WBCHSE Class 12 Physics Notes For Refraction Of Light Prism

WBCHSE Class 12 Physics Prism Notes

Optics Refraction Of Light Prism Of Some Definitions

Prism:

A prism is a portion of a transparent medium confined using two plane faces inclined to each other. In the DEFGHK is a prism and it is confined by the two planes DEHK and DFGK.

Refracting face:

The two plane faces inclined to each P S other at some angle by which the prism is bound are called the refracting faces. In, DEHK and DFGK are the refracting faces. BL

Edge:

The line along which the two refracting faces meet is called the edge of the prism. In Rg. 2.42, the line DK is the ‘ edge of the prism.

Refracting angle or angle of the prism:

The angle included between the two refracting faces is called the refract¬ ing angle or simply the angle of the prism. In the figure, ZEDF is the angle of the prism.

Side face and base:

In general, besides two refracting surfaces, a prism further on is enclosed by three more surfaces. Among these surfaces, two surfaces are triangular and are
placed perpendicular to the edge of the prism.

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In these two surfaces are DEF and KHG. These are the side faces of the prism. Another one is rectangular and situated perpendicular to the side face. In the surface is EFGH. It is called the base of the prism

Principal section:

The triangular cross-section cut by a perpendicular plane at right angles to the edge of a prism is called a principal section of the prism In ABC is a principal section of the prism

Light can enter or emerge from the prism through its refracting surfaces as those surfaces are plain and smooth In some cases, the base and side faces of a prism are made rough so that no light passes through it A prism is usually represented by its principal section

Refraction of Light along the Principal Section of a Prism

In ABC is the principal section of a prism. AB and AC are the refracting faces and BC is the base of the prism. A ray PQ is an incident on the face AB at Q where NQO is the normal. The prism is supposed to be an optically denser medium with

Respect to its surroundings. So, after refraction on plane AB, the refracted ray QR bends towards the normal NQO. The refracted ray QR is then incident on the face AC at H where N’RO is normal. The emergent ray RS bends av/ay from the normal N’RO after refraction.

So RQRS is the whole path of the ray of light. The angle between the direction of the incident ray and the direction of the’ emergent ray gives the angle of deviation. In, the angle of deviation, δ = ∠MTS.

WBCHSE class 12 physics prism notes Expression for the angle of deviation:

Let the angle of the prism = ∠BAC – A. For refraction on the face AB, the angle of incidence =∠PQN = i₁, and the angle of refraction = ∠RQO = r₁  for refraction on the face AC, the angle of incidence of QR = ∠QRO =r_2, and the angle of refraction. = ∠N’RS =i₂ .

Now we get from the ∠QRT, angle of deviation,

δ = ∠MTR = ∠TQR + ∠TRQ

= (i₁ ~ r₁) + (i₂– r₂) = i₁ + i₂– (r₁+ r₂) Now, from the quadrilateral AQOR, as NO and N’O are normals on AB and AC respectively, so A + ∠QOR = 180°.

Again, from the triangle QOR

∠QOR + r₁ + r₂ = 180°

∴ A = r₁ + r₂ and ……………………….(1)

δ  = i₁ + i₂-A ……………………….(2)

So the angle of deviation δ depends on the angle of incidence. From the principle of reversibility of light rays follows that, if a ray enters the face A C along SR at an angle of incidencei₂ it will emerge along QP at an angle i₁ and suffer the same deviation. In other words, the same deviation occurs for two values of i.

Angle of Minimum Deviation

‘The angle of deviation of a ray of light passing through a prism depends on the angle of incidence because in the relation

WBBSE Class 12 Refraction Through Prism Notes

δ= i₁ + i₂– A, A is constant and i₂ depends on i₂ . A graph is drawn taking the angle of incidence z as abscissa and the angle of deviation o as ordinateThe graph indicates that the deviation decreases at first with the increase in the angle of incidence and then attains a minimum (δ_m) for a particular value of the angle of incidence i₀. Then with a further increase in the angle of incidence, the; deviation also increases.

So for every prism, there is a fixed angle of incidence for which the deviation suffered by a ray of light traversing the prism minimal. This angle of deviation is called the angle of minimum deviation of a prism. The position of the prism for which the value of the deviation becomes minimum is called the position of minimum deviation of the prism. This position of the prism is unique i.e., for only, one position of the particular prism, the deviation becomes minimal.

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Analysis of (i -δ) graph and condition of minimum deviation:

A line AB parallel to the i -i-axis is drawn. The coordinates of points, A and B are respectively (i₁,  ) and (i₂, δ ). The angle of deviation is o for an angle of incidence i₁ or i₂. We have seen that for angle of incidence i1 the emergent angle is z’2. The length of the line AB = (i₂– i₁). As the line AB moves parallel to itself downwards, the value of o decreases and the length of the line ( i₂– i₁) also decreases gradually. At point C, the length of the line AB becomes zero and o also becomes minimum. Then i₂– i₁ = 0 Or, i₂= i₁

So we can say that during refraction through a prism, when the angle of incidence and the angle of emergence are equal, the angle of deviation becomes minimal.

Condition of minimum deviation by calculus:

We know for refraction through a prism, the angle of deviation of a ray of light

Condition of minimum deviation by calculus: we know for refraction through a prism, the angle of deviation of a ray of light

δ = i₁+i₂-A ………………..(1)

And angle of the prism

A= r₁ +r₂ ………………..(2)

The angle of deviation 8 depends on the angle of incidence i₁.

For minimum deviation δ

⇒ \(\frac{d}{d i_1}(\delta)=0\)

∴ \(\frac{d}{d i_1}\left(i_1+i_2-A\right)\) =0

Or, 1\(\frac{d i_2}{d i_1}\)

∴ \(\frac{d i_2}{d i_1}=-1\) ………………..(3)

Differentiating equation (2) we have,

⇒ \(\frac{d}{d r_1}(A)=\frac{d}{d r_1}\left(r_1+r_2\right)\)

⇒ \(0=1+\frac{d r_2}{d r_1} \quad \text { or, } \frac{d r_2}{d r_1}=-1\) ……………………(4)

For refraction at Q and R , according to Snell’s law we have, sin i₁ = μ sin r₁ and sin i₂ = μ sin r₂; where μ = refractive index of the. material of the prism.

Differentiating the above equations we have

⇒ \(\cos i_1 d i_1=\mu \cos r_1 d r_1 \quad \text { and } \cos i_2 d i_2=\mu \cos r_2 d r_2\)

∴  \(\frac{\cos i_1}{\cos i_2} \cdot \frac{d i_1}{d i_2}=\frac{\cos r_1}{\cos r_2} \cdot \frac{d r_1}{d r_2}\)  ………………………………(5)

From equations (3), (4) and (5) we have

⇒ \(\frac{\cos i_1}{\cos i_2}=\frac{\cos r_1}{\cos r_2^2} \quad \text { or, } \frac{\cos ^2 i_1}{\cos ^2 i_2}=\frac{\cos ^2 r_1}{\cos ^2 r_2}\)

Or, \(\frac{1-\sin ^2 i_1}{1-\sin ^2 i_2}=\frac{1-\sin ^2 r_1}{1-\sin ^2 r_2}\)

Or, \(\frac{1-\sin ^2 i_1}{1-\sin ^2 i_2}=\frac{\mu^2-\mu^2 \sin ^2 r_1}{\mu^2-\mu^2 \sin ^2 r_2}=\frac{\mu^2-\sin ^2 i_1}{\mu^2-\sin ^2 i_2}\)

Or, \(\frac{1-\sin ^2 i_1}{1-\sin ^2 i_2}=\frac{\left(\mu^2-\sin ^2 i_1\right)-\left(1-\sin ^2 i_1\right)}{\left(\mu^2-\sin ^2 i_2\right)-\left(1-\sin ^2 i_2\right)}\)

\(\frac{a}{b}=\frac{c}{d}=\frac{c-a}{d-b}\)

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

= 1

∴ \(1-\sin ^2 i_1=1-\sin ^2 i_2 \quad \text { or, } \sin ^2 i_1=\sin ^2 i_2\)

Or, i₁ =  i₂

So the deviation is minimal when the angle of incidence (i₁) is equal to the angle of emergence (i₂)

The brightness of a ray of light at minimum deviation:

From the graph of the angle of incidence i and angle of deviation δ, it is found that generally if the angle of incidence is different, the angle of deviation is also different.

But if the angle of deviation attains its minimum value δ_m, then it is observed that for a wider range of angle of incidence (from i₁ to  i₂ vide, the angle of deviation of the ray becomes almost equal to the angle of minimum deviation δ_m, i.e., all the rays within this range emerge from the prism making a minimum angle of deviation δ_m.

So it can be said that in case of minimum deviation, the brightness of the emergent ray increases considerably. For this characteristic, the minimum deviation of a prism is of great importance

Path of Ray through a Prism for Minimum Deviation

Suppose a ray of light passes through a prism with minimum deviation along the path PQRS According to the condition of minimum deviation ix = i2. If the refractive index of the material of the prism is then

μ = \(\frac{\sin i_1}{\sin r_1}=\frac{\sin i_2}{\sin r_2}\)

r₁ = r₂

Since ∠AQR = 90°- r₁ and ∠ARQ = 90°- r₂ ,

So ∠AQR = ∠ARQ

Since, [r₁ = r₂]

So the triangle AQR is an isosceles triangle having AQ = AR.

So, for the minimum deviation of a ray, the point of incidence, Q and the point of emergence, R are equidistant from the vertex A of the prism.

So it is evident that when the deviation of a ray in a prism is minimum, the path of the ray through the prism becomes symmetrical.

Suppose, for the prism, AB = AC

∴ AQ= AR

∴ \(\frac{A Q}{A B}=\frac{A R}{A C}\)

i.e., the line QR is parallel to BC.

So, for an isosceles prism, when the deviation is minimum the path of the ray through the prism becomes parallel to the base of the prism.

Again the deviation of the ray due to refraction at AB = ( i₁ – i₂) and the deviation of the ray due to refraction at AC = ( i₂ – i₁). At the position of the minimum deviation of the prism,  i₁= i₂ and r₁ = r₂. So the deviations stated earlier become equal. Therefore, we can say that at the minimum deviation position of the prism, the total deviation is divided equally between the two refracting faces of the prism.

Short Notes on Dispersion by Prism

Refractive Index And Angle Of Minimum Deviation

We know that In the case of refraction of a prism, the angle of deviation of a ray δ = i₁+ i₂ = And the angle of the prism  A = r_{1 +} r₂ 

For minimum deviation r₁ = r₂. , Again when i₁ =  i₂ then r₁+ r₂

So angle of minimum deviation

δ_m = i₁+i₂– A   = 2i₁– A

Or, \(=\frac{A+\delta_{m}}{2}\)

Again, A = r₁+r₂ = 2r₁

Now considering refraction at the face AB we have, angle of incidence = i1 and angle of refraction = r₁

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

⇒ \(\frac{\sin i_1}{\sin r_1}=\frac{\sin \frac{A+\delta_m}{2}}{\sin \frac{A}{2}}\)

So, If we know the values of the angle of the prism μ, and the angle of minimum deviation δ , we can determine the value of the refractive Index of the material of the prism

Refraction Through Prism Class 12 Notes

Refraction Of Light Prism Of Some Definitions Numerical Examples

Example 1.  The refractive index of the material of a prism Is \(\sqrt{\frac{3}{2}}\) and the refracting angle Is 90°. of the angle of minimum deviation of the prism and the angle of Incidence at the minimum deviation position.
Solution:

Here A = 90°: Let the angle of the minimum deviation be m

We know, μ = \(\frac{\sin \frac{A+\delta_m}{2}}{\sin \frac{A}{2}}=\frac{\sin i_1}{\sin r_1}\)

⇒ \(\sqrt{\frac{3}{2}}=\frac{\sin \frac{90^{\circ}+\delta_m}{2}}{\sin \frac{90^{\circ}}{2}}=\frac{\sin \frac{90^{\circ}+\delta_m}{2}}{\sin 45^{\circ}}\)

Or, \(\sin \frac{90^{\circ}+\delta_m}{2}=\sqrt{\frac{3}{2}} \times \sin 45^{\circ}\)

= \(\frac{\sqrt{3}}{\sqrt{2}} \times \frac{1}{\sqrt{2}}=\frac{\sqrt{3}}{2}\)

⇒ \(i_1=\frac{90^{\circ}+\delta_m}{2}=60^{\circ}\)

Or, 90° + δ_m = 120°

δ_m = 120°- 90°

δ_m = 30°

⇒ \(r_1=\frac{A}{2}=\frac{90^{\circ}}{2}\)

= 45°

Class 12 physics prism refraction notes 

Example 2. The refractive of the material of a prism is \(\sqrt{2}\) and the angle of minimum deviation is 30. Calculate the value of the refracting angle of the prism.
solution:

Let the refracting angle of the prism be A We know, \(\)

⇒  \(\frac{\sin \frac{A+\delta_m}{2}}{\sin \frac{A}{2}}\)

Or, = \(\sqrt{2}=\frac{\sin \left(\frac{A+30^{\circ}}{2}\right)}{\sin \frac{A}{2}}\)

Or, \(\sqrt{2} \cdot \sin \frac{A}{2}=\sin \left(15^{\circ}+\frac{A}{2}\right)\)

= \(\sin 15^{\circ} \cdot \cos \frac{A}{2}+\cos 15^{\circ} \cdot \sin \frac{A}{2}\)

⇒ \(\frac{\sin \frac{A}{2}}{\cos \frac{A}{2}}=\frac{\sin 15^{\circ}}{\sqrt{2}-\cos 15^{\circ}}\)

Now, \(\sin 15^{\circ}=\sin \left(45^{\circ}-30^{\circ}\right)\)

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

Similarly , Cos 15°  \(\frac{\sqrt{3}+1}{2 \sqrt{2}}\)

∴ \(\tan \frac{A}{2}=\frac{\frac{\sqrt{3}-1}{2 \sqrt{2}}}{\sqrt{2}-\frac{\sqrt{3}+1}{2 \sqrt{2}}}\)

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

= tan 30°

\(\frac{\Lambda}{2}\) = 30°

Or, A= 60°

Common Questions on Light Refraction in Prisms

Example 3. The angle of minimum deviation is the same as the angle of a glass prism of refractive index = \(\sqrt{3}\). What is the angle of the prism
Solution:

According to the question,  δ_m = A

⇒ \(\frac{\sin \frac{A+\delta_m}{2}}{\sin \frac{A}{2}}\)

Or, \(=\frac{\sin \frac{A+A}{2}}{\sin \frac{A}{2}}=\frac{\sin A}{\sin \frac{A}{2}}\)

= \(\frac{2 \sin \frac{A}{2} \cos \frac{A}{2}}{\sin \frac{A}{2}}=2 \cos \frac{A}{2}\)

μ = \(\sqrt{3}\)

So, \(2 \cos \frac{A}{2}=\sqrt{3}\)

Or, \(\cos \frac{A}{2}=\frac{\sqrt{3}}{2}\) = cos 30°

Or, \(\frac{A}{2}=30^{\circ}\)

A = 30° × 2

A = 60°

Example 4. What- will be the angle of emergence of a ray of light -; n through a prism for an angle of incidence of 45°? The angle of the prism = 60°; refractive index of the prism = \(\sqrt{3}\)
Solution:

Here A = 60 ,μ =  \(\sqrt{3}\)

For refraction at the first face

μ = \(\frac{\sin i_1}{\sin r_1}\)

Or, \(\frac{\sin i_1}{\sin r_1}\)

Or, \(\sqrt{2}\)

Or, r₁ =  30°

We know, that A = r₁+r₂

r₂= A-r₁ = 60°-30° = 30° = r₁

So, the angle of emergence (i₂) = angle of incidence (i₁) = 45° The angle of emergence of the ray of light from the second face is 45°

Example 5. A ray of— light is incident at an angle- of- incidence- 40°—- on a prism having a refractive index of 1.6. What should be the value of the angle of the prism for minimum deviation? Given sin 40° = 0.6428; sin23°42; = 0.4018
Solution:

Here  μ = 1.6

The angle of incidence at the first face = i₁ = 40

Let the angle of emergence at the second fac

For minimum deviation i₁ = i₂= 40

⇒ \(\delta_m=i_1+i_2-A=40^{\circ}+40^{\circ}-A\)

or, \(A+\delta_{m_1}=80^{\circ}\)

We know, \(\mu=\frac{\sin \frac{A+\delta_m}{2}}{\sin \frac{A}{2}} \text { or, } 1.6=\frac{\sin \frac{80^{\circ}}{2}}{\sin \frac{A}{2}}\)

or, \(\sin \frac{A}{2}=\frac{\sin 40^{\circ}}{1.6}=\frac{0.6428}{1.6}\)

= 0.4018= sin 23°42′

\(\frac{A}{2}\) = 23°42′

A = 23°42′ × 2

Or, A = 47° 24′

Practice Problems on Prism Refraction Angles

Example 6. The refracting angle of the n glass prism is 60° and the refractive Index of glass is 1.6 The angle of incidence of a ray of light on the first refracting surface is 45°. Calculate the angle of deviation of the ray. Given that sin 26°14′ = 0.4419, sin 33°46′ = 0.5558 and sin 62°47′ = 0.8893 .
Solution:

Here A = 60°; ft – 1.6

The angle of incidence on the first face = i₁ = 45

For refraction on the first face, ft = \(\frac{\sin i_1}{\sin r_1}\)

⇒ \(\sin r_1=\frac{\sin t_1}{\mu}=\frac{\sin 45^{\circ}}{1.6}\)

= \(\frac{1}{\sqrt{2} \times 1.6}\)

= 0.4419

= sin 26.23°

Or, r₁ = 26.23°

We, know A = r₁ +r₂

r₂ = A- r₁ = 60°- 26.23° = 33.77°

For refraction at the second face

⇒ \(\frac{\sin i_2}{\sin r_2}\)

Or, sin =1.6 \(\frac{\sin i_2}{\sin 33.77^{\circ}}\)

Or, sin i₂ = 1.6 × sin 33.77°

= 1.6 × 0.5559 = 0.8894

= sin 62.8°

i₂ = 62.8°

So, the angle of deviation,

δ = i₁ + i₂-A = 45° + 62.8°- 60°

= 47.8°

Example 7.  If the refracting angle of a prism is A, the refractive index of its material is fi and the angle of deviation of a ray of light Incident normally on the first refracting face is 8, then prove that ft
Solution:

The angle of deviation for refraction in the prism,

δ = i₁ + i₂  -A and A = r₁ + r₂

For normal incidence ix = 0 and rx = 0

∴ δ  = i₂-A Or, i₂ = A + δ

A = r₂

The refractive index of the prism,

μ = \(\frac{\sin i_2}{\sin r_2}=\frac{\sin (A+\delta)}{\sin A}\)

Refraction through prism class 12 notes 

Example 8.  A ray of light Is an Incident at an angle of 60° a prism with a refracting angle 30°. If any emerges from the other face and makes an angle of 30° with the Incident ray then, show that the emergent ray passes perpendicularly through the refracting surface. Determine the refractive Index of the material of the prism.
Solution:

According to question’, δ = 30°, i₁=6 0° and

A = 30°.

We know, δ = i₁ + i₂-A

∴ 30° = 60° + i₁– 30° or, i₂ = 0

The emergent ray Is perpendicular to the refracting surface.

Again, A = r₂ + r₁

As, i₁ = 0 , so angle of incidence of the second face, r2 = 0

r₁ = A = 30°

∴ The refractive index of the material of the prism

μ = \(\frac{\sin i_1}{\sin r_1}=\frac{\sin 60^{\circ}}{\sin 30^{\circ}}=\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}\)

= \(\sqrt{3}\)

Refraction of light prism class 12 Important Definitions Related to Prism Refraction

Example 9. A glass prism with a refracting angle of 60° and of refractive index 1.6, is immersed in water (refractive Index is 1.33). What is its angle of minimum deviation? [sin 36.87° = 0.6
Solution:

Here, A = 60°; _aμ_b = 1.6; _aμ_w =1.33

⇒ \({ }_w \mu_g=\frac{{ }_a \mu_g}{{ }_a \mu_w}=\frac{1.6}{1.33}\)

= 1.2

Let the angle of minimum deviation for the immersed prism is

⇒ \(w^{\mu_g}=\frac{\sin \left(\frac{A+\delta_m}{2}\right)}{\sin \frac{A}{2}}]\)

Or, \(1.2=\frac{\sin \left(\frac{60^{\circ}+\delta_m}{2}\right)}{\sin \frac{60^{\circ}}{2}}\)

Or, \(1.2 \times \sin 30^{\circ}=\sin \frac{60^{\circ}+\delta_m}{2}\)

Or, \(\sin \frac{60^{\circ}+\delta_m}{2}=1.2 \times \frac{1}{2}\)

= 0.6 = sin 36. 87

Or, \(\frac{60^{\circ}+\delta_m}{2}\) = 36.87

Or, \(\delta_m=73.74^{\circ}-60^{\circ}=13.74^{\circ}\)

Example 10. A ray of light passes through an equilateral prism In such a way triangle of incidence becomes equal to the angle of emergence and each of these angles is \(\frac{3}{4}\) of the angle of deviation. Determine the angle of deviation.
Solution:

Here, A= 60 and i₁= i₂= \(\frac{3}{4} \delta\)

Now \(\)

= \(\frac{3}{2} \delta-60^{\circ}\)

Or, \(\frac{1}{2} \delta=60^{\circ}\)

Or = 120

∴ Angle od deviation = 120

Refraction of light prism class 12 Examples of Applications of Prisms in Optics

Example. 11  The angle of minimum deviation of a glass prism of refracting angle 60° is 30°. If the velocity of light In a vacuum is 3 × 10^-8m s^-1, then determine Its velocity In the glass
Solution:

The refractive index of the material of the prism,

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

μ = \(\frac{\sin \frac{60^{\circ}+30^{\circ}}{2}}{\sin \frac{60^{\circ}}{2}}=\frac{\sin 45^{\circ}}{\sin 30^{\circ}}\)

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

Again , \(\mu=\frac{\text { velocity of light in vacuum }}{\text { velocity of light in glass }\left(\nu_g\right)}\)

⇒ \(\sqrt{2}=\frac{3 \times 10^8}{v_g}\)

⇒  \(v_g=\frac{3 \times 10^8}{\sqrt{2}}=2.12 \times 10^8 \mathrm{~m} \cdot \mathrm{s}^{-1}\)