WBCHSE Class 12 Physics Notes For Types of Reflecting Prisms

Optics

Refraction Of Light Types of Reflecting Prisms

Total Reflecting Prism:

Total internal reflection of light can easily take place in a prism made of crown glass and whose principle cross-section is a right-angled isosceles triangle. So this type of prism is called a total reflecting prism.

ABC is die principal section of a total reflecting prism { Its sides.AS and BC are equal and ∠ABC = 90′. If the ray PQ is incident perpendicularly on the face AS, it is incident on the face AC at an angle of 45° which is greater than the critical angle of glass and air, about 42°. So it is totally
reflected and passes along RST perpendicular to the side SC. Thus the hypotenuse face

AC of the prison acts as a plane mirror. So it is called a total reflecting prism. it is to be noted that in this case, the deviation of the ray is 90°

Class 12 Physics Unit 6 Optics Chapter 2 Refraction Of Light Reflecting Prism

Advantages of a total reflecting prism:

  1. In the case of a plane mirror multiple images may be formed due to several reflections from the front and the back surfaces of the mirror and the images thus formed is also bright.
  2. But in the case of a total reflecting prism, only one bright image is formed due to total internal reflection.
  3.  When the mercury coating of the back surface of a plane mirror is damaged, the image becomes indistinct.
  4. But such a problem arises in a totally reflecting prism mirror is damaged, and the image becomes indistinct. But such a problem arises in a totally reflecting prism

Disadvantages of a total reflecting prism:

  1. A total-reflecting prism is more costly than a plane mirror.
  2. If the glass of the prism is not completely homogeneous, the image becomes less distinct

Effecting prism: The inverted image of an object can be made erect with the help of a total reflecting prism.

Erecting of the image by deviating a ray through 0°:

ABC is an isosceles right-angled prism, where ∠A = 90° uniting side of prism for no emergent ray j and ∠B = ∠C = 45°. Suppose that QP is the inverted image of an object. The ray coming from Q after refraction on the face AB is incident on the face BC at an angle greater than the critical angle of glass and air. So the ray is totally reflected from this face and emerging from the face AC forms its image at Q1.

Similarly, a ray starting from P comes to the point P1. So this P1Q1 is inverted with respect to QP, thus erecting an inverted object. can type of prism is called a total reflecting prism.

Class 12 Physics Unit 6 Optics Chapter 2 Refraction Of Light Isosceles Right Angled Prism

In this case, no deviation of the ray has taken place.

In instruments like telescopes, binoculars, etc., total reflecting

RST is perpendicular to the side SC. Thus the hypotenuse face, prism is therefore used to erect an inverted image.

Erecting of the image by deviating 180°:

To get an erect Image, the above-mentioned prism can be used in another way. QP is the Inverted Image of an object. The hypotenuse faces BC of the prism ABC Is held In front of it. A ray of PX from P is incident normally on the face BC and enters the prism.

After refraction, it Is incident at Y on the face AB. ‘I lie angle of incidence of the ray AT at the face AH is 45° which Is greater than the critical angle (nearly 42) of glass and air. So the ray moves along YZ after total reflection at Y and is incident on the face AC.

For The same reason, the ray moves along ZR after total reflection at Z and incident normally on the face BC. It emergences RP’ and comes to the point P’

Class 12 Physics Unit 6 Optics Chapter 2 Refraction Of Light Inverted Image And The Hypotenuse Face

Similarly, a ray coming from Q comes to the point  Q’ obviously according to the figure, the image P’Q’ becomes inverted relative to PQ. So the erect image P’Q’ of an inverted object Is thus formed.

Each incident ray bends twice at 90°, thus producing a total deviation of 180°. So, an inverted image can be made erect by deviating a ray through 180°.

Prism periscope:

Total reflecting prisms are commonly used nowadays in good-quality periscopes instead of inclined parallel mirrors. This type of periscope is called prism periscope and the image formed by a prism periscope is more bright than that formed by a simple periscope.

The periscope tube contains two right-angled isosceles prisms P1 and P2. P1 is fixed at the top in such a way that rays of light coming from a.Distant object enter the prism through a window and after total internal reflec¬ tion goes downwards. The hypotenuse face P2, which is fixed at the bottom receives these rays and reflects them totally In n horizontal direction through the ohworvinlon window. Thu observer thus sees an exact Image of the distant object.

Class 12 Physics Unit 6 Optics Chapter 2 Refraction Of Light Prism Periscope

Unit 6 Optics Chapter 2 Refraction Of Light Types of Reflecting Prisms Numerical Examples

Example 1. A man with a telescope can just observe point A on the circumference of the base of an empty cylindrical vessel.   When the vessel is filled completely with a liquid of refractive index 1.5 the man can just observe the middle point B of the base of the vessel without moving either the vessel or the telescope If The diameter of the base of the vessel is 10 cm, what is the height of the vessel.
Solution:

When the vessel Is empty, a light from point A enters the telescope ‘l’ following the straight path AO. When the vessel Is filled with the liquid, a ray of light from point II moves along BO and after refraction in air enters the telescope. Let h be the height of the vessel.

Class 12 Physics Unit 6 Optics Chapter 2 Refraction Of Light Empty Vessel

Here

According to the figure

⇒ \(\frac{\sin l}{\sin r}=\frac{1}{\mu} \quad \text { or, } \mu=\frac{\sin r}{\sin t}\)

Or, \(1.5=\frac{\frac{A C}{A O}}{\frac{B C}{B O}}=\frac{A C}{A O} \times \frac{B O}{B C}=\frac{A C}{B C} \times \frac{B O}{A O}\)

Or,\(1.5=\frac{10}{5} \times \frac{\sqrt{B C^2+C O^2}}{\sqrt{A C^2+C O^2}}=2 \times \frac{\sqrt{25+h^2}}{\sqrt{100+h^2}}\)

Or,\(2.25=\frac{4\left(25+h^2\right)}{100+h^2}\)

h = 8.45 cm

Example 2. A post remains above is dipped the water straight of the in a pond.“The rays of the sun are Inclined at an angle of 45° to the surface of water what will be the length of the shadow of the post at the bottom of the pond? The refractive index of water p
Solution:

ABC is the post and BFG is the surface of water

The length of the shadow at the bottom of the pond = CD

Class 12 Physics Unit 6 Optics Chapter 2 Refraction Of Light Pond

Let CE = BF= x = ED = yellow

According to the figure

∠AFN = i= angle of incidence

∠BAF = i

∠EFD = r = angle of refraction

We know, \(\mu=\frac{\sin i}{\sin r}\)

⇒ \(\frac{4}{3}=\frac{\sin 45^{\circ}}{\sin r}\)

Since i = 45°

Or, \(\frac{3}{4} \times \frac{1}{\sqrt{2}}=\frac{3}{4 \sqrt{2}}\)

From the

tan i= \(\frac{x}{1}\)

Or, \(\tan 45=\frac{x}{1}\)

Or, x = 1cm

tan 45° = \(\frac{x}{1}\)

Again, \(\sin r=\frac{y}{\sqrt{y^2+9}} \quad \text { or, } \frac{3}{4 \sqrt{2}}=\frac{y}{\sqrt{y^2+9}}\)

Or, y²× 32 = 9(y² + 9) or, 23y² = 81

Or, \(\frac{81}{23}\)

y²= 1.879 m

The length of the shadow of the post at the bottom of the pond = x+y= 1+1.876

= 2.876 m

Example 3. There is a point object at a height above the surface of water in a tank. If the bottom of the tank acts as a plane mirror where will be the Imago formed? If on observer looks from the air at the surface of the water normally, calculate the distance of the image front the surface of the water of the tank formed by the mirror-like bottom surface of the tank. Refractive Index of water = \(\frac{4}{3}\)
Solution:

Q is a point source. For refraction in water, the apparent position of Q is Q’

Class 12 Physics Unit 6 Optics Chapter 2 Refraction Of Light Refractive Index In Water

So, \(\mu=\frac{\text { apparent height }}{\text { real height }}=\frac{P Q^{\prime}}{P Q}\)

= \(\frac{P Q^{\prime}}{h}\)

Or, PQ’ = μh =  \(\frac{4}{3}\)

Therefore the distance of Q’ from the bottom of the tank

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

So the image of Q’ will be formed at a distance of (d +\(\frac{4}{3}\)) from the bottom of the tank

= d + d +\(\frac{4}{3}\)h = 2d +\(\frac{4}{3}\)h

⇒ \(\frac{4}{3}\) = \(\frac{2 d+\frac{4}{3} h}{x}\)

=  \(x\frac{3}{4}\left(2 d+\frac{4}{3} h\right)=\frac{3}{2} d+h\)

Example 4.  A rectangular glass slab of thickness 3 cm and of refractive Index 1.5 is placed in front of a concave mirror, perpendicular to Its principal axis. The radius of curvature of the mirror Is 10 cm. Where Is an object to be placed on the principal axis so that Its image will he formed on the object?
Solution:

The rectangular glass slab is placed perpendicular to the principal axis of the concave mirror M1M2 Suppose that if an object is placed at O on the principal axis, its image will be formed at
O. OABM1 is the path of the ray. Tiie ray after reflection at M1 retraces the path and forms an image at O. So the ray BM1 must be incident on the mirror at M1  perpendicularly. If Af, B must be incident on the mirror at Af, perpendicularly. If Af, B O’ is the centre of curvature of the concave mirror.

PO’ = 10 cm

Now OO’ = \(t\left(1-\frac{1}{\mu}\right)=3\left(1-\frac{1}{1.5}\right)\)

= 3-2 = 1cm

So, the distance of O from the concave mirror

= PO’ + OO’ = 10 + 1 = 11 cm

Class 12 Physics Unit 6 Optics Chapter 2 Refraction Of Light Rectangular Glass Slab

Example 5. A cross mark at the bottom of an empty vessel is focused with the help of a vertical microscope. Now! water (refractive Index = \(\frac{4}{3}\) ) Is poured into the vessel.  The height of water in the vessel Is 4 cm. Another lighter liquid which does not mix with water and has  a refractive index is, is poured into the water. The height of the liquid is 2 cm. How much should the microscope be raised vertically to focus the mark again?
Solution:

The real depth of the combination of water and the

liquid = 4 + 2 = 6 cm

The apparent depth of the cross-mark

⇒ \(\frac{4}{\frac{4}{3}}+\frac{2}{\frac{3}{2}}=3+\frac{4}{3}\)

= 4.33 cm

So, to focus the cross mark, the microscope is to be raised vertically through a height (6-4.33) = 1.67 cm.

WBCHSE Class 12 Physics Notes For Types Of Reflecting Prisms

Example 6. A concave mirror with a radius of curvature of 1 m Is placed at the bottom In a reservoir of water. When the sun Is situated directly over the head, the mirror forms an Image of the sun. If the depth of water Is  80 cm and  40 cm, calculate the Image distances from the mirror, f Given \(\mu_w=\frac{4}{3}\)

Solution:

The sun is an object situated at infinity. So its image will be formed by the concave mirror at its focus i.e. \(\frac{100}{2}\) = 50 cm above the mirror.

When the depth of water in the reservoir is 80 cm the image is formed at n distance of 50 cm inside water from the mirror. : § But when the depth of water is 40 cm the image will be formed in air. Light rays will be refracted while passing from water t0 ain So> the refracted rays wiU converge at O’ and an image will be formed at O’

Displacement of the image

= \(O O^{\prime}=t\left(1-\frac{1}{\mu}\right)=10\left(1-\frac{3}{4}\right)\)

= \(10 \times \frac{1}{4}\)

= 2.5 cm

Distance of image from the mirror

= PO’ = 50 – 2.5

= 47.5 cm

Class 12 Physics Unit 6 Optics Chapter 2 Refraction Of Light Refracted Rays

Example 7. The width of a rectangular glass slab is 5 cm. From a point on Its bottom surface, light rays are incident on its top face and after total reflection form a circle of light of radius 8 cm. What is the refractive Index of the
Solution:

Let o be a bright point at the bottom face of the rectangular slab. light rays starting from

Return to the bottom face after total reflection from the upper face of the slab.

As a result, a circle of light of radius OA = OB = 8 cm formed on the bottom face. So the angle of incidence on the upper face is P(. (critical angle) glass slab

According to the

OP = \(\sqrt{O C^2+P C^2}=\sqrt{(4)^2+(5)^2}\)

= \(\sqrt{41} \mathrm{~cm}\)

Refractive index \(\)

\(\mu=\frac{1}{\sin \theta_c}=\frac{1}{\frac{O C}{O P}}\) \(=\frac{O P}{O C}=\frac{\sqrt{41}}{4}\)

= 1.6

Class 12 Physics Unit 6 Optics Chapter 2 Refraction Of Light Bottom Space Of Rectangular Slab

Example 8.  A glass sphere having a centre at 0 and two perpendicular diameters AOB and COD. A ray parallel to AOB is incident on the sphere at P where AP = PC and emerges from the sphere at B. Calculate the refractive index of glass and the deviation of the emergent ray
Solution:

Class 12 Physics Unit 6 Optics Chapter 2 Refraction Of Light Deviation Of The Emergent Ray

According to the  arc AP = arc PC

∠AOP = ∠POC = 45° = i,

∠POB = 45° + 90° = 135°

From the triangle POB,

r+r+ 135° = 180° or, 2r = 45° or, r = 22.5°

= \(\frac{\sin i}{\sin r}=\frac{\sin 45^{\circ}}{\sin 22.5^{\circ}}\)

= 2 cos 22.5° = 2 × 0. 924 = 1.85

Angle of deviation

δ  =i-r+i-r = 2 (i-r) = 2(45°-22.5°)

= 2 × 22.5° = 45°

Example 9. A glass slab is placed on a page of a book kept horizontally. What should be the value of the minimum refractive index of the glass slab so that the printed letters of the page will not be visible from any vertical side of the slab? 
Solution:

It is assumed that there is a thin layer of air between the page of the book and the glass slab. So any ray coming from

any portion of the page of the hook Is Incident on the glass slab According to the figure, at an angle of almost

Class 12 Physics Unit 6 Optics Chapter 2 Refraction Of Light A Glass Slab

If the angle of refraction Is Φ and the refractive index of glass Is μ, S. then, sin Φ = \(\frac{1}{\mu}\)

If θC is the critical angle, then sin θC = \(\frac{1}{\mu}\)

This refracted ray is incident on any vertical l side of the slab at an angle θ  (say) So. θ + Φ = 90°.

If  . θ is greater than Φ, a total internal reflection of light takes place and the printed letters of the page will not be visible from any vertical side, μ will be minimum when. θ = Φ

2Φ = 90° , Or, Φ = 45°

∴ \(\mu_{\min }=\frac{1}{\sin \phi}=\frac{1}{\sin 45^{\circ}}\)

= \(\sqrt{2}\)

= 1.414

Example 10. A surface of a prism having a refractive index 1.5 is covered with n liquid of refractive index \(\frac{3 \sqrt{2}}{4}\) . What should be the minimum angle of incidence of an incident ray so that on the other surface of the prism tire ray will be totally reflected from the surface covered with liquid? The refracting angle of tho prism =75° [sin48°36′ = 0.75].
Solution:

Let the critical angle between the prism and the liquid be θC. If the ray of light is totally reflected from the surface covered with liquid then the angle of incidence of the ray on the surface is θC

Class 12 Physics Unit 6 Optics Chapter 2 Refraction Of Light Refracting Angle Of The Prism

So, 1.5 sin θC = \(\frac{3 \sqrt{2}}{4}\)

Or, \(\frac{3 \sqrt{2}}{4}\)

Or, \(\frac{3 \sqrt{2}}{4}\) x \(\frac{10}{15}\)

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

Again , r1+ θC = A

Or,  r1+ 45° = 75°  Or, = 30°

Now , sin i1 = μ sin r1 = 1.5 sin 30° = 0.75 Or, i1= 48° 36′

Example 11. A ray of light Is Incident normally on one, side of on isosceles right-angled prism and Is totally reflected on the other side.

  1. What Is the value of minimum refractive lodes of the material of the prism?
  2. If the prism Is Immersed In water, draw the diagram showing the direction of the emergent ray. In the diagram, point out the values of the angles,μ of water \(\frac{4}{3}\)

Solution:

1. ABC Is an Isosceles light-angled prism and Its sides are AB= BC The light lay PQ Is incident on the face AR normally and enters the prism. The ray Is an Incident at R on the face AC.

It Is evident from the figure that angle of Incidence of the ray at R Is 45° . Now if the ray Is to be totally reflected from R then this angle of incidence should he greater than the critical angle of the material of the prism μ maximum critical angle will be θC = 45° . If H be the minimum refractive Index of the material of the prism then,

sinθC  = \(\frac{1}{\mu}\)

Or, in 45° = \(\frac{1}{\mu}\)

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

= 1.414

Class 12 Physics Unit 6 Optics Chapter 2 Refraction Of Light Right Angled Prism And Totally Reflected From The Other Side

2. If the prism is immersed in water, then the refractive Index of glass relative to water is,

⇒ \({ }_u^{\mu_E}=\frac{\mu_S}{\mu_i}=\frac{\sqrt{2}}{\frac{4}{3}}=\frac{3 \sqrt{2}}{4}\)

If the critical angle between the prism and water is  θC

⇒ \(\theta_c^{\prime}=\sin ^{-1} \frac{1}{w^{\mu_g}}=\sin ^{-1}\left(\frac{4}{3 \sqrt{2}}\right)\)

= 70.53°

But the angle of Incidence at R is 45° [Fig. 2.64(b)] and it Is less than the critical angle. So at R, light rays will not be totally reflected. It will be refracted and will enter water. If the angle of refraction is r, then

⇒ \(\sqrt{2} \sin 45^{\circ}=\frac{4}{3} \sin r\)

sin r= \(\sqrt{2} \times \frac{3}{4} \times \frac{1}{\sqrt{2}}\)

= 0.75 = sin 48.59°

Or, r = 48.59°

Example 12.  A ray of light Is incident grazing the refracting surface of a prism having refracting angle A. It emerges the other refracting surface making an angle θ with the normal to the surface. Prove that the refractive index  of the material of the prism is given by \(\mu=\left[1+\left(\frac{\cos A+\sin \theta}{\sin A}\right)^2\right]^{1 / 2}\)
Solution:

For grazing incidence on the refracting surface AB of the prism, i1 = 90°

Class 12 Physics Unit 6 Optics Chapter 2 Refraction Of Light Grazing The Refacting Surface Of Prism

So, in this case r1 = dc (critical angle)

Now, A = r1 + r2 or, r2= A-r1 = A – θC

Considering refraction at the second refracting surface, AC

We get \(\mu=\frac{\sin \theta}{\sin r_2}\)

Or, \(\sin \theta=\mu \sin r_2=\mu \sin \left(A-\theta_c\right)\)

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

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

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

⇒ \(\sin A \sqrt{\beta^2-1}-\cos A\)

Or, \(\sin \theta+\cos A=\sin A \sqrt{\mu^2-1}\)

Or, \(\frac{\sin \theta+\cos A}{\sin A}=\sqrt{\mu^2-1}\)

Or,\(\mu^2-1=\left(\frac{\sin \theta+\cos A}{\sin A}\right)^2\)

Or,\(\mu^2=1+\left(\frac{\sin \theta+\cos A}{\sin A}\right)^2\)

Or,\(\mu=\left[1+\left(\frac{\sin \theta+\cos A}{\sin A}\right)^2\right]^{\frac{1}{2}}\)

Example 13. A ray of light incident normally on a refracting surface of the prism it totally reflated from the other refracting surface. If the prism it Immersed In water how will the ray act? The refractive index of a glass of =1.5; the refractive index of water = 1.33.
Solution:

At the ray of light it incident on the face of the prism normally so it goes straight through the surface. The ray it incident j on the second refracting face and is totally reflected. So the angle i or, of Incidence of the ray at the second face is greater than the critical  angle θC

⇒ \(\sin \theta_C=\frac{1}{a^{\mu_E}}=\frac{1}{1.5}=0.667\)

= sin 41.8°

θC = 41.8°

Now, if the prism is immersed in water, the refractive index of glass j with respect to water is,

wμg =aμg /aμw

= \(\frac{1.5}{1.33}\)

= 1.128

In this case, if the critical angle is  θ’C then

⇒ \(\sin \theta_C^{\prime}=\frac{1}{u^{\mu_g}}\)

= \(\frac{1}{1.128}\)

θ’= sin 62.44°

So the angle of incidence of the ray at the second face (41.8) is less than the critical angle (62.44). Hence, instead of being . totally reflected from the second face, the ray is refracted through it.

Example 14. A tank height of 33.25 cm is completely filled with liquid (μ= 1.33).An object is placed at the bottom of the tank on the axis of a concave mirror.   Image of the object is formed 25 cm below the surface of the liquid . What is the focal length of the mirror?

Class 12 Physics Unit 6 Optics Chapter 2 Refraction Of Light Height Of The Tank On The Axis Of Concave Mirror
Solution:

Here, R is the actual position of the object and A is the apparent position of the image

We, know, \(\mu=\frac{\text { real depth }}{\text { apparent depth }}\)

Or, 1.33  \(=\frac{33.25}{x_a}\)

Therefore, the apparent depth of the object

Here, object distance , u = \(-\left(15+\frac{33.25}{1.33}\right)\)

And image distance, v= -(15+25)cm = -40 cm

Applying the mirror equation, we get

⇒ \(-\frac{1}{40}-\frac{1}{40}=\frac{1}{f}\)

Or, f = \(-\left(\frac{40 \times 40}{40+40}\right)\) = -20

Therefore, the required focal length is 20 cm.

Example 15. What should be the minimum value of the refractive index of a right-angled isosceles prism to that the is F prism can deviate a ray through 180′ by total internal reflection

Class 12 Physics Unit 6 Optics Chapter 2 Refraction Of Light Rigt Angled Isosceles Prism

Solution:  ABC is a right-angled isosceles triangle, whose ∠ABC is 90 and AB = BC

Here light ray is incident along MN on the side AB and is reflected back along NO. Now, the reflected ray NO is incident on side BC and gets reflected along the path OP.

Therefore, light rays suffer total internal reflection when incident ic onsides AB and BC

It is clearly seen that the value of incident angle ig on sides AB and BC is 45° or less than 45°

ic ≤ 45°

sin-1 1/μ <45°

Or, 1/μ < sin 45°

Or,μ > \(\sqrt{2}\)

Therefore, the refractive index of the material of the prism is at least Jz so that the prism can deviate a ray through 180° by total internal reflection

Example 16. The face of the prism of refracting angle A is coated with silver. A light ray after first being Incident at an angle of Incidence 2A on the first face of the prism, is refracted and is then reflected from the second face, retracing its path. Calculate the value of the refractive index of the prism
Solution:

Let PQR be a glass prism. The refracting surface PR of the prism is coated with mercury

Class 12 Physics Unit 6 Optics Chapter 2 Refraction Of Light Prism Is Coated With Mercury

In the, incident ray DB is refracted along BC and the light ray returns following the same path after getting reflected from the surface PR

∴ ∠PCB = 90°

And ∠PBC = a = (90°-A) ………………………………… (1)

Where BPC= refracting angle of the prism = A-\theta_c\right

Again, from the a+r = 90°

∴ a =  (90°-r) ……………………………….(2)

Comparing equations (1)and (2), we may write

Where , \(=\frac{\sin i}{\sin r}=\frac{\sin 2 A}{\sin A}\)

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

μ = 2cos A

Therefore, the required refractive index is 2 cosA.

Example 17. A ray of light falls normally on one side other than the hypotenuse of a right-angled isosceles prism of refractive index 1.5. From which side will the ray emerge from the prism? Find the deviation of the incident ray
Solution:

ABC is a right-angled isosceles triangle i.e., Angle of prism A = 90

Class 12 Physics Unit 6 Optics Chapter 2 Refraction Of Light Right Angled Isosceles Triangle

∠ABC = ∠ACB = 45°

AB. The ray of light is incident normally on the side AC and is inci¬ dent on the side BC at an angle 45°

.As a result, after total internal reflection, the ray passes along QR perpendicular to the side

Now, the angle of deviation = ∠SQR = δ = 90°

Example 18. A vessel contains a liquid of refractive index \(\frac{5}{3}\). Inside j the liquid, S is a point source which is observed from above the liquid. An opaque disc ot radius 1 cn, is floating on the liquid such that its centre is just the source, at this circumstance liquid of the vessel is leaving gradually through a hole. What is the depth of the liquid, so that the source no more remains visible from above
Solution:

Let x be the required dÿepth obviously, at this position light rays from the source S must be incident at a critical angle (dc) at the edge of the disc

Class 12 Physics Unit 6 Optics Chapter 2 Refraction Of Light Critical Angle At Edge Of The Disc

So, \(\sin \theta_c=\frac{1}{\sqrt{1+x^2}}\)

Also \(\sin \theta_c=\frac{1}{\mu}=\frac{3}{5}\)

\(\frac{3}{5}=\frac{1}{\cos \sqrt{1+x^2}}\)

Or, \(\)

Or, \(1+x^2=\frac{25}{9}\)

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

= 1.33 cm

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