GEOMETRICAL OPTICS
EXERCISES- Q11 to Q20
11. A concave mirror of radius R is kept on a horizontal table (figure 18-E2). Water (refractive index = µ) is poured into it up to a height h. Where should an object be placed so that its image is formed on itself?
Figure for Q-11
ANSWER: When there is no water, an object placed at the center of curvature C will have its image at the same place. When water is put into the mirror up to a height of h, the new position of such point becomes O which we assume x above the water surface. Now the object placed at O gets its image at O itself though its apparent position will be C. (See the diagram below).
Diagram for Q-11
We know that,
Real depth / Apparent Depth = µ
→x/(R-h) = 1/µ (with respect to the mirror)
→x = (R-h)/µ above the water surface.
12. A point source S is placed midway between two converging mirrors having equal focal length f as shown in figure (18-E3). Find the values of d for which only one image is formed.
Figure for Q-12
ANSWER: The conditions for only one image are as follows:-
Case - 1
We know that the image of the object placed at 2f from the mirror is at the same place. Since both of the mirrors have equal f if an object is placed at 2f distance from each mirror and this place is midway between the mirrors, then only one image at that point will be formed. For this
2f + 2f = d
→d = 4f
Case - 2
If the object is at an equal distance of f from each mirror, the rays after reflection from one mirror will become parallel and these parallel rays after reflection from the other mirror will form the image at f distance from it. So only one image will be formed. In this case
Diagram for Q-12
f + f = d
→d =2f
13. A converging mirror M₁, a point source S, and a diverging mirror M₂ are arranged as shown in figure (18-E4). The source is placed at a distance of 30 cm from M₁. The focal length of each of the mirrors is 20 cm. Consider only the images formed by a maximum of two reflections. It is found that one image is formed on the source itself.
(a) Find the distance between the two mirrors.
(b) Find the location of the image formed by the single reflection from M₂.
Figure for Q-13
ANSWER: For M₁,
u = 30 cm, f = 20 cm, v =?
From the mirror formula,
1/v + 1/u = 1/f
→1/v = 1/20 - 1/30 =(3-2)/60 = 1/60
→v = 60 cm (on the same side that of object).
Let the distance between the mirrors = x. For the reflection from the mirror M₂,
u = (60-x) cm, f = 20 cm, v = -(x-30) cm. From the mirror formula
-1/(x-30) + 1/(60-x) = 1/20
→(x-30)-(60-x) =(x-30)(60-x)/20
→(2x-90)*20 = -x² -1800 + 90x
→40x - 1800 = -x² -1800 + 90x
→x² -50x = 0
→x(x-50) = 0
Since x ≠ 0,
x - 50 = 0
→x = 50 cm
For the single reflection from M₂,
u = -(x-30) cm =-(50-30) cm =-20 cm
f = 20 cm, from the mirror formula
1/v + 1/u = 1/f
→1/v - 1/20 = 1/20
→1/v = 1/20 + 1/20 = 2/20 = 1/10
→v = 10 cm
The positive sign says that the image is on the right side of the diverging mirror. So the image is 10 cm from the diverging mirror farther from the converging mirror.
14. A light ray falling at an angle of 45° with the surface of a clean slab of ice of thickness 1.00 m is refracted into it at an angle of 30°. Calculate the time taken by the light rays to cross the slab. Speed of light in vacuum = 3x10⁸ m/s.
ANSWER: Let us draw a diagram as follows.
Diagram for Q-14
Assume the distance traveled in the ice = x. Given the thickness of the slab d = 1 m.
Hence x= (1 m)*Sec 30° = 2/√3 m
Speed of light in vacuum = vₒ = 3x10⁸ m/s.
If the speed of light in ice be vᵢ,
Then the refractive index µ = vₒ/vᵢ
But also, µ = sin i/ sin r = sin 45°/sin 30°
=(1/√2)/(1/2) =√2, hence
vₒ/vᵢ = √2
→vᵢ = vₒ/√2 = 3x10⁸/√2 m/s =2.121x10⁸ m/s
Time taken by the light ray to cross the ice slab
=x/vᵢ =(2/√3)/2.121x10⁸ s =0.544x10⁻⁸*10⁹ ns
=5.44 ns.
15. A pole of length 1.00 m stands half dipped in a swimming pool with water level 50.0 cm higher than the bed. The refractive index of water is 1.33 and sunlight is coming at an angle of 45° with the vertical. Find the length of the shadow of the pole on the bed.
ANSWER: Let us draw a diagram as follows.
Diagram for Q-15
The hight of the pole AB = 1 m = 100 cm.
Depth of water BC = 50 cm.
Length of the shadow on the water surface CD
= AC*tan 45° = 50*1 cm = 50 cm.
Let the angle of refraction = r
Then sin 45°/sin r =µ
→sin r = 1/√2µ = 1/(√2*1.33) =3/4√2
We know tan r = sin r/cos r =sin r/√(1-sin² r)
→tan r = (3/4√2)/√(1-9/32) =0.75*4√2/(√23*√2)
=3/√23 =0.626
The length of shadow on the bed =BE =BF+FE
=DC + DF*tan r
=50 + 50*0.626 =81.3 cm
16. A small piece of wood is floating on the surface of a 2.5 m deep lake. Where does the shadow form on the bottom when the sun is just setting? Refractive index of water =4/3.
ANSWER: When the sun is setting, the rays are parallel to the water surface. Hence the angle of incidence i = 90°.
µ = sin 90°/sin r
→sin r = 1/µ = 3/4
Diagram for Q-16
If the wood is floating at A, its shadow forms at B. Distance of the shadow from the position directly below the piece of wood = BC (see the diagram above)
=AC* tan r
=2.5*sin r/√(1-sin² r)
=2.5*3/4√(1-9/16)
=7.5/√7 m
=2.83 m
17. An object P is focused by a microscope M. A glass slab of thickness 2.1 cm is introduced between P and M. If the refractive index of the slab is 1.5, by what distance should the microscope be shifted to focus the object again?
ANSWER: Thickness of the glass slab, t = 2.1 cm.
The refractive index µ = 1.5
Hence the shift for the microscope
Δt =(1-1/µ)t
=(1-1/1.5)*2.1 cm
=0.70 cm
18. A vessel contains water up to a height of 20 cm and above it an oil up to another 20 cm. The refractive index of the water and the oil are 1.33 and 1.30 respectively. Find the apparent depth of the vessel when viewed from above.
ANSWER: The depth of the vessel d = 20+20 =40 cm
The shift due to water, Δ₁ = (1-1/1.33)*20 cm
=0.33*20/1.33 cm
=20*3/(4*3) =5 cm
The shift due to oil, Δ₂ = (1-1/1.3)*20 cm
=0.3*20/1.3 cm
=4.6 cm
Total shift due to oil and water =Δ₁ + Δ₂
=5 + 4.6 =9.6 cm
Hence the apparent depth of the vessel
=d - (Δ₁ + Δ₂)
=40-9.6 cm
=30.4 cm
19. Locate the image of the point P as seen by the eye in the figure (18-E5).
Figure for Q-19
ANSWER: Total shift of the image will be the sum of shifts due to each of the slabs.
Δ = Δ₁ + Δ₂ + Δ₃
=(1-1/1.4)*0.4 + (1-1/1.3)*0.3 + (1-1/1.2)*0.2
=0.4²/1.4 + 0.3²/1.3 + 0.2²/1.2
=0.11 + 0.07 + 0.03
=0.21 cm
Hence the image of P will be 0.21 cm above P.
20. k transparent slabs are arranged one over another. The refractive indices of the slabs are µ₁, µ₂, µ₃, .... µₖ and the thicknesses are t₁, t₂, t₃, ..... tₖ. An object is seen through this combination with nearly perpendicular light. Find the equivalent refractive index of the system which will allow the image to be formed at the same place.
ANSWER: Total shifts due to k transparent slabs
Δ = Δ₁ + Δ₂ +Δ₃ + ..... + Δₖ
=(1-1/µ₁)t₁+(1-1/µ₂)t₂+(1-1/µ₃)t₃+ ..... +(1-1/µₖ)tₖ
=(t₁+t₂+t₃+ ... +tₖ)-(t₁/µ₁+t₂/µ₂+t₃/µ₃+ .... +tₖ/µₖ)
=t - (t₁/µ₁+t₂/µ₂+t₃/µ₃+ .... +tₖ/µₖ)
ₖ
Where t =Σtᵢ
ⁱ⁼¹
Let the equivalent refractive index of the system =µ
Then shift due to it will also be Δ
Here Δ = (1-1/µ)*t = t - t/µ
equating the two values of Δ
t - (t₁/µ₁+t₂/µ₂+t₃/µ₃+ .... +tₖ/µₖ) = t - t/µ
→ t₁/µ₁+t₂/µ₂+t₃/µ₃+ .... +tₖ/µₖ = t/µ
→µ = t/(t₁/µ₁+t₂/µ₂+t₃/µ₃+ .... +tₖ/µₖ)
ₖ
Σtᵢ
→µ = ⁱ⁼¹
ₖ
Σ(tᵢ/µᵢ)
ⁱ⁼¹
11. A concave mirror of radius R is kept on a horizontal table (figure 18-E2). Water (refractive index = µ) is poured into it up to a height h. Where should an object be placed so that its image is formed on itself?
ANSWER: When there is no water, an object placed at the center of curvature C will have its image at the same place. When water is put into the mirror up to a height of h, the new position of such point becomes O which we assume x above the water surface. Now the object placed at O gets its image at O itself though its apparent position will be C. (See the diagram below).
We know that,
Real depth / Apparent Depth = µ
→x/(R-h) = 1/µ (with respect to the mirror)
→x = (R-h)/µ above the water surface.
12. A point source S is placed midway between two converging mirrors having equal focal length f as shown in figure (18-E3). Find the values of d for which only one image is formed.
ANSWER: The conditions for only one image are as follows:-
Case - 1
We know that the image of the object placed at 2f from the mirror is at the same place. Since both of the mirrors have equal f if an object is placed at 2f distance from each mirror and this place is midway between the mirrors, then only one image at that point will be formed. For this
2f + 2f = d
→d = 4f
Case - 2
If the object is at an equal distance of f from each mirror, the rays after reflection from one mirror will become parallel and these parallel rays after reflection from the other mirror will form the image at f distance from it. So only one image will be formed. In this case
f + f = d
→d =2f
13. A converging mirror M₁, a point source S, and a diverging mirror M₂ are arranged as shown in figure (18-E4). The source is placed at a distance of 30 cm from M₁. The focal length of each of the mirrors is 20 cm. Consider only the images formed by a maximum of two reflections. It is found that one image is formed on the source itself.
(a) Find the distance between the two mirrors.
(b) Find the location of the image formed by the single reflection from M₂.
ANSWER: For M₁,
u = 30 cm, f = 20 cm, v =?
From the mirror formula,
1/v + 1/u = 1/f
→1/v = 1/20 - 1/30 =(3-2)/60 = 1/60
→v = 60 cm (on the same side that of object).
Let the distance between the mirrors = x. For the reflection from the mirror M₂,
u = (60-x) cm, f = 20 cm, v = -(x-30) cm. From the mirror formula
-1/(x-30) + 1/(60-x) = 1/20
→(x-30)-(60-x) =(x-30)(60-x)/20
→(2x-90)*20 = -x² -1800 + 90x
→40x - 1800 = -x² -1800 + 90x
→x² -50x = 0
→x(x-50) = 0
Since x ≠ 0,
x - 50 = 0
→x = 50 cm
For the single reflection from M₂,
u = -(x-30) cm =-(50-30) cm =-20 cm
f = 20 cm, from the mirror formula
1/v + 1/u = 1/f
→1/v - 1/20 = 1/20
→1/v = 1/20 + 1/20 = 2/20 = 1/10
→v = 10 cm
The positive sign says that the image is on the right side of the diverging mirror. So the image is 10 cm from the diverging mirror farther from the converging mirror.
14. A light ray falling at an angle of 45° with the surface of a clean slab of ice of thickness 1.00 m is refracted into it at an angle of 30°. Calculate the time taken by the light rays to cross the slab. Speed of light in vacuum = 3x10⁸ m/s.
ANSWER: Let us draw a diagram as follows.
Assume the distance traveled in the ice = x. Given the thickness of the slab d = 1 m.
Hence x= (1 m)*Sec 30° = 2/√3 m
Speed of light in vacuum = vₒ = 3x10⁸ m/s.
If the speed of light in ice be vᵢ,
Then the refractive index µ = vₒ/vᵢ
But also, µ = sin i/ sin r = sin 45°/sin 30°
=(1/√2)/(1/2) =√2, hence
vₒ/vᵢ = √2
→vᵢ = vₒ/√2 = 3x10⁸/√2 m/s =2.121x10⁸ m/s
Time taken by the light ray to cross the ice slab
=x/vᵢ =(2/√3)/2.121x10⁸ s =0.544x10⁻⁸*10⁹ ns
=5.44 ns.
15. A pole of length 1.00 m stands half dipped in a swimming pool with water level 50.0 cm higher than the bed. The refractive index of water is 1.33 and sunlight is coming at an angle of 45° with the vertical. Find the length of the shadow of the pole on the bed.
ANSWER: Let us draw a diagram as follows.
The hight of the pole AB = 1 m = 100 cm.
Depth of water BC = 50 cm.
Length of the shadow on the water surface CD
= AC*tan 45° = 50*1 cm = 50 cm.
Let the angle of refraction = r
Then sin 45°/sin r =µ
→sin r = 1/√2µ = 1/(√2*1.33) =3/4√2
We know tan r = sin r/cos r =sin r/√(1-sin² r)
→tan r = (3/4√2)/√(1-9/32) =0.75*4√2/(√23*√2)
=3/√23 =0.626
The length of shadow on the bed =BE =BF+FE
=DC + DF*tan r
=50 + 50*0.626 =81.3 cm
16. A small piece of wood is floating on the surface of a 2.5 m deep lake. Where does the shadow form on the bottom when the sun is just setting? Refractive index of water =4/3.
ANSWER: When the sun is setting, the rays are parallel to the water surface. Hence the angle of incidence i = 90°.
µ = sin 90°/sin r
→sin r = 1/µ = 3/4
If the wood is floating at A, its shadow forms at B. Distance of the shadow from the position directly below the piece of wood = BC (see the diagram above)
=AC* tan r
=2.5*sin r/√(1-sin² r)
=2.5*3/4√(1-9/16)
=7.5/√7 m
=2.83 m
17. An object P is focused by a microscope M. A glass slab of thickness 2.1 cm is introduced between P and M. If the refractive index of the slab is 1.5, by what distance should the microscope be shifted to focus the object again?
ANSWER: Thickness of the glass slab, t = 2.1 cm.
The refractive index µ = 1.5
Hence the shift for the microscope
Δt =(1-1/µ)t
=(1-1/1.5)*2.1 cm
=0.70 cm
18. A vessel contains water up to a height of 20 cm and above it an oil up to another 20 cm. The refractive index of the water and the oil are 1.33 and 1.30 respectively. Find the apparent depth of the vessel when viewed from above.
ANSWER: The depth of the vessel d = 20+20 =40 cm
The shift due to water, Δ₁ = (1-1/1.33)*20 cm
=0.33*20/1.33 cm
=20*3/(4*3) =5 cm
The shift due to oil, Δ₂ = (1-1/1.3)*20 cm
=0.3*20/1.3 cm
=4.6 cm
Total shift due to oil and water =Δ₁ + Δ₂
=5 + 4.6 =9.6 cm
Hence the apparent depth of the vessel
=d - (Δ₁ + Δ₂)
=40-9.6 cm
=30.4 cm
19. Locate the image of the point P as seen by the eye in the figure (18-E5).
ANSWER: Total shift of the image will be the sum of shifts due to each of the slabs.
Δ = Δ₁ + Δ₂ + Δ₃
=(1-1/1.4)*0.4 + (1-1/1.3)*0.3 + (1-1/1.2)*0.2
=0.4²/1.4 + 0.3²/1.3 + 0.2²/1.2
=0.11 + 0.07 + 0.03
=0.21 cm
Hence the image of P will be 0.21 cm above P.
20. k transparent slabs are arranged one over another. The refractive indices of the slabs are µ₁, µ₂, µ₃, .... µₖ and the thicknesses are t₁, t₂, t₃, ..... tₖ. An object is seen through this combination with nearly perpendicular light. Find the equivalent refractive index of the system which will allow the image to be formed at the same place.
ANSWER: Total shifts due to k transparent slabs
Δ = Δ₁ + Δ₂ +Δ₃ + ..... + Δₖ
=(1-1/µ₁)t₁+(1-1/µ₂)t₂+(1-1/µ₃)t₃+ ..... +(1-1/µₖ)tₖ
=(t₁+t₂+t₃+ ... +tₖ)-(t₁/µ₁+t₂/µ₂+t₃/µ₃+ .... +tₖ/µₖ)
=t - (t₁/µ₁+t₂/µ₂+t₃/µ₃+ .... +tₖ/µₖ)
ₖ
Where t =Σtᵢ
ⁱ⁼¹
Let the equivalent refractive index of the system =µ
Then shift due to it will also be Δ
Here Δ = (1-1/µ)*t = t - t/µ
equating the two values of Δ
t - (t₁/µ₁+t₂/µ₂+t₃/µ₃+ .... +tₖ/µₖ) = t - t/µ
→ t₁/µ₁+t₂/µ₂+t₃/µ₃+ .... +tₖ/µₖ = t/µ
→µ = t/(t₁/µ₁+t₂/µ₂+t₃/µ₃+ .... +tₖ/µₖ)
ₖ
Σtᵢ
→µ = ⁱ⁼¹
ₖ
Σ(tᵢ/µᵢ)
ⁱ⁼¹
 |
| Figure for Q-11 |
ANSWER: When there is no water, an object placed at the center of curvature C will have its image at the same place. When water is put into the mirror up to a height of h, the new position of such point becomes O which we assume x above the water surface. Now the object placed at O gets its image at O itself though its apparent position will be C. (See the diagram below).
 |
| Diagram for Q-11 |
We know that,
Real depth / Apparent Depth = µ
→x/(R-h) = 1/µ (with respect to the mirror)
→x = (R-h)/µ above the water surface.
12. A point source S is placed midway between two converging mirrors having equal focal length f as shown in figure (18-E3). Find the values of d for which only one image is formed.
 |
| Figure for Q-12 |
ANSWER: The conditions for only one image are as follows:-
Case - 1
We know that the image of the object placed at 2f from the mirror is at the same place. Since both of the mirrors have equal f if an object is placed at 2f distance from each mirror and this place is midway between the mirrors, then only one image at that point will be formed. For this
2f + 2f = d
→d = 4f
Case - 2
If the object is at an equal distance of f from each mirror, the rays after reflection from one mirror will become parallel and these parallel rays after reflection from the other mirror will form the image at f distance from it. So only one image will be formed. In this case
|
| Diagram for Q-12 |
f + f = d
→d =2f
13. A converging mirror M₁, a point source S, and a diverging mirror M₂ are arranged as shown in figure (18-E4). The source is placed at a distance of 30 cm from M₁. The focal length of each of the mirrors is 20 cm. Consider only the images formed by a maximum of two reflections. It is found that one image is formed on the source itself.
(a) Find the distance between the two mirrors.
(b) Find the location of the image formed by the single reflection from M₂.
 |
| Figure for Q-13 |
ANSWER: For M₁,
u = 30 cm, f = 20 cm, v =?
From the mirror formula,
1/v + 1/u = 1/f
→1/v = 1/20 - 1/30 =(3-2)/60 = 1/60
→v = 60 cm (on the same side that of object).
Let the distance between the mirrors = x. For the reflection from the mirror M₂,
u = (60-x) cm, f = 20 cm, v = -(x-30) cm. From the mirror formula
-1/(x-30) + 1/(60-x) = 1/20
→(x-30)-(60-x) =(x-30)(60-x)/20
→(2x-90)*20 = -x² -1800 + 90x
→40x - 1800 = -x² -1800 + 90x
→x² -50x = 0
→x(x-50) = 0
Since x ≠ 0,
x - 50 = 0
→x = 50 cm
For the single reflection from M₂,
u = -(x-30) cm =-(50-30) cm =-20 cm
f = 20 cm, from the mirror formula
1/v + 1/u = 1/f
→1/v - 1/20 = 1/20
→1/v = 1/20 + 1/20 = 2/20 = 1/10
→v = 10 cm
The positive sign says that the image is on the right side of the diverging mirror. So the image is 10 cm from the diverging mirror farther from the converging mirror.
14. A light ray falling at an angle of 45° with the surface of a clean slab of ice of thickness 1.00 m is refracted into it at an angle of 30°. Calculate the time taken by the light rays to cross the slab. Speed of light in vacuum = 3x10⁸ m/s.
ANSWER: Let us draw a diagram as follows.
 |
| Diagram for Q-14 |
Assume the distance traveled in the ice = x. Given the thickness of the slab d = 1 m.
Hence x= (1 m)*Sec 30° = 2/√3 m
Speed of light in vacuum = vₒ = 3x10⁸ m/s.
If the speed of light in ice be vᵢ,
Then the refractive index µ = vₒ/vᵢ
But also, µ = sin i/ sin r = sin 45°/sin 30°
=(1/√2)/(1/2) =√2, hence
vₒ/vᵢ = √2
→vᵢ = vₒ/√2 = 3x10⁸/√2 m/s =2.121x10⁸ m/s
Time taken by the light ray to cross the ice slab
=x/vᵢ =(2/√3)/2.121x10⁸ s =0.544x10⁻⁸*10⁹ ns
=5.44 ns.
15. A pole of length 1.00 m stands half dipped in a swimming pool with water level 50.0 cm higher than the bed. The refractive index of water is 1.33 and sunlight is coming at an angle of 45° with the vertical. Find the length of the shadow of the pole on the bed.
ANSWER: Let us draw a diagram as follows.
 |
| Diagram for Q-15 |
The hight of the pole AB = 1 m = 100 cm.
Depth of water BC = 50 cm.
Length of the shadow on the water surface CD
= AC*tan 45° = 50*1 cm = 50 cm.
Let the angle of refraction = r
Then sin 45°/sin r =µ
→sin r = 1/√2µ = 1/(√2*1.33) =3/4√2
We know tan r = sin r/cos r =sin r/√(1-sin² r)
→tan r = (3/4√2)/√(1-9/32) =0.75*4√2/(√23*√2)
=3/√23 =0.626
The length of shadow on the bed =BE =BF+FE
=DC + DF*tan r
=50 + 50*0.626 =81.3 cm
16. A small piece of wood is floating on the surface of a 2.5 m deep lake. Where does the shadow form on the bottom when the sun is just setting? Refractive index of water =4/3.
ANSWER: When the sun is setting, the rays are parallel to the water surface. Hence the angle of incidence i = 90°.
µ = sin 90°/sin r
→sin r = 1/µ = 3/4
 |
| Diagram for Q-16 |
If the wood is floating at A, its shadow forms at B. Distance of the shadow from the position directly below the piece of wood = BC (see the diagram above)
=AC* tan r
=2.5*sin r/√(1-sin² r)
=2.5*3/4√(1-9/16)
=7.5/√7 m
=2.83 m
17. An object P is focused by a microscope M. A glass slab of thickness 2.1 cm is introduced between P and M. If the refractive index of the slab is 1.5, by what distance should the microscope be shifted to focus the object again?
ANSWER: Thickness of the glass slab, t = 2.1 cm.
The refractive index µ = 1.5
Hence the shift for the microscope
Δt =(1-1/µ)t
=(1-1/1.5)*2.1 cm
=0.70 cm
18. A vessel contains water up to a height of 20 cm and above it an oil up to another 20 cm. The refractive index of the water and the oil are 1.33 and 1.30 respectively. Find the apparent depth of the vessel when viewed from above.
ANSWER: The depth of the vessel d = 20+20 =40 cm
The shift due to water, Δ₁ = (1-1/1.33)*20 cm
=0.33*20/1.33 cm
=20*3/(4*3) =5 cm
The shift due to oil, Δ₂ = (1-1/1.3)*20 cm
=0.3*20/1.3 cm
=4.6 cm
Total shift due to oil and water =Δ₁ + Δ₂
=5 + 4.6 =9.6 cm
Hence the apparent depth of the vessel
=d - (Δ₁ + Δ₂)
=40-9.6 cm
=30.4 cm
19. Locate the image of the point P as seen by the eye in the figure (18-E5).
 |
| Figure for Q-19 |
ANSWER: Total shift of the image will be the sum of shifts due to each of the slabs.
Δ = Δ₁ + Δ₂ + Δ₃
=(1-1/1.4)*0.4 + (1-1/1.3)*0.3 + (1-1/1.2)*0.2
=0.4²/1.4 + 0.3²/1.3 + 0.2²/1.2
=0.11 + 0.07 + 0.03
=0.21 cm
Hence the image of P will be 0.21 cm above P.
20. k transparent slabs are arranged one over another. The refractive indices of the slabs are µ₁, µ₂, µ₃, .... µₖ and the thicknesses are t₁, t₂, t₃, ..... tₖ. An object is seen through this combination with nearly perpendicular light. Find the equivalent refractive index of the system which will allow the image to be formed at the same place.
ANSWER: Total shifts due to k transparent slabs
Δ = Δ₁ + Δ₂ +Δ₃ + ..... + Δₖ
=(1-1/µ₁)t₁+(1-1/µ₂)t₂+(1-1/µ₃)t₃+ ..... +(1-1/µₖ)tₖ
=(t₁+t₂+t₃+ ... +tₖ)-(t₁/µ₁+t₂/µ₂+t₃/µ₃+ .... +tₖ/µₖ)
=t - (t₁/µ₁+t₂/µ₂+t₃/µ₃+ .... +tₖ/µₖ)
ₖ
Where t =Σtᵢ
ⁱ⁼¹
Let the equivalent refractive index of the system =µ
Then shift due to it will also be Δ
Here Δ = (1-1/µ)*t = t - t/µ
equating the two values of Δ
t - (t₁/µ₁+t₂/µ₂+t₃/µ₃+ .... +tₖ/µₖ) = t - t/µ
→ t₁/µ₁+t₂/µ₂+t₃/µ₃+ .... +tₖ/µₖ = t/µ
→µ = t/(t₁/µ₁+t₂/µ₂+t₃/µ₃+ .... +tₖ/µₖ)
ₖ
Σtᵢ
→µ = ⁱ⁼¹
ₖ
Σ(tᵢ/µᵢ)
ⁱ⁼¹
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CHAPTER- 14 - Fluid Mechanics
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CHAPTER- 4 - The Forces
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CHAPTER- 2 - "Vector related Problems"
CHAPTER- 2 - "Vector related Problems"
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