GEOMETRICAL OPTICS
EXERCISES- Q31 to Q40
31. Light falls from glass (µ=1.5) to air. Find the angle of incidence for which the angle of deviation is 90°.
ANSWER: Glass is a denser medium than air. When the light gets refracted from denser to the rarer medium it deviated away from the normal at the point of incidence. Hence the angle of refraction is larger than the angle of incidence. The angle of deviation is the difference between the angle of refraction and the angle of incidence. But the angle of deviation will not be 90° in the case of refraction because the maximum angle of refraction is itself 90°. At this stage, if the angle of incidence is increased, total internal reflection takes place. When it happens, the angle (α) between the reflected ray and the surface is the same as between the surface and the projected incident ray. So the angle of deviation =2α.
Diagram for Q-31
For the given condition,
2α = 90°
→α = 45°
So the angle of incidence = angle of reflection
= 90° - α
=90° - 45°
=45°
32. A point source is placed at a depth h below the surface of the water (refractive index =µ). (a) Show that the light escapes through a circular area on the water surface with its center directly above the point source. (b) Find the angle subtended by a radius of the area on the source.
ANSWER: Let us draw a vertical line PBA through the point source P which meets the surface at point B which is directly above the source.
(a) The light will escape if it is refracted. The maximum angle of incidence for the refraction will be the critical angle C for which the angle of refraction = 90°. The angle between the vertical line and the extreme refracted ray is also =C. See the diagram below.
Diagram for Q-32
In a three dimensional space around the vertical line, this angle C will be the same and it will make an inverted cone with vertex at the source and its circular base DBE at the water surface. Obviously, this circle is the area through which the light can escape.
(b) DB = BE is the radius of the circular area which subtends an angle C on the source.
But Sin C/Sin 90° =1/µ
→Sin C = 1/µ
→C = sin⁻¹(1/µ)
33. A container contains water up to a height of 20 cm and there is a point source at the center of the bottom of the container. A rubber ring of radius r floats centrally on the water. The ceiling of the room is 2.0 m above the water surface.
(a) Find the radius of the shadow of the ring formed on the ceiling if r = 15 cm.
(b) Find the maximum value of r for which the shadow of the ring is formed on the ceiling. Refractive index of water =4/3.
ANSWER: Let S = point source. ACB is the floating ring with its center C directly above the source S. Radius of the ring AC = BC = r. DFE is the shadow of the ring on the ceiling and DF is the radius of this shadow. The shadow making ray is the refracted ray BE for which from the Snell's Law
Sin i/Sin r =1/µ
Diagram for Q-33 (a)
(a) For radius r = 15 cm =0.15 m
Sin i = BC/BS = 0.15/√(0.20²+0.15²) =0.15/0.25
=3/5
µ = 4/3
Hence Sin i/Sin r = 1/µ = 3/4
→Sin r = 4*(3/5)/3 =4/5 =0.8
Hence tan r = sin r/√(1-sin² r) =0.8/√(1-0.64)
=0.8/0.6 =4/3
From the triangle BGE,
GE/BG = tan r = 4/3
→GE = 2.0*4/3 = 8/3 m = 2.66 m
The radius of the shadow = FE =FG+GE =0.15+2.66
=2.81 m
(b) The maximum value of r will be corresponding to the critical angle C as in the case of problem 32(a).
Sin C/Sin 90° = 1/µ
→Sin C = 3/4
Hence the maximum value of r = CS*tan C
=20*sin C/√(1-sin²C)
=20*(3/4)/√(1-9/16)
=20*(3/4)/(√7/4)
=60/√7
=22.6 cm
34. Find the angle of minimum deviation for an equilateral prism made of a material of refractive index 1.732. What is the angle of incidence for this deviation?
ANSWER: Let the angle of minimum deviation = δₘ
Here given µ = 1.732 and A = 60° (Due to equilateral prism). Hence,
µ = Sin{(A+δₘ)/2}/Sin(A/2)
→1.732 = Sin{(δₘ+60°)/2}/sin30°
→Sin{(δₘ+60°)/2}/(½) = 1.732
→Sin{(δₘ+60°)/2} = 0.866
→(δₘ+60°)/2 = 60°
→δₘ = 120° - 60° = 60°
For minimum deviation, i = i'; and
δₘ = 2i - A
→60° = 2i - 60°
→2i = 120°
→i = 60°
35. Find the angle of deviation suffered by the light ray shown in the figure (18-E10). The refractive index µ = 1.5 for the prism material.
Figure for Q-35
ANSWER: Let ABC be the given prism in which A = 4°.
Diagram for Q-35
Since side AB of the prism is perpendicular to the incident ray, it will be refracted into the prism without deviation. It will then be refracted through the face AC. Here, by the geometry, i = 4°. Given µ = 1.5, hence
Sin i/Sin r = 1/µ
→Sin 4°/Sin r = 1/ 1.5
→Sin r = 1.5*Sin 4° =0.104 = Sin 6°
→r = 6°
Hence the angle of deviation δ = r - i =6°-4° =2°.
36. A light ray going through the prism with the angle of prism 60°, is found to deviate by 30°. What limit on the refractive index can be put from these data?
ANSWER: Given A = 60°. Since for the given ray, the deviation is 30°, the angle of minimum deviation will be either less than 30° or equal to 30°. The angle of minimum deviation, δₘ is given as
µ = Sin{(A+δₘ)/2}/Sin(A/2)
Since in this relation µ increases with increase in δₘ, hence
µ ≤ Sin{(60°+30°)/2}/Sin(60°/2)
→µ ≤ Sin 45°/Sin 30°
→µ ≤ (1/√2)/(1/2)
→µ ≤ 2/√2
→µ ≤ √2
37. Locate the image formed by refraction in the situation shown in figure (18-E11).
The figure for Q-37
ANSWER: For refraction at spherical surfaces,
µ₂/v - µ₁/u =(µ₂-µ₁)/R
Here, µ₁ = 1, µ₂ = 1.5, R = 20 cm, u = -25 cm. hence,
1.5/v + 1/25 = (1.5-1.0)/20 = 0.5/20
→1.5/v = 1/40 - 1/25 =(5 - 8)/200 =-3/200
→v/1.5 = -200/3
→v = -200*1.5/3 = -100 cm.
Negative sign shows that the image is on the side of the source. Hence the image is 100 cm from the surface on the side of S.
38. A spherical surface of radius 30 cm separates two transparent media A and B with refractive indices 1.33 and 1.48 respectively. Medium A is on the convex side of the surface. Where should a point object be placed in medium A so that the paraxial rays become parallel after refraction at the surface?
ANSWER: Let us assume that the medium A is on the left side of the spherical surface. So the object distance in medium A will have a negative value (-u). Here,
µ₁ = 1.33, µ₂ = 1.48, R = 30 cm. Since the paraxial rays are parallel after refraction, the image is at infinity, v = ∞.
From the relation,
µ₂/v - µ₁/u =(µ₂-µ₁)/R
→1.48/∞ + 1.33/u = (1.48-1.33)/30 =0.15/30
→u/1.33 = 30/0.15
→u = 1.33*200 =266 cm.
So the object is 266.0 cm away from the separating surface.
39. Figure (18-E12) shows a transparent hemisphere of radius 3.0 cm made of a material of refractive index 2.0. (a) A narrow beam of parallel rays is incident on the hemisphere as shown in the figure. Are the rays totally reflected at the plane surface?
(b) Find the image formed by the refraction at the first surface.
(c) Find the image formed by the reflection or refraction at the plane surface.
(d) Trace qualitatively the final rays as they come out of the hemisphere.
The figure for Q-39
ANSWER: (a) Since the rays along the radial direction they are perpendicular to the outer surface. These rays will pass undeviated and fall on the plane surface at an angle of incidence = 45°.
The refractive index of the material, µ = 2.0, hence the critical angle C = Sin⁻¹(1/µ) = Sin⁻¹(½) = 30°. Since the angle of incidence is greater than the critical angle, the total internal reflection will take place. So the rays are fully reflected at the plane surface.
(b) Let us consider the positive X-axis along the concave medium.
µ₁ = 1.0, µ₂ = 2.0, R = 3.0 cm, u = -∞, v =?
µ₂/v - µ₁/u = (µ₂ - µ₁)/R
→2/v + 1/∞ =(2-1)/3.0
→2/v =1/3
→v = 3*2 = 6 cm
So the image is formed 6 cm inside the transparent medium from A. It means if the sphere is completed the image forms at the point diametrically opposite to A.
(c) As we saw that the rays are fully reflected at the plane surface the image forming rays will intersect after reflection. And the image will be the mirror image of the image formed as in part (b). In fact, it will also be the mirror image of point A in BC.
(d) When the rays come out, they will be diverging.
Diagram for Q-39(d)
40. A small object is embedded in a glass sphere (µ=1.5) of radius 5.0 cm at a distance of 1.5 cm left to the center. Locate the image of the object as seen by an observer standing (a) to the left of the sphere and (b) to the right of the sphere.
ANSWER: (a) For this case we take A as the origin and AB negative X-axis. Now, u = -(5-1.5) =-3.5 cm, R = -5 cm, µ₁ = 1.5, µ₂ = 1.0, v = ?
µ₂/v - µ₁/u =(µ₂ - µ₁)/R
→1/v + 1.5/3.5 = -(1.0-1.5)/5 =1/10
→1/v = -3/7 + 1/10 =(-30 + 7)/70 =-23/70
→v = -70/23 =-3.0 cm
So the image is 3 cm from A i.e, 2 cm left to the center.
Diagram for Q-40
(b) When the observer is to the right of the sphere, the refraction is from the face near B. Taking B as the origin. u = -(5+1.5) =-6.5 cm. Hence
1/v - 1.5/(-6.5) =-(1.0-1.5)/5 =1/10
→1/v =1/10 - 3/13 =(13-30)/130 =-17/130
→v = -130/17 =-7.65 cm
So the image is 7.65 cm left of B i.e. 7.65-5 =2.65 cm left to the center.
31. Light falls from glass (µ=1.5) to air. Find the angle of incidence for which the angle of deviation is 90°.
ANSWER: Glass is a denser medium than air. When the light gets refracted from denser to the rarer medium it deviated away from the normal at the point of incidence. Hence the angle of refraction is larger than the angle of incidence. The angle of deviation is the difference between the angle of refraction and the angle of incidence. But the angle of deviation will not be 90° in the case of refraction because the maximum angle of refraction is itself 90°. At this stage, if the angle of incidence is increased, total internal reflection takes place. When it happens, the angle (α) between the reflected ray and the surface is the same as between the surface and the projected incident ray. So the angle of deviation =2α.
For the given condition,
2α = 90°
→α = 45°
So the angle of incidence = angle of reflection
= 90° - α
=90° - 45°
=45°
32. A point source is placed at a depth h below the surface of the water (refractive index =µ). (a) Show that the light escapes through a circular area on the water surface with its center directly above the point source. (b) Find the angle subtended by a radius of the area on the source.
ANSWER: Let us draw a vertical line PBA through the point source P which meets the surface at point B which is directly above the source.
(a) The light will escape if it is refracted. The maximum angle of incidence for the refraction will be the critical angle C for which the angle of refraction = 90°. The angle between the vertical line and the extreme refracted ray is also =C. See the diagram below.
In a three dimensional space around the vertical line, this angle C will be the same and it will make an inverted cone with vertex at the source and its circular base DBE at the water surface. Obviously, this circle is the area through which the light can escape.
(b) DB = BE is the radius of the circular area which subtends an angle C on the source.
But Sin C/Sin 90° =1/µ
→Sin C = 1/µ
→C = sin⁻¹(1/µ)
33. A container contains water up to a height of 20 cm and there is a point source at the center of the bottom of the container. A rubber ring of radius r floats centrally on the water. The ceiling of the room is 2.0 m above the water surface.
(a) Find the radius of the shadow of the ring formed on the ceiling if r = 15 cm.
(b) Find the maximum value of r for which the shadow of the ring is formed on the ceiling. Refractive index of water =4/3.
ANSWER: Let S = point source. ACB is the floating ring with its center C directly above the source S. Radius of the ring AC = BC = r. DFE is the shadow of the ring on the ceiling and DF is the radius of this shadow. The shadow making ray is the refracted ray BE for which from the Snell's Law
Sin i/Sin r =1/µ
(a) For radius r = 15 cm =0.15 m
Sin i = BC/BS = 0.15/√(0.20²+0.15²) =0.15/0.25
=3/5
µ = 4/3
Hence Sin i/Sin r = 1/µ = 3/4
→Sin r = 4*(3/5)/3 =4/5 =0.8
Hence tan r = sin r/√(1-sin² r) =0.8/√(1-0.64)
=0.8/0.6 =4/3
From the triangle BGE,
GE/BG = tan r = 4/3
→GE = 2.0*4/3 = 8/3 m = 2.66 m
The radius of the shadow = FE =FG+GE =0.15+2.66
=2.81 m
(b) The maximum value of r will be corresponding to the critical angle C as in the case of problem 32(a).
Sin C/Sin 90° = 1/µ
→Sin C = 3/4
Hence the maximum value of r = CS*tan C
=20*sin C/√(1-sin²C)
=20*(3/4)/√(1-9/16)
=20*(3/4)/(√7/4)
=60/√7
=22.6 cm
34. Find the angle of minimum deviation for an equilateral prism made of a material of refractive index 1.732. What is the angle of incidence for this deviation?
ANSWER: Let the angle of minimum deviation = δₘ
Here given µ = 1.732 and A = 60° (Due to equilateral prism). Hence,
µ = Sin{(A+δₘ)/2}/Sin(A/2)
→1.732 = Sin{(δₘ+60°)/2}/sin30°
→Sin{(δₘ+60°)/2}/(½) = 1.732
→Sin{(δₘ+60°)/2} = 0.866
→(δₘ+60°)/2 = 60°
→δₘ = 120° - 60° = 60°
For minimum deviation, i = i'; and
δₘ = 2i - A
→60° = 2i - 60°
→2i = 120°
→i = 60°
35. Find the angle of deviation suffered by the light ray shown in the figure (18-E10). The refractive index µ = 1.5 for the prism material.
ANSWER: Let ABC be the given prism in which A = 4°.
Since side AB of the prism is perpendicular to the incident ray, it will be refracted into the prism without deviation. It will then be refracted through the face AC. Here, by the geometry, i = 4°. Given µ = 1.5, hence
Sin i/Sin r = 1/µ
→Sin 4°/Sin r = 1/ 1.5
→Sin r = 1.5*Sin 4° =0.104 = Sin 6°
→r = 6°
Hence the angle of deviation δ = r - i =6°-4° =2°.
36. A light ray going through the prism with the angle of prism 60°, is found to deviate by 30°. What limit on the refractive index can be put from these data?
ANSWER: Given A = 60°. Since for the given ray, the deviation is 30°, the angle of minimum deviation will be either less than 30° or equal to 30°. The angle of minimum deviation, δₘ is given as
µ = Sin{(A+δₘ)/2}/Sin(A/2)
Since in this relation µ increases with increase in δₘ, hence
µ ≤ Sin{(60°+30°)/2}/Sin(60°/2)
→µ ≤ Sin 45°/Sin 30°
→µ ≤ (1/√2)/(1/2)
→µ ≤ 2/√2
→µ ≤ √2
37. Locate the image formed by refraction in the situation shown in figure (18-E11).
ANSWER: For refraction at spherical surfaces,
µ₂/v - µ₁/u =(µ₂-µ₁)/R
Here, µ₁ = 1, µ₂ = 1.5, R = 20 cm, u = -25 cm. hence,
1.5/v + 1/25 = (1.5-1.0)/20 = 0.5/20
→1.5/v = 1/40 - 1/25 =(5 - 8)/200 =-3/200
→v/1.5 = -200/3
→v = -200*1.5/3 = -100 cm.
Negative sign shows that the image is on the side of the source. Hence the image is 100 cm from the surface on the side of S.
38. A spherical surface of radius 30 cm separates two transparent media A and B with refractive indices 1.33 and 1.48 respectively. Medium A is on the convex side of the surface. Where should a point object be placed in medium A so that the paraxial rays become parallel after refraction at the surface?
ANSWER: Let us assume that the medium A is on the left side of the spherical surface. So the object distance in medium A will have a negative value (-u). Here,
µ₁ = 1.33, µ₂ = 1.48, R = 30 cm. Since the paraxial rays are parallel after refraction, the image is at infinity, v = ∞.
From the relation,
µ₂/v - µ₁/u =(µ₂-µ₁)/R
→1.48/∞ + 1.33/u = (1.48-1.33)/30 =0.15/30
→u/1.33 = 30/0.15
→u = 1.33*200 =266 cm.
So the object is 266.0 cm away from the separating surface.
39. Figure (18-E12) shows a transparent hemisphere of radius 3.0 cm made of a material of refractive index 2.0. (a) A narrow beam of parallel rays is incident on the hemisphere as shown in the figure. Are the rays totally reflected at the plane surface?
(b) Find the image formed by the refraction at the first surface.
(c) Find the image formed by the reflection or refraction at the plane surface.
(d) Trace qualitatively the final rays as they come out of the hemisphere.
ANSWER: (a) Since the rays along the radial direction they are perpendicular to the outer surface. These rays will pass undeviated and fall on the plane surface at an angle of incidence = 45°.
The refractive index of the material, µ = 2.0, hence the critical angle C = Sin⁻¹(1/µ) = Sin⁻¹(½) = 30°. Since the angle of incidence is greater than the critical angle, the total internal reflection will take place. So the rays are fully reflected at the plane surface.
(b) Let us consider the positive X-axis along the concave medium.
µ₁ = 1.0, µ₂ = 2.0, R = 3.0 cm, u = -∞, v =?
µ₂/v - µ₁/u = (µ₂ - µ₁)/R
→2/v + 1/∞ =(2-1)/3.0
→2/v =1/3
→v = 3*2 = 6 cm
So the image is formed 6 cm inside the transparent medium from A. It means if the sphere is completed the image forms at the point diametrically opposite to A.
(c) As we saw that the rays are fully reflected at the plane surface the image forming rays will intersect after reflection. And the image will be the mirror image of the image formed as in part (b). In fact, it will also be the mirror image of point A in BC.
(d) When the rays come out, they will be diverging.
40. A small object is embedded in a glass sphere (µ=1.5) of radius 5.0 cm at a distance of 1.5 cm left to the center. Locate the image of the object as seen by an observer standing (a) to the left of the sphere and (b) to the right of the sphere.
ANSWER: (a) For this case we take A as the origin and AB negative X-axis. Now, u = -(5-1.5) =-3.5 cm, R = -5 cm, µ₁ = 1.5, µ₂ = 1.0, v = ?
µ₂/v - µ₁/u =(µ₂ - µ₁)/R
→1/v + 1.5/3.5 = -(1.0-1.5)/5 =1/10
→1/v = -3/7 + 1/10 =(-30 + 7)/70 =-23/70
→v = -70/23 =-3.0 cm
So the image is 3 cm from A i.e, 2 cm left to the center.
(b) When the observer is to the right of the sphere, the refraction is from the face near B. Taking B as the origin. u = -(5+1.5) =-6.5 cm. Hence
1/v - 1.5/(-6.5) =-(1.0-1.5)/5 =1/10
→1/v =1/10 - 3/13 =(13-30)/130 =-17/130
→v = -130/17 =-7.65 cm
So the image is 7.65 cm left of B i.e. 7.65-5 =2.65 cm left to the center.
ANSWER: Glass is a denser medium than air. When the light gets refracted from denser to the rarer medium it deviated away from the normal at the point of incidence. Hence the angle of refraction is larger than the angle of incidence. The angle of deviation is the difference between the angle of refraction and the angle of incidence. But the angle of deviation will not be 90° in the case of refraction because the maximum angle of refraction is itself 90°. At this stage, if the angle of incidence is increased, total internal reflection takes place. When it happens, the angle (α) between the reflected ray and the surface is the same as between the surface and the projected incident ray. So the angle of deviation =2α.
 |
| Diagram for Q-31 |
For the given condition,
2α = 90°
→α = 45°
So the angle of incidence = angle of reflection
= 90° - α
=90° - 45°
=45°
32. A point source is placed at a depth h below the surface of the water (refractive index =µ). (a) Show that the light escapes through a circular area on the water surface with its center directly above the point source. (b) Find the angle subtended by a radius of the area on the source.
ANSWER: Let us draw a vertical line PBA through the point source P which meets the surface at point B which is directly above the source.
(a) The light will escape if it is refracted. The maximum angle of incidence for the refraction will be the critical angle C for which the angle of refraction = 90°. The angle between the vertical line and the extreme refracted ray is also =C. See the diagram below.
 |
| Diagram for Q-32 |
In a three dimensional space around the vertical line, this angle C will be the same and it will make an inverted cone with vertex at the source and its circular base DBE at the water surface. Obviously, this circle is the area through which the light can escape.
(b) DB = BE is the radius of the circular area which subtends an angle C on the source.
But Sin C/Sin 90° =1/µ
→Sin C = 1/µ
→C = sin⁻¹(1/µ)
33. A container contains water up to a height of 20 cm and there is a point source at the center of the bottom of the container. A rubber ring of radius r floats centrally on the water. The ceiling of the room is 2.0 m above the water surface.
(a) Find the radius of the shadow of the ring formed on the ceiling if r = 15 cm.
(b) Find the maximum value of r for which the shadow of the ring is formed on the ceiling. Refractive index of water =4/3.
ANSWER: Let S = point source. ACB is the floating ring with its center C directly above the source S. Radius of the ring AC = BC = r. DFE is the shadow of the ring on the ceiling and DF is the radius of this shadow. The shadow making ray is the refracted ray BE for which from the Snell's Law
Sin i/Sin r =1/µ
 |
| Diagram for Q-33 (a) |
(a) For radius r = 15 cm =0.15 m
Sin i = BC/BS = 0.15/√(0.20²+0.15²) =0.15/0.25
=3/5
µ = 4/3
Hence Sin i/Sin r = 1/µ = 3/4
→Sin r = 4*(3/5)/3 =4/5 =0.8
Hence tan r = sin r/√(1-sin² r) =0.8/√(1-0.64)
=0.8/0.6 =4/3
From the triangle BGE,
GE/BG = tan r = 4/3
→GE = 2.0*4/3 = 8/3 m = 2.66 m
The radius of the shadow = FE =FG+GE =0.15+2.66
=2.81 m
(b) The maximum value of r will be corresponding to the critical angle C as in the case of problem 32(a).
Sin C/Sin 90° = 1/µ
→Sin C = 3/4
Hence the maximum value of r = CS*tan C
=20*sin C/√(1-sin²C)
=20*(3/4)/√(1-9/16)
=20*(3/4)/(√7/4)
=60/√7
=22.6 cm
34. Find the angle of minimum deviation for an equilateral prism made of a material of refractive index 1.732. What is the angle of incidence for this deviation?
ANSWER: Let the angle of minimum deviation = δₘ
Here given µ = 1.732 and A = 60° (Due to equilateral prism). Hence,
µ = Sin{(A+δₘ)/2}/Sin(A/2)
→1.732 = Sin{(δₘ+60°)/2}/sin30°
→Sin{(δₘ+60°)/2}/(½) = 1.732
→Sin{(δₘ+60°)/2} = 0.866
→(δₘ+60°)/2 = 60°
→δₘ = 120° - 60° = 60°
For minimum deviation, i = i'; and
δₘ = 2i - A
→60° = 2i - 60°
→2i = 120°
→i = 60°
35. Find the angle of deviation suffered by the light ray shown in the figure (18-E10). The refractive index µ = 1.5 for the prism material.
 |
| Figure for Q-35 |
ANSWER: Let ABC be the given prism in which A = 4°.
 |
| Diagram for Q-35 |
Since side AB of the prism is perpendicular to the incident ray, it will be refracted into the prism without deviation. It will then be refracted through the face AC. Here, by the geometry, i = 4°. Given µ = 1.5, hence
Sin i/Sin r = 1/µ
→Sin 4°/Sin r = 1/ 1.5
→Sin r = 1.5*Sin 4° =0.104 = Sin 6°
→r = 6°
Hence the angle of deviation δ = r - i =6°-4° =2°.
36. A light ray going through the prism with the angle of prism 60°, is found to deviate by 30°. What limit on the refractive index can be put from these data?
ANSWER: Given A = 60°. Since for the given ray, the deviation is 30°, the angle of minimum deviation will be either less than 30° or equal to 30°. The angle of minimum deviation, δₘ is given as
µ = Sin{(A+δₘ)/2}/Sin(A/2)
Since in this relation µ increases with increase in δₘ, hence
µ ≤ Sin{(60°+30°)/2}/Sin(60°/2)
→µ ≤ Sin 45°/Sin 30°
→µ ≤ (1/√2)/(1/2)
→µ ≤ 2/√2
→µ ≤ √2
37. Locate the image formed by refraction in the situation shown in figure (18-E11).
 |
| The figure for Q-37 |
ANSWER: For refraction at spherical surfaces,
µ₂/v - µ₁/u =(µ₂-µ₁)/R
Here, µ₁ = 1, µ₂ = 1.5, R = 20 cm, u = -25 cm. hence,
1.5/v + 1/25 = (1.5-1.0)/20 = 0.5/20
→1.5/v = 1/40 - 1/25 =(5 - 8)/200 =-3/200
→v/1.5 = -200/3
→v = -200*1.5/3 = -100 cm.
Negative sign shows that the image is on the side of the source. Hence the image is 100 cm from the surface on the side of S.
38. A spherical surface of radius 30 cm separates two transparent media A and B with refractive indices 1.33 and 1.48 respectively. Medium A is on the convex side of the surface. Where should a point object be placed in medium A so that the paraxial rays become parallel after refraction at the surface?
ANSWER: Let us assume that the medium A is on the left side of the spherical surface. So the object distance in medium A will have a negative value (-u). Here,
µ₁ = 1.33, µ₂ = 1.48, R = 30 cm. Since the paraxial rays are parallel after refraction, the image is at infinity, v = ∞.
From the relation,
µ₂/v - µ₁/u =(µ₂-µ₁)/R
→1.48/∞ + 1.33/u = (1.48-1.33)/30 =0.15/30
→u/1.33 = 30/0.15
→u = 1.33*200 =266 cm.
So the object is 266.0 cm away from the separating surface.
39. Figure (18-E12) shows a transparent hemisphere of radius 3.0 cm made of a material of refractive index 2.0. (a) A narrow beam of parallel rays is incident on the hemisphere as shown in the figure. Are the rays totally reflected at the plane surface?
(b) Find the image formed by the refraction at the first surface.
(c) Find the image formed by the reflection or refraction at the plane surface.
(d) Trace qualitatively the final rays as they come out of the hemisphere.
 |
| The figure for Q-39 |
ANSWER: (a) Since the rays along the radial direction they are perpendicular to the outer surface. These rays will pass undeviated and fall on the plane surface at an angle of incidence = 45°.
The refractive index of the material, µ = 2.0, hence the critical angle C = Sin⁻¹(1/µ) = Sin⁻¹(½) = 30°. Since the angle of incidence is greater than the critical angle, the total internal reflection will take place. So the rays are fully reflected at the plane surface.
(b) Let us consider the positive X-axis along the concave medium.
µ₁ = 1.0, µ₂ = 2.0, R = 3.0 cm, u = -∞, v =?
µ₂/v - µ₁/u = (µ₂ - µ₁)/R
→2/v + 1/∞ =(2-1)/3.0
→2/v =1/3
→v = 3*2 = 6 cm
So the image is formed 6 cm inside the transparent medium from A. It means if the sphere is completed the image forms at the point diametrically opposite to A.
(c) As we saw that the rays are fully reflected at the plane surface the image forming rays will intersect after reflection. And the image will be the mirror image of the image formed as in part (b). In fact, it will also be the mirror image of point A in BC.
(d) When the rays come out, they will be diverging.
 |
| Diagram for Q-39(d) |
40. A small object is embedded in a glass sphere (µ=1.5) of radius 5.0 cm at a distance of 1.5 cm left to the center. Locate the image of the object as seen by an observer standing (a) to the left of the sphere and (b) to the right of the sphere.
ANSWER: (a) For this case we take A as the origin and AB negative X-axis. Now, u = -(5-1.5) =-3.5 cm, R = -5 cm, µ₁ = 1.5, µ₂ = 1.0, v = ?
µ₂/v - µ₁/u =(µ₂ - µ₁)/R
→1/v + 1.5/3.5 = -(1.0-1.5)/5 =1/10
→1/v = -3/7 + 1/10 =(-30 + 7)/70 =-23/70
→v = -70/23 =-3.0 cm
So the image is 3 cm from A i.e, 2 cm left to the center.
 |
| Diagram for Q-40 |
(b) When the observer is to the right of the sphere, the refraction is from the face near B. Taking B as the origin. u = -(5+1.5) =-6.5 cm. Hence
1/v - 1.5/(-6.5) =-(1.0-1.5)/5 =1/10
→1/v =1/10 - 3/13 =(13-30)/130 =-17/130
→v = -130/17 =-7.65 cm
So the image is 7.65 cm left of B i.e. 7.65-5 =2.65 cm left to the center.
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Links to the Chapters
Links to the Chapters
CHAPTER- 18 - Geometrical Optics
CHAPTER- 17 - Light Waves
CHAPTER- 16 - Sound Waves
OBJECTIVE-I
OBJECTIVE-II
EXERCISES - Q-1 TO Q-10
EXERCISES - Q-11 TO Q-20
EXERCISES -Q-21 TO Q-30
EXERCISES -Q-31 TO Q-40
EXERCISES -Q-41 TO Q-50
EXERCISES -Q-51 TO Q-60
EXERCISES -Q-61 TO Q-70
EXERCISES - Q-71 TO Q-80
EXERCISES - Q-81 TO Q-89
CHAPTER- 15 - Wave Motion and Waves on a String
OBJECTIVE-II
EXERCISES - Q-1 TO Q-10
EXERCISES - Q-11 TO Q-20
EXERCISES - Q-21 TO Q-30
EXERCISES - Q-31 TO Q-40
EXERCISES - Q-41 TO Q-50
CHAPTER- 14 - Fluid Mechanics
CHAPTER- 13 - Fluid Mechanics
CHAPTER- 12 - Simple Harmonic Motion
CHAPTER- 11 - Gravitation
CHAPTER- 10 - Rotational Mechanics
CHAPTER- 9 - Center of Mass, Linear Momentum, Collision
CHAPTER- 18 - Geometrical Optics
CHAPTER- 16 - Sound Waves
OBJECTIVE-I
OBJECTIVE-II
EXERCISES - Q-1 TO Q-10
EXERCISES - Q-11 TO Q-20
EXERCISES -Q-21 TO Q-30
EXERCISES -Q-31 TO Q-40
EXERCISES -Q-41 TO Q-50
EXERCISES -Q-51 TO Q-60
EXERCISES -Q-61 TO Q-70
EXERCISES - Q-71 TO Q-80
EXERCISES - Q-81 TO Q-89
CHAPTER- 15 - Wave Motion and Waves on a String
EXERCISES - Q-1 TO Q-10
EXERCISES - Q-11 TO Q-20
EXERCISES - Q-21 TO Q-30
EXERCISES - Q-31 TO Q-40
EXERCISES - Q-41 TO Q-50
CHAPTER- 14 - Fluid Mechanics
CHAPTER- 13 - Fluid Mechanics
CHAPTER- 12 - Simple Harmonic Motion
CHAPTER- 11 - Gravitation
CHAPTER- 10 - Rotational Mechanics
CHAPTER- 9 - Center of Mass, Linear Momentum, Collision
CHAPTER- 8 - Work and Energy
Click here for → Question for Short Answers
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Click here for → Exercise(43-54)
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Click here for → Exercise(43-54)
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CHAPTER- 7 - Circular Motion
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CHAPTER- 6 - Friction
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CHAPTER- 6 - Friction
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For more practice on problems on friction solve these- "New Questions on Friction".
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CHAPTER- 5 - Newton's Laws of Motion
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Click here for → Newton's Laws of Motion-Exercises(Q. No. 1 to 12)
Click here for→Newton's Laws of Motion,Exercises(Q.No. 13 to 27)
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CHAPTER- 4 - The Forces
The Forces-
"Questions for short Answers"
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CHAPTER- 3 - Kinematics - Rest and Motion
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Click here for EXERCISES (Question number 11 to 20)
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Click here for EXERCISES (Question number 31 to 40)
Click here for EXERCISES (Question number 41 to 52)
CHAPTER- 2 - "Vector related Problems"
CHAPTER- 2 - "Vector related Problems"
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