\text { the surface of separation }(d)}{\text { apparent depth of the object from the surface of separartion}}\)

= \(\frac{\mu_2}{\mu_1}\)

2. If the object Is In the rarer medium and the observer is in a denser medium: The refractive index of the denser medium concerning the rarer medium apparent height of the object from

= \(\frac{\text { the surface of separation }\left(d^{\prime}\right)}{\text { real height of the object from the surface of separation (d)}}\)

16. The refractive index of glass

=\(\frac{\text { real thickness of a thick mirror }}{\text { apparent thickness of a thick mirror }}\)

⇒ \(\frac{1}{b^{\mu_a}}=\frac{\mu_b}{\mu_a}\)

[where, A = r₁+r₂]

17. Deviation of a ray of light after refraction through a prism,

[i = angle of incidence on the first face of the prism, i’μ_a = angle of emergence from the second face to the prism, r₁

= angle of refraction In the first face, = angle of incidence in the second face, A = refracting angle of the prism]

18. In case of minimum deviation, i₁ = i₂ = i (say)

r₁ = r₂ = r (say)

19. In that case, the minimum deviation, δ

m = 2i-r and A = 2r.

20. Relation between refractive index (fi) and minimum deviation (8m ) :

μ\(\frac{\sin \left(\frac{A+\delta_m}{2}\right)}{\sin \frac{A}{2}}\)

21. For the normal and nearly normal incidence of a ray of light, deviation in the case of a thin prism,

⇒ δ = (μ-1)A

22. Limiting the angle of a prism for no emergent light from a prism

⇒ \( \sin ^{-1}\left(\frac{\sin i_1}{\mu}\right)+\sin ^{-1}\left(\frac{1}{\mu}\right)\)

When iy = 0 , then A>θ_c; when iy = 90°, then A >2θ_c.

Limiting value of the angle of incidence in case of definite prism for no emergent light from it,

⇒\(i_1=\sin ^{-1}\left[\sin A \sqrt{\mu^2-1}-\cos A\right]\)

23. A ray of light passes through n number of media of refractive indices, μ₁,  μ₂, μ₃, ……………….. μ_n respectively. The planes of the media are parallel. If the emergent ray from the n -th medium is parallel to the ray incident on the first medium, then μ₁ = μ₂

24. For a hollow prism, the angle of the prism is A ≠ 0 but the angle of deviation is δ = 0.

25. A container of depth 2d is half filled with a liquid of refractive index μ₁ and μ₂    the remaining half is filled with another liquid of refractive index When seen from the top, the apparent depth of the container is

⇒ \(d\left(\frac{1}{\mu_1}+\frac{1}{\mu_2}\right)\)

26. An observer is situated at a depth h in a water body. The water surface appears as a porous circular mirror to

 27. If a parallel plane glass slab of thickness t is kept in the path of a beam of converging rays, then the displacement of the intersecting point of the rays,

⇒ \(O O^{\prime}=x=\left(1-\frac{1}{\mu}\right) t\)

To transmit through a glass slab of thickness t and refractive index μ, time taken by light is, where c is the velocity of light in vacuum \(\frac{\mu t}{c}\)

Optics

Refraction Of Light Very Short Questions

Question 1. What is the angle of deviation due to the refraction of a ray of light incident perpendicularly on a refracting surface?
Answer: Zero

Question 2. Arrange the following media according to the increasing optical density Air, diamond, glass, water, glycerine.
Answer:

Air < Water < Glycerine < Glass < Diamond]

Question 3. Can the value of absolute refractive index of a medium be
Answer: No

Question 4. State the relation of velocity of light with refractive index
Answer:

⇒ \(\left[\mu=\frac{c}{v}\right]\)

Question 5. The refractive index of a medium depends on temperature’—is the statement correct or wrong?
Answer: Correct

Question 6. Arrange the following media according increasing velocity of light through them: vacuum, diamond, water, air, glass
Answer:

Diamond < Glass < Water < Air < Vacuum

Question 7. Does the velocity of light in a vacuum depend on

1. Wavelength of light
2. Frequency of light
3. Intensity of light?

Answer: No

Question 8 If water is heated, how refractive index it will change?
Answer: Decrease

Question 9. The refractive index of glass is 1.5. What is the velocity of light
Answer: [2 × 10⁸m/s].

Question 10. Can the relative refractive index of a medium concerning another be less than unity?
Answer: Yes

Question 11. The refractive index of a medium is a physical quantity having no dimension and no unit—is the statement true or
Answer: True

Question 12. A light ray of wavelength 4500 A enters a glass slab of refractive index 1.5, from the vacuum. What is the wavelength of light inside the slab?
Answer: 30000 A°

Question 13. For which colour of light is the refractive index of glass the minimum?
Answer: Red

Question 14. If the frequency of light increases will there be any change in the refractive index of the medium?
Answer: No

Question 15. For which colour of light is the refractive index of glass a maximum?
Answer: Violet

Question 16. Does the refractive index of glass depend on the colour of light? If so, how?
Answer: Yes, the refractive index increases with the decrease of wavelength

Question 18. When a ray of light is incident on a plane normally, then what is the value of the angle of refraction?
Answer: Zero

Question 17. If for refractive index of water relative to air is \(\), what will be the refractive index of air relative to water
Answer:

⇒ \(\frac{3}{4}\)

Question 18. When light travels from air to glass, how does its wavelength change? less than 1
Answer: Wavelength decreases

Question 19. The absolute refractive index of water and glass are \(\frac{4}{3}\) and \(\frac{3}{2}\). What is the ratio of velocity of light in glass and water?
Answer: [8:9]

Question 20. what is the value of the product of the refractive index of relative of the first medium and that of the first-second medium relative to the second medium?
Answer: 1

Question 21. For what angle of incidence the lateral shift produced by a parallel-sided glass plate is zero?
Answer: For i = 0

Question 22. For what angle of incidence the lateral shift produced by a parallel-sided glass plate is maximum?
Answer: For i = 90°

Question 23. what is the maximum lateral shift produced by a parallel-sided glass plate of thickness t?
Answer: t

Question 24. The bird descends vertically downwards in the direction of a pond during its flight. To a fish which is underwater and directly below the bird, what will be the apparent position of the bird?
Answer:

The bird’s apparent position will be slightly above its actual position i.e., the bird will appear higher up than it is

Question 25. An object lying inside a pond is viewed by a man from above (air) along a horizontal plane. Now if the man moves away from the object keeping his eyes along the same horizontal plane, how will the apparent depth of the object change?
Answer: Decrease along the caustic curve

Question 26. A transparent cube of glass of refractive index n and thickness ‘t is placed on a spot of ink drawn on white paper. When the spot is viewed from above (air) normally, the spot appears to shift through a distance of At towards the observer. What is the value of Δt?
Answer:

⇒ \(\left[\left(l-\frac{1}{\mu}\right) t\right]\)

Question 27. If a straight rod is held obliquely in water how does the immersed portion of the rod appear?
Answer: Refraction of light

Question 28. Multiple images are formed in a thick mirror. Which Image looks brightest?
Answer: Second images

Question 29. If a completely transparent object has to be made Invisible In a vacuum, what should be the value of its relative Index?
Answer: 1

Question 30. What Is the critical angle of light when passing from water \(\left(\mu \text { of water }=\frac{4}{3}\right)\)
Answer: 49

Question 31. For which colour of light is the critical angle between glass and air minimum?
Answer: Violet

Question 32. What are the factors of light which are responsible for creating mirages?
Answer: Refraction and total internal reflection]

Question 33. In which direction do we have to look to see the setting sun if we are underwater? (μ_w = 1.33)
Answer: At an angle of 49° with the normal drawn at the surface of the water]

Question 34. From sunrise to sunset, the sun subtends an angle of 180’ to our eyes. What will be the value of this angle to an observer underwater?
Answer: 98°

Question 35. Light passes through an optical fibre following which physical phenomenon?
Answer: Total internal reflection

Question 36. What is the approximate value of the refractive index of a diamond
Answer: 2.42

Question 37. what kind of image is formed in a desert by the formation
Answer: Virtual

Does critical depend on the colour of life

Question 38. The angular altitude at which we see a star is not its actual angular altitude’—is the statement true or false?
Answer: True

Question 39. The sun appears to be elliptical during sunset. What is the reason behind it?
Answer: Refraction of light

Question 40. Under what condition the angle of deviation of a refracted ray through a prism will be minimal?
Answer: If the angle of incidence and the angle of emergence are equal

Question 41. If the medium surrounding the four faces of a prism is denser an the entire material of the prism, then in which direction will the light rays bend when emerging from the prism?
Answer: It will bend upwards, towards the prism apex

Question 42. How can an inverted image be made erect by using a total reflecting prism?
Answer: By total internal reflection on the hypotenuse of a right-angled

Question 43. ‘When light travels from a rarer medium to a prism which is a denser medium, a light ray bends towards the base of the prism. Is the statement true or false?
Answer: True

Question 44. At the position of minimum deviation, what is the nature of the path of a light ray through a prism?
Answer: Symmetrical

Question 45. For a thin prism on what factor does the magnitude of the angle of deviation of light rays depend?
Answer: Angle of incidence

Question 47. what will be the angle of deviation of a ray of light incident normally on any smaller side of a total reflecting prism?
Answer: 90°c

Optics

Refraction Of Light Fill In The Blanks

Question 1. Which of the following_________________wavelength, frequency or velocity does not change during the refraction of light?
Answer: Frequency

Question 2. For any particular medium, the refractive index is greater If the wavelength of light is____________________ 
Answer: Small

Question 3. If μ_a,  μ_b and μ_c are the absolute refractive indices of three media then _aμ_b × _bμ_c  _________________________
Answer: _aμ_c

Question 4. When a man stands inside a shallow pond the depth of water at that place appears___________________places appear comparatively___________________ and the depth of other 
Answer: Maximum, Less

Question 5. Critical angle of which pair of medium is lesser___________air and water or air and diamond
Answer: Air and Diamond

Question 6. We see the sun ___________________ a few minutes ___________________ sunset or sunrise
Answer: After, Before

Question 7. If a ray of light moves from a ___________________ medium to a ___________________, total internal reflection does not take place
Answer: Rarer, Denser

 Optics

Refraction Of Light Assertion Reason Type

Direction: These questions have statement 1 and statement 2 of the four characters given below, choose t=the one that best describes the two statements

1. Statement 1 is true, statement 2 is true, and statement 2 is a correct explanation for statement 1.
2. Statement 1 is true, statement 2 is true, and statement 2 is not a correct explanation for statement 1
3. Statement I is true, and statement 2 is false
4. Statement 1 is false statement 2 is true

Question 1. 

Statement 1: The Greater the refractive index of a medium or denser the medium, the lesser the velocity of light In that medium

Statement 2: Refractive index is inversely proportional to velocity.

Answer: 1.  Statement 1 is true, statement 2 is true, and statement 2 is a correct explanation for statement 1.

Question 2.

Statement 1: The critical angle of the light passing from glass to is minimal for violet colour

Statement 2: The wavelength of violet light is greater than the light of other colours

Answer: 3. Statement I is true, and statement 2 is false

Question 3.

Statement 1: The twinkling of the star is due to the reflection of light

Statement 2: The velocity of light changes while going from one medium to the other.

Answer: 4. Statement 1 is false statement 2 is true

Question 4.

Statement 1: The relative refractive index of a medium can be less than unity.

Statement 2: The angle of incidence is equal to the angle of refraction

Answer: 3. Statement I is true, and statement 2 is false

Question 5.

Starement 1: When a ra> of light enters glass from atr. Its frequency changes

Sutrmem 2: The velocity of light in glass is less than that in oil.

Answer: 4. Statement 1 is false statement 2 is true

Question 6.

Statement 1: The refractive index of a medium Is Inversely proportional to temperature

Statement 2: Refractive index U is directly proportional to the density of the medium.

Answer: 2. Statement 1 is true, statement 2 is true, and statement 2 is not a correct explanation for statement 1

Question 7. 

Statement 1: The velocity of light rays* of different colours is the same. But the velocity of the light rays is different for anv another medium.

Statement 2: If v = velocity of light in the respective

Answer: 1.  Statement 1 is true, statement 2 is true, and statement 2 is a correct explanation for statement 1.

Question 8.

Statement 1: The linages formed the total lniern.il reflections are much brighter than those formed by lenses.

Statement 2: There Is no loss of Intensity during total Internal reflection

Answer: 1.  Statement 1 is true, statement 2 is true, and statement 2 is a correct explanation for statement 1.

Optics

Refraction Of Light Match The Columns

Question 1. 

Answer:  1- A, C, 2-A, C, 3- B, 4- 4, 5- A, C

Question 2.

Answer: 1-C, D, 2- A, 3- B, E

Question 3.

Answer: 1- D, 2- A, 3- B, 4. C

μ₁, μ₂ and μ₃ are the refractive indices of the first, second and third medium respectively

Question 4.

Answer: 1- B, 2-1, 3 -C