GEOMETRICAL OPTICS
EXERCISES- Q21 to Q30
21. A cylindrical vessel of diameter 12 cm contains 800π cm³ of water. A cylindrical glass piece of diameter 8.0 cm and height 8.0 cm is placed in the vessel. If the bottom of the vessel under the glass piece is seen by the paraxial rays (see figure 18-E6), locate its image. The index of refraction of glass is 1.50 and that of the water is 1.33.
The figure for Q-21
ANSWER: We need to know the final height of water above the glass piece so that the total shift due to glass and water can be calculated.
The volume of the glass piece =πr²h
=π*(8/2)²*8 cm³
=128π cm³
The total volume of the water and the glass piece
=128π + 800π cm³
=928π cm³ ------------ (i)
Let the height of water above the glass piece = x
The volume of the vessel up to the water height
=π*(12/2)²*(x+8)
=36πx + 288π -------------(ii)
Equating (i) and (ii)
36πx + 288π = 928π
→x = 640/36 =17.78 cm
The shift of the image due to glass
Δ=(1-1/1.50)*8 cm
=2.67 cm
The shift of the image due to water
Δ'=(1-1/1.33)*17.78 cm
=4.44 cm
The total shift of the image = Δ+Δ'=2.67+4.44 =7.08 cm
≈7.01 cm above the bottom.
22. Consider the situation in figure (18-E7). The bottom of the pot is a reflecting plane mirror. S is a small fish and T is a human eye. Refractive index of water is µ. (a) At what distance(s) from itself will the fish see the image(s) of the eye? (b) At what distance(s) from itself will the eye see the image(s) of the fish?
Figure for Q-22
ANSWER: (a) Since the eye is in a rarer medium than the fish, the fish will see the eye farther than the real distance. Let the apparent distance of the image of the eye be 'h' above the water surface.
Real depth/apparent depth =H/h =1/µ
→h = µH
The distance of the image seen by the fish =H/2 + h
=H/2 + µH
=H(µ+½) above itself.
Another image will be in the mirror below. The distance of this image seen by the fish = (H+h)+H/2
=H+µH+H/2
=µH+3H/2
=H(µ+3/2) below itself.
(b) For the eye, the direct distance of the fish image will be closer than the real. Let x = apparent depth.
Real depth of the fish/Apparent depth of the fish =µ
→(H/2)/x =µ
→x = H/2µ
Hence the distance of the image seen by the eye =H+x
=H+H/2µ
=H(1+1/2µ) below itself.
Another image of the fish will be seen in the mirror. The image of the fish will be at H/2 from the bottom. Real depth of this image from the water surface
=H+H/2 =3H/2
Let X =apparent depth
Real depth/apparent depth =µ
→(3H/2)/X =µ
→X = 3H/2µ
Hence the distance of this image seen by the eye
=H+X
=H+3H/2µ
=H(1+3/2µ) below itself.
23. A small object is placed at the center of the bottom of a cylindrical vessel of radius 3 cm and height 4 cm filled completely with water. Consider the ray leaving the vessel through a corner. Suppose this ray and the ray along the axis of the vessel are used to trace the image. Find the apparent depth of the image and the ratio of real depth to the apparent depth under the assumptions taken. Refractive index of water = 1.33.
ANSWER: Given µ = 1.33, let us draw a diagram as below.
Diagram for Q-23
From the figure, AD =√(3²+4²) =5
Sin i = 3/5
Sin r = CD/BD =3/BD
Here Sin i/Sin r = 1/µ
= (3/5)/(3/BD)=1/1.33
→BD/5 =1/1.33
→BD = 3.76 cm
BC² = BD²-CD² =3.75² - 3² = 5.06
→BC = 2.25 cm
Hence the apparent depth of the image = 2.25 cm
The ratio, Real depth/apparent depth =AC/BC
=4/2.25
=1.78
24. A cylindrical vessel, whose diameter and height both are equal to 30 cm, is placed on a horizontal surface and a small particle P is placed in it at a distance of 5.0 cm from the center. An eye is placed at a position such that the edge of the bottom is just visible (see figure 18-E8). The particle P is in the plane of the drawing. Up to what minimum height should water be poured in the vessel to make the particle P visible?
The figure for Q-24
ANSWER: When the refracted ray BE coming from P is along the ray CBE, the particle P will be just visible.
Diagram for Q-24
Suppose at this instant the water level is d. Draw a normal BD at the point of refraction B. Let PD = x.
Now, Sin i/Sin r = 1/µ = 3/4
→(PD/PB)/(Sin 45°) = 3/4
→PD/PB = 3/4√2
→x/√(x²+d²) = 3/4√2
→x²/(x²+d²) =9/32
→32x² = 9x²+9d²
→23x² = 9d²
→x = 3d/√23
In right angle triangle BCD,
BD/CD = tan45° = 1
→BD = CD
→d = 15+(x-5) =x+10 =3d/√23+10
→d(1-3/√23) =10
→d*0.37 =10
→d = 10/0.37 = 26.7 cm
25. A light ray is incident at an angle of 45° with the normal to a √2 cm thick plate (µ=2.0). Find the shift in the path of the light as it emerges out from the plate.
ANSWER: ABF is the original path of the ray. When the plate is kept in its way, the path is ABCD. The shift in the path is CE =x (say).
Diagram for Q-25
Sin i/Sin r = µ
→Sin 45°/sin r = 2
→Sin r =1/2√2 =0.353 =20.7°
BG/BC = cos r
→BC =BG/cos r
CE/BC = sin (i-r)
→x = BG*sin(i-r)/cos r =√2*sin(45°-20.7°)/cos20.7°
→x = √2*0.41/0.94 =0.62 cm
26. An optical fiber (µ=1.72) is surrounded by a glass coating (µ=1.5). Find the critical angle for the total internal reflection at the fiber-glass interface.
ANSWER: Let the critical angle for total internal reflection at the fiber-glass interface = C
Since Sin i/Sin r = µ₂/µ₁ = Sin C/Sin 90° =Sin C
Given, µ₁ = 1.72 and µ₂ = 1.5
Hence, Sin C = 1.5/1.72 =150/172 =75/86
→C = Sin⁻¹(75/86)
27. A light ray is incident normally on the face AB of a right-angled prism ABC (µ=1.50) as shown in figure (18-E9). What is the largest angle ⲫ for which the light ray is totally reflected at the surface AC?
The figure for Q-27
ANSWER: From the figure, it is clear that
Diagram for Q-27
angle, i + ⲫ =90°
For maximum ⲫ, i should be equal to the critical angle C.
→C + ⲫ = 90°
→C =90° - ⲫ
So, Sin i/sin r = 1/µ
→Sin C/Sin90° = 1/1.5
→Sin (90°-ⲫ) =2/3
→Cos ⲫ = 2/3
→ⲫ = cos⁻¹(2/3)
28. Find the maximum angle of refraction when a light ray is refracted from glass (µ=1.50) to air.
ANSWER: It is clear that glass is a denser medium than the air. So when the ray comes out from glass to air it is refracted away from the normal at the point of incidence. Thus the angle of refraction r (in the air) is greater than the angle of incidence i (in glass). when i is increased such that r =90°, the value of this i is called the critical angle C, because for i>C the Total Internal Reflection occurs and refraction is not possible. Hence the maximum possible angle of refraction = 90°.
29. Light is incident from glass(µ=1.5) to air. Sketch the variation of the angle of deviation δ with the angle of incident i for 0 < i < 90°.
ANSWER: When light is refracted from glass to air, the angle of refraction r is greater than the angle of incidence i. The angle of deviation δ is the angle between refracted or reflected ray and the original path of the ray. δ = r-i. Let us study the variation of δ by drawing a diagram as below.
Diagram for Q-29
When light is incident normally at the glass/air surface it emerges out straight, i = r = 0.
Hence when i = 0, δ = 0.
When i is increased δ = r - i increases up to the critical condition when r = 90°. For this critical angle
Sin i/Sin 90° = 1/µ
→Sin C = 1/1.5 = 0.67
→C = 41.8°
For i = C = 41.8°, δ = 90°-41.8° =48.2°
When i is just increased, total internal reflection occurs. For which δ = 2(90°-i) =2(90°-41.8°) =83.6°
So here is a sudden jump in the δ.
But if now i is increased δ starts decreasing. For i = 90°, again δ = 0.
With this data, we can now plot the i-δ graph below.
i - δ graph
Between 0° to 41.8°, δ increases with increase in i, but the variation is not linear. However, the variation of δ in the total internal reflection part is linear.
30. Light is incident from glass(µ=1.50) to water (µ=1.33). Find the range of the angle of deviation for which there are two angles of incidence.
ANSWER: First we calculate the critical angle.
Sin i/Sin r = µ₂/µ₁
→Sin C/Sin 90° = 1.33/1.50
→Sin C = 4/4.5 = 8/9
→C = Sin⁻¹(8/9) =62.37°
For the critical angle, deviation δ = 90°-Sin⁻¹(8/9)
=Cos⁻¹(8/9)
{=90°-62.37°
= 27.63°}
Let us plot the variation of δ with respect to i as in the previous problem.
i-δ variation
In this graph, if we draw a line AB parallel to the i-axis, it intersects the graph at two places for δ=0 to Cos⁻¹(8/9). Which means there are two values of i for a single value of δ. Hence the range of δ is 0 to cos⁻¹(8/9) for which there are two values of i.
21. A cylindrical vessel of diameter 12 cm contains 800π cm³ of water. A cylindrical glass piece of diameter 8.0 cm and height 8.0 cm is placed in the vessel. If the bottom of the vessel under the glass piece is seen by the paraxial rays (see figure 18-E6), locate its image. The index of refraction of glass is 1.50 and that of the water is 1.33.
ANSWER: We need to know the final height of water above the glass piece so that the total shift due to glass and water can be calculated.
The volume of the glass piece =πr²h
=π*(8/2)²*8 cm³
=128π cm³
The total volume of the water and the glass piece
=128π + 800π cm³
=928π cm³ ------------ (i)
Let the height of water above the glass piece = x
The volume of the vessel up to the water height
=π*(12/2)²*(x+8)
=36πx + 288π -------------(ii)
Equating (i) and (ii)
36πx + 288π = 928π
→x = 640/36 =17.78 cm
The shift of the image due to glass
Δ=(1-1/1.50)*8 cm
=2.67 cm
The shift of the image due to water
Δ'=(1-1/1.33)*17.78 cm
=4.44 cm
The total shift of the image = Δ+Δ'=2.67+4.44 =7.08 cm
≈7.01 cm above the bottom.
22. Consider the situation in figure (18-E7). The bottom of the pot is a reflecting plane mirror. S is a small fish and T is a human eye. Refractive index of water is µ. (a) At what distance(s) from itself will the fish see the image(s) of the eye? (b) At what distance(s) from itself will the eye see the image(s) of the fish?
ANSWER: (a) Since the eye is in a rarer medium than the fish, the fish will see the eye farther than the real distance. Let the apparent distance of the image of the eye be 'h' above the water surface.
Real depth/apparent depth =H/h =1/µ
→h = µH
The distance of the image seen by the fish =H/2 + h
=H/2 + µH
=H(µ+½) above itself.
Another image will be in the mirror below. The distance of this image seen by the fish = (H+h)+H/2
=H+µH+H/2
=µH+3H/2
=H(µ+3/2) below itself.
(b) For the eye, the direct distance of the fish image will be closer than the real. Let x = apparent depth.
Real depth of the fish/Apparent depth of the fish =µ
→(H/2)/x =µ
→x = H/2µ
Hence the distance of the image seen by the eye =H+x
=H+H/2µ
=H(1+1/2µ) below itself.
Another image of the fish will be seen in the mirror. The image of the fish will be at H/2 from the bottom. Real depth of this image from the water surface
=H+H/2 =3H/2
Let X =apparent depth
Real depth/apparent depth =µ
→(3H/2)/X =µ
→X = 3H/2µ
Hence the distance of this image seen by the eye
=H+X
=H+3H/2µ
=H(1+3/2µ) below itself.
23. A small object is placed at the center of the bottom of a cylindrical vessel of radius 3 cm and height 4 cm filled completely with water. Consider the ray leaving the vessel through a corner. Suppose this ray and the ray along the axis of the vessel are used to trace the image. Find the apparent depth of the image and the ratio of real depth to the apparent depth under the assumptions taken. Refractive index of water = 1.33.
ANSWER: Given µ = 1.33, let us draw a diagram as below.
From the figure, AD =√(3²+4²) =5
Sin i = 3/5
Sin r = CD/BD =3/BD
Here Sin i/Sin r = 1/µ
= (3/5)/(3/BD)=1/1.33
→BD/5 =1/1.33
→BD = 3.76 cm
BC² = BD²-CD² =3.75² - 3² = 5.06
→BC = 2.25 cm
Hence the apparent depth of the image = 2.25 cm
The ratio, Real depth/apparent depth =AC/BC
=4/2.25
=1.78
24. A cylindrical vessel, whose diameter and height both are equal to 30 cm, is placed on a horizontal surface and a small particle P is placed in it at a distance of 5.0 cm from the center. An eye is placed at a position such that the edge of the bottom is just visible (see figure 18-E8). The particle P is in the plane of the drawing. Up to what minimum height should water be poured in the vessel to make the particle P visible?
ANSWER: When the refracted ray BE coming from P is along the ray CBE, the particle P will be just visible.
Suppose at this instant the water level is d. Draw a normal BD at the point of refraction B. Let PD = x.
Now, Sin i/Sin r = 1/µ = 3/4
→(PD/PB)/(Sin 45°) = 3/4
→PD/PB = 3/4√2
→x/√(x²+d²) = 3/4√2
→x²/(x²+d²) =9/32
→32x² = 9x²+9d²
→23x² = 9d²
→x = 3d/√23
In right angle triangle BCD,
BD/CD = tan45° = 1
→BD = CD
→d = 15+(x-5) =x+10 =3d/√23+10
→d(1-3/√23) =10
→d*0.37 =10
→d = 10/0.37 = 26.7 cm
25. A light ray is incident at an angle of 45° with the normal to a √2 cm thick plate (µ=2.0). Find the shift in the path of the light as it emerges out from the plate.
ANSWER: ABF is the original path of the ray. When the plate is kept in its way, the path is ABCD. The shift in the path is CE =x (say).
Sin i/Sin r = µ
→Sin 45°/sin r = 2
→Sin r =1/2√2 =0.353 =20.7°
BG/BC = cos r
→BC =BG/cos r
CE/BC = sin (i-r)
→x = BG*sin(i-r)/cos r =√2*sin(45°-20.7°)/cos20.7°
→x = √2*0.41/0.94 =0.62 cm
26. An optical fiber (µ=1.72) is surrounded by a glass coating (µ=1.5). Find the critical angle for the total internal reflection at the fiber-glass interface.
ANSWER: Let the critical angle for total internal reflection at the fiber-glass interface = C
Since Sin i/Sin r = µ₂/µ₁ = Sin C/Sin 90° =Sin C
Given, µ₁ = 1.72 and µ₂ = 1.5
Hence, Sin C = 1.5/1.72 =150/172 =75/86
→C = Sin⁻¹(75/86)
27. A light ray is incident normally on the face AB of a right-angled prism ABC (µ=1.50) as shown in figure (18-E9). What is the largest angle ⲫ for which the light ray is totally reflected at the surface AC?
ANSWER: From the figure, it is clear that
angle, i + ⲫ =90°
For maximum ⲫ, i should be equal to the critical angle C.
→C + ⲫ = 90°
→C =90° - ⲫ
So, Sin i/sin r = 1/µ
→Sin C/Sin90° = 1/1.5
→Sin (90°-ⲫ) =2/3
→Cos ⲫ = 2/3
→ⲫ = cos⁻¹(2/3)
28. Find the maximum angle of refraction when a light ray is refracted from glass (µ=1.50) to air.
ANSWER: It is clear that glass is a denser medium than the air. So when the ray comes out from glass to air it is refracted away from the normal at the point of incidence. Thus the angle of refraction r (in the air) is greater than the angle of incidence i (in glass). when i is increased such that r =90°, the value of this i is called the critical angle C, because for i>C the Total Internal Reflection occurs and refraction is not possible. Hence the maximum possible angle of refraction = 90°.
29. Light is incident from glass(µ=1.5) to air. Sketch the variation of the angle of deviation δ with the angle of incident i for 0 < i < 90°.
ANSWER: When light is refracted from glass to air, the angle of refraction r is greater than the angle of incidence i. The angle of deviation δ is the angle between refracted or reflected ray and the original path of the ray. δ = r-i. Let us study the variation of δ by drawing a diagram as below.
When light is incident normally at the glass/air surface it emerges out straight, i = r = 0.
Hence when i = 0, δ = 0.
When i is increased δ = r - i increases up to the critical condition when r = 90°. For this critical angle
Sin i/Sin 90° = 1/µ
→Sin C = 1/1.5 = 0.67
→C = 41.8°
For i = C = 41.8°, δ = 90°-41.8° =48.2°
When i is just increased, total internal reflection occurs. For which δ = 2(90°-i) =2(90°-41.8°) =83.6°
So here is a sudden jump in the δ.
But if now i is increased δ starts decreasing. For i = 90°, again δ = 0.
With this data, we can now plot the i-δ graph below.
Between 0° to 41.8°, δ increases with increase in i, but the variation is not linear. However, the variation of δ in the total internal reflection part is linear.
30. Light is incident from glass(µ=1.50) to water (µ=1.33). Find the range of the angle of deviation for which there are two angles of incidence.
ANSWER: First we calculate the critical angle.
Sin i/Sin r = µ₂/µ₁
→Sin C/Sin 90° = 1.33/1.50
→Sin C = 4/4.5 = 8/9
→C = Sin⁻¹(8/9) =62.37°
For the critical angle, deviation δ = 90°-Sin⁻¹(8/9)
=Cos⁻¹(8/9)
{=90°-62.37°
= 27.63°}
Let us plot the variation of δ with respect to i as in the previous problem.
In this graph, if we draw a line AB parallel to the i-axis, it intersects the graph at two places for δ=0 to Cos⁻¹(8/9). Which means there are two values of i for a single value of δ. Hence the range of δ is 0 to cos⁻¹(8/9) for which there are two values of i.
 |
| The figure for Q-21 |
ANSWER: We need to know the final height of water above the glass piece so that the total shift due to glass and water can be calculated.
The volume of the glass piece =πr²h
=π*(8/2)²*8 cm³
=128π cm³
The total volume of the water and the glass piece
=128π + 800π cm³
=928π cm³ ------------ (i)
Let the height of water above the glass piece = x
The volume of the vessel up to the water height
=π*(12/2)²*(x+8)
=36πx + 288π -------------(ii)
Equating (i) and (ii)
36πx + 288π = 928π
→x = 640/36 =17.78 cm
The shift of the image due to glass
Δ=(1-1/1.50)*8 cm
=2.67 cm
The shift of the image due to water
Δ'=(1-1/1.33)*17.78 cm
=4.44 cm
The total shift of the image = Δ+Δ'=2.67+4.44 =7.08 cm
≈7.01 cm above the bottom.
22. Consider the situation in figure (18-E7). The bottom of the pot is a reflecting plane mirror. S is a small fish and T is a human eye. Refractive index of water is µ. (a) At what distance(s) from itself will the fish see the image(s) of the eye? (b) At what distance(s) from itself will the eye see the image(s) of the fish?
 |
| Figure for Q-22 |
ANSWER: (a) Since the eye is in a rarer medium than the fish, the fish will see the eye farther than the real distance. Let the apparent distance of the image of the eye be 'h' above the water surface.
Real depth/apparent depth =H/h =1/µ
→h = µH
The distance of the image seen by the fish =H/2 + h
=H/2 + µH
=H(µ+½) above itself.
Another image will be in the mirror below. The distance of this image seen by the fish = (H+h)+H/2
=H+µH+H/2
=µH+3H/2
=H(µ+3/2) below itself.
(b) For the eye, the direct distance of the fish image will be closer than the real. Let x = apparent depth.
Real depth of the fish/Apparent depth of the fish =µ
→(H/2)/x =µ
→x = H/2µ
Hence the distance of the image seen by the eye =H+x
=H+H/2µ
=H(1+1/2µ) below itself.
Another image of the fish will be seen in the mirror. The image of the fish will be at H/2 from the bottom. Real depth of this image from the water surface
=H+H/2 =3H/2
Let X =apparent depth
Real depth/apparent depth =µ
→(3H/2)/X =µ
→X = 3H/2µ
Hence the distance of this image seen by the eye
=H+X
=H+3H/2µ
=H(1+3/2µ) below itself.
23. A small object is placed at the center of the bottom of a cylindrical vessel of radius 3 cm and height 4 cm filled completely with water. Consider the ray leaving the vessel through a corner. Suppose this ray and the ray along the axis of the vessel are used to trace the image. Find the apparent depth of the image and the ratio of real depth to the apparent depth under the assumptions taken. Refractive index of water = 1.33.
ANSWER: Given µ = 1.33, let us draw a diagram as below.
|
| Diagram for Q-23 |
From the figure, AD =√(3²+4²) =5
Sin i = 3/5
Sin r = CD/BD =3/BD
Here Sin i/Sin r = 1/µ
= (3/5)/(3/BD)=1/1.33
→BD/5 =1/1.33
→BD = 3.76 cm
BC² = BD²-CD² =3.75² - 3² = 5.06
→BC = 2.25 cm
Hence the apparent depth of the image = 2.25 cm
The ratio, Real depth/apparent depth =AC/BC
=4/2.25
=1.78
24. A cylindrical vessel, whose diameter and height both are equal to 30 cm, is placed on a horizontal surface and a small particle P is placed in it at a distance of 5.0 cm from the center. An eye is placed at a position such that the edge of the bottom is just visible (see figure 18-E8). The particle P is in the plane of the drawing. Up to what minimum height should water be poured in the vessel to make the particle P visible?
 |
| The figure for Q-24 |
ANSWER: When the refracted ray BE coming from P is along the ray CBE, the particle P will be just visible.
 |
| Diagram for Q-24 |
Suppose at this instant the water level is d. Draw a normal BD at the point of refraction B. Let PD = x.
Now, Sin i/Sin r = 1/µ = 3/4
→(PD/PB)/(Sin 45°) = 3/4
→PD/PB = 3/4√2
→x/√(x²+d²) = 3/4√2
→x²/(x²+d²) =9/32
→32x² = 9x²+9d²
→23x² = 9d²
→x = 3d/√23
In right angle triangle BCD,
BD/CD = tan45° = 1
→BD = CD
→d = 15+(x-5) =x+10 =3d/√23+10
→d(1-3/√23) =10
→d*0.37 =10
→d = 10/0.37 = 26.7 cm
25. A light ray is incident at an angle of 45° with the normal to a √2 cm thick plate (µ=2.0). Find the shift in the path of the light as it emerges out from the plate.
ANSWER: ABF is the original path of the ray. When the plate is kept in its way, the path is ABCD. The shift in the path is CE =x (say).
 |
| Diagram for Q-25 |
Sin i/Sin r = µ
→Sin 45°/sin r = 2
→Sin r =1/2√2 =0.353 =20.7°
BG/BC = cos r
→BC =BG/cos r
CE/BC = sin (i-r)
→x = BG*sin(i-r)/cos r =√2*sin(45°-20.7°)/cos20.7°
→x = √2*0.41/0.94 =0.62 cm
26. An optical fiber (µ=1.72) is surrounded by a glass coating (µ=1.5). Find the critical angle for the total internal reflection at the fiber-glass interface.
ANSWER: Let the critical angle for total internal reflection at the fiber-glass interface = C
Since Sin i/Sin r = µ₂/µ₁ = Sin C/Sin 90° =Sin C
Given, µ₁ = 1.72 and µ₂ = 1.5
Hence, Sin C = 1.5/1.72 =150/172 =75/86
→C = Sin⁻¹(75/86)
27. A light ray is incident normally on the face AB of a right-angled prism ABC (µ=1.50) as shown in figure (18-E9). What is the largest angle ⲫ for which the light ray is totally reflected at the surface AC?
 |
| The figure for Q-27 |
ANSWER: From the figure, it is clear that
 |
| Diagram for Q-27 |
angle, i + ⲫ =90°
For maximum ⲫ, i should be equal to the critical angle C.
→C + ⲫ = 90°
→C =90° - ⲫ
So, Sin i/sin r = 1/µ
→Sin C/Sin90° = 1/1.5
→Sin (90°-ⲫ) =2/3
→Cos ⲫ = 2/3
→ⲫ = cos⁻¹(2/3)
28. Find the maximum angle of refraction when a light ray is refracted from glass (µ=1.50) to air.
ANSWER: It is clear that glass is a denser medium than the air. So when the ray comes out from glass to air it is refracted away from the normal at the point of incidence. Thus the angle of refraction r (in the air) is greater than the angle of incidence i (in glass). when i is increased such that r =90°, the value of this i is called the critical angle C, because for i>C the Total Internal Reflection occurs and refraction is not possible. Hence the maximum possible angle of refraction = 90°.
29. Light is incident from glass(µ=1.5) to air. Sketch the variation of the angle of deviation δ with the angle of incident i for 0 < i < 90°.
ANSWER: When light is refracted from glass to air, the angle of refraction r is greater than the angle of incidence i. The angle of deviation δ is the angle between refracted or reflected ray and the original path of the ray. δ = r-i. Let us study the variation of δ by drawing a diagram as below.
 |
| Diagram for Q-29 |
Hence when i = 0, δ = 0.
When i is increased δ = r - i increases up to the critical condition when r = 90°. For this critical angle
Sin i/Sin 90° = 1/µ
→Sin C = 1/1.5 = 0.67
→C = 41.8°
For i = C = 41.8°, δ = 90°-41.8° =48.2°
When i is just increased, total internal reflection occurs. For which δ = 2(90°-i) =2(90°-41.8°) =83.6°
So here is a sudden jump in the δ.
But if now i is increased δ starts decreasing. For i = 90°, again δ = 0.
With this data, we can now plot the i-δ graph below.
 |
| i - δ graph |
30. Light is incident from glass(µ=1.50) to water (µ=1.33). Find the range of the angle of deviation for which there are two angles of incidence.
ANSWER: First we calculate the critical angle.
Sin i/Sin r = µ₂/µ₁
→Sin C/Sin 90° = 1.33/1.50
→Sin C = 4/4.5 = 8/9
→C = Sin⁻¹(8/9) =62.37°
For the critical angle, deviation δ = 90°-Sin⁻¹(8/9)
=Cos⁻¹(8/9)
{=90°-62.37°
= 27.63°}
Let us plot the variation of δ with respect to i as in the previous problem.
 |
| i-δ variation |
In this graph, if we draw a line AB parallel to the i-axis, it intersects the graph at two places for δ=0 to Cos⁻¹(8/9). Which means there are two values of i for a single value of δ. Hence the range of δ is 0 to cos⁻¹(8/9) for which there are two values of i.
===<<<O>>>===
===<<<O>>>===
My Channel on YouTube → SimplePhysics with KK
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
Click here for → OBJECTIVE-I
Click here for → OBJECTIVE-II
Click here for → Exercises (1-10)
Click here for → Question for Short Answers
Click here for → OBJECTIVE-I
Click here for → OBJECTIVE-II
Click here for → Exercises (1-10)
Click here for → Exercises (11-20)
Click here for → Exercises (21-30)
Click here for → Exercises (31-42)
Click here for → Exercise(43-54)
Click here for → Exercises (21-30)
Click here for → Exercises (31-42)
Click here for → Exercise(43-54)
Click here for → Exercises (31-42)
Click here for → Exercise(43-54)
CHAPTER- 7 - Circular Motion
Click here for → Questions for Short Answer
Click here for → OBJECTIVE-I
Click here for → OBJECTIVE-II
Click here for → EXERCISES (1-10)
Click here for → EXERCISES (11-20)
Click here for → EXERCISES (21-30)
CHAPTER- 6 - Friction
Click here for → Questions for Short Answer
Click here for → OBJECTIVE-I
Click here for → OBJECTIVE-II
Click here for → EXERCISES (1-10)
Click here for → EXERCISES (11-20)
Click here for → EXERCISES (21-30)
Click here for → OBJECTIVE-I
Click here for → OBJECTIVE-II
Click here for → EXERCISES (1-10)
Click here for → EXERCISES (11-20)
Click here for → EXERCISES (21-30)
CHAPTER- 6 - Friction
Click here for → Questions for Short Answer
Click here for → Questions for Short Answer
Click here for → OBJECTIVE-I
Click here for → Friction - OBJECTIVE-II
Click here for → EXERCISES (1-10)
Click here for → Exercises (11-20)
Click here for → EXERCISES (21-31)
For more practice on problems on friction solve these- "New Questions on Friction".
Click here for → OBJECTIVE-I
Click here for → Friction - OBJECTIVE-II
Click here for → EXERCISES (1-10)
Click here for → Exercises (11-20)
Click here for → EXERCISES (21-31)
---------------------------------------------------------------------------------
---------------------------------------------------------------------------------
CHAPTER- 5 - Newton's Laws of Motion
Click here for → QUESTIONS FOR SHORT ANSWER
Click here for → QUESTIONS FOR SHORT ANSWER
Click here for→ Newton's laws of motion - Objective - I
Click here for → Newton's Laws of Motion - Objective -II
Click here for → Newton's Laws of Motion-Exercises(Q. No. 1 to 12)
Click here for→ Newton's laws of motion - Objective - I
Click here for → Newton's Laws of Motion - Objective -II
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)
-------------------------------------------------------------------------------
-------------------------------------------------------------------------------
CHAPTER- 4 - The Forces
The Forces-
"Questions for short Answers"
Click here for "The Forces" - OBJECTIVE-I
Click here for "The Forces" - OBJECTIVE-II
Click here for "The Forces" - Exercises
--------------------------------------------------------------------------------------------------------------
--------------------------------------------------------------------------------------------------------------
CHAPTER- 3 - Kinematics - Rest and Motion
Click here for "Questions for short Answers"
Click here for "OBJECTIVE-I"
Click here for EXERCISES (Question number 1 to 10)
Click here for EXERCISES (Question number 11 to 20)
Click here for EXERCISES (Question number 21 to 30)
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"
