GEOMETRICAL OPTICS
EXERCISES- Q41 to Q50
41. A biconvex thick lens is constructed with glass (µ=1.50). Each of the surfaces has a radius of 10 cm and the thickness at the middle is 5 cm. Locate the image of an object placed far away from the lens.
ANSWER: For the refraction at the first surface, the origin is at the intersection of the axis with the surface. So,
u = -∞, R = 10 cm, µ₁ = 1, µ₂ = 1.5, v = v' =?
From, µ₂/v - µ₁/u = (µ₂ - µ₁)/R
1.5/v' - 1/(-∞) = (1.5 - 1)/10 =0.5/10 =1/20
→1.5/v' = 1/20
→v' = 20*1.5 =30 cm
Diagram for Q-41
For the refraction at the farther surface, the origin is at the intersection of the axis with this surface. So, here object is the image for the first surface which distance from the new origin is lesser by the thickness of the lens. It is the virtual object.
u = v' = 30 - 5 = 25 cm. R = -10 cm, µ₁ = 1.5, µ₂ = 1.0, v =?
From, µ₂/v - µ₁/u = (µ₂ - µ₁)/R
1/v - 1.5/25 = -(1-1.5)/10 = 1/20
→1/v = 1/20 + 3/50 = (5+6)/100 =11/100
→v = 100/11 =9.1 cm.
It is on the positive side. Hence the final image is 9.1 cm from the farther surface on the other side
42. A narrow pencil of parallel light is incident normally on a solid transparent sphere of radius r. What should be the refractive index if the pencil is to be focussed (a) at the surface of the sphere, (b) at the center of the sphere?
ANSWER: (a) Let the radius of the sphere = r. Here u = -∞, v = 2r, R = r, µ₁ = 1, µ₂ = ? From,
µ₂/v - µ₁/u = (µ₂ - µ₁)/R
→µ₂/2r - 1/(-∞) = (µ₂-1)/r
→µ₂/2 = µ₂-1
→µ₂ = 2µ₂-2
→µ₂ = 2
(b) For the image at the center of the sphere,
v = r. Now the equation is
µ₂/r -1/(-∞) = (µ₂ -1)/r
→µ₂ = µ₂ - 1
It is not possible.
But, if µ₂ is large so that µ₂ ≈ µ₂-1, then the pencil will be focused close to the center.
43. One end of a cylindrical glass rod (µ=1.5) of radius 1.0 cm is rounded in the shape of a hemisphere. The rod is immersed in water (µ=4/3) and an object is placed in the water along the axis of the rod at a distance of 8.0 cm from the rounded edge. Locate the image of the object.
ANSWER: Let us see the rod horizontally. Assume the object to the left of the rod. Here, u = -8.0 cm, R=1.0 cm, µ₁ = 4/3, µ₂ = 1.5, v = ?
µ₂/v + µ₁/8 =(1.5-1.33)/1
→1.5/v +4/24 =0.17
→1.5/v = 0.17-1/6 =0.17 - 0.17 = 0
→v = ∞
So the image of the object is at the infinity.
44. A paperweight in the form of a hemisphere of radius 3.0 cm is used to hold down a printed page. An observer looks at the page vertically through the paperweight. At what height above the page will the printed letters near the center appear to the observer?
ANSWER: The first refraction will be on the plane surface for which, u = 0, R =∞, µ₁ =1, µ₂ = 1.5, v = v' =?
From µ₂/v - µ₁/u = (µ₂ - µ₁)/R
→1.5/v' - 1/0 =(1.5-1)/3
→1.5/v'-∞=1/6
→1.5/v' = ∞
→v' =0
So the image is on the flat surface near the object.
For the refraction at the curved surface, this image is the object. Taking the origin at the intersection of the vertical line through the object and the curved surface, u = - 3 cm, R = -3 cm, µ₁ = 1.5, µ₂ = 1, v =?. From µ₂/v - µ₁/u = (µ₂ - µ₁)/R
→1/v + 1.5/3 = -(1-1.5)/3
→1/v = 0.5/3 - 0.5 = (0.5 - 1.5)/3 = -1/3
→v = - 3 cm = u.
Hence no shift is observed.
45. Solve the previous problem if the paperweight is inverted at its place so that the spherical surface touches the paper.
ANSWER: For the first refraction at the spherical surface, u = 0, R = 3 cm, µ₁ = 1, µ₂ = 1.5, v =v' =?
From the µ₂/v - µ₁/u = (µ₂ - µ₁)/R
1.5/v' -1/0 =(1.5-1)/3
→1.5/v' = ∞
→v' = 0
So again the image is at the curved surface near the object. For the refraction through the flat surface, this image position is the object distance which is here 3 cm. Now the problem is like an object under the water surface. Let the image be at h distance below the water, then
h/3 = 1/µ
→h = 3/1.5 = 2 cm
Thus the letters will appear 3 cm-2 cm = 1 cm above the page.
46. A hemispherical portion of the surface of a solid glass sphere (µ=1.5)of radius r is silvered to make the inner side reflecting. An object is placed on the axis of the hemisphere at a distance 3r from the center of the sphere. The light from the object is refracted at the unsilvered part, then reflected from the silvered part and again refracted at the unsilvered part. Locate the final image formed.
ANSWER: We take the point of intersection of the axis and the unsilvered surface as the origin. For the refraction at this surface,
u = -(3r-r) = -2r, R = r, µ₁ = 1, µ₂ = 1.5, v = ?
From the µ₂/v - µ₁/u = (µ₂ - µ₁)/R
1.5/v + 1/2r = (1.5-1)/r
→1.5/v = 0.5/r -0.5/r = 0
→v = ∞
For the reflection at the mirror, this is the object distance. So u = ∞, R = -r, v =? From the mirror formula
1/v + 1/u = 2/R
1/v + 1/∞ =-2/r
→v = -r/2
So the image will be r/2 in front of the center of the silvered surface, which is the focal length of the mirror.
Diagram for Q-46
For the refraction again from the unsilvered surface, let us take the direction of the rays as along positive x-axis. So, u = -(r+r/2) = -3r/2
R = -r, µ₁ = 1.5, µ₂ = 1,
So from µ₂/v - µ₁/u = (µ₂ - µ₁)/R
1/v - 1.5/(-3r/2) = (1-1.5)/(-r)
→1/v + 1/r = 0.5/r
→1/v = 0.5/r - 1/r =-0.5/r
→v = -2r.
The negative sign says that the image is on side of the object which is inside the sphere. 2r is the diameter of the sphere. The image is diametrically opposite to the origin at the unsilvered surface i.e. at the reflecting silvered surface of the sphere.
47. The convex surface of a thin concavo-convex lens of the glass of refractive index 1.5 has a radius of curvature 20 cm. The concave surface has a radius of curvature 60 cm. The convex side is silvered and placed on a horizontal surface as shown in figure (18-E13), (a) where should a pin be placed on the axis so that its image is formed at the same place? (b) If the concave part is filled with water (µ=4/3), find the distance through which the pin should be moved so that the image of the pin again coincides with the pin.
The figure for Q-47
ANSWER: (a) If the object is placed at the center of curvature of a mirror, the image is also formed at that place. Due to the glass lens the center of curvature of the composite system changes. Let the equivalent focal length of the system = F.
The focal length of the concavo-convex lens (f) can be found out with the help of the lens maker's formula,
1/f =(µ₂/µ₁ - 1)(1/R₁-1/R₂)
Here µ₁ = 1, µ₂ = 1.5, R₁ = -60 cm, R₂ =-20 cm
→1/f = (1.5-1)(-1/60 + 1/20)
→1/f = 0.5*2/60 =1/60
→f = 60 cm
To form the final image the rays from the object passes through the lens then reflects from the mirror and again passes through the lens and come out. Thus the equivalent focal length F can be calculated as
1/F = 1/f + 1/fₘ + 1/f
(fₘ = focal length of the mirror =R/2 = 20/2 =10 cm)
→1/F = 2/f + 1/fₘ
→1/F = 2/60 + 1/10 =1/30 + 1/10
→1/F = (1+3)/30 =4/30 =2/15
→F = 15/2 =7.5 cm
So the equivalent radius of curvature of the system =2F =2*7.5 =15 cm
So the pin should be placed at 15 cm from the lens on the axis.
(b) When the concave part of the mirror is filled with water it acts as a lens. Let the focal length of this lens =fᵢ. For it R₁=∞, R₂ = -60 cm, µ₂ = 4/3, µ₁ = 1.
1/fᵢ = (4/3 -1)(1/∞ + 1/60) =(1/3)*1/60 =1/180
→fᵢ =180 cm
Now the equivalent focal length F is
1/F =2/fᵢ + 2/f +1/fₘ
→1/F =2/180 + 2/60 + 1/10
→1/F = (2+6+18)/180 =26/180
→F = 180/26 =6.93 cm
Equivalent radius of curvature R = 2F =2*6.93 =13.86 cm.
If the pin is kept at 13.86 cm from the lens the image will coincide with it. Thus the pin has to be moved 15 - 13.86 =1.14 cm towards the lens.
48. A double convex lens has a focal length of 25 cm. The radius of curvature of one of the surface is double of the other. Find the radii, if the refractive index of the material of the lens is 1.5.
ANSWER: Given, f =25 cm, R₁ = r (say), R₂ =-2r, µ₂=1.5, µ₁ = . From the lens maker's formula
1/f =(µ₂/µ₁ - 1)(1/R₁-1/R₂)
→1/25 =(1.5-1)(1/r+1/2r)
→1/25 =0.5*3/2r =3/4r
→4r = 75
→r = 75/4 =18.75 cm =R₁
And R₂ = 2*18.75 =37.5 cm
49. The radii of the curvature of the lens are +20 cm and +30 cm. The material of the lens has a refracting index of 1.6. Find the focal length of the lens (a) if it is placed in the air, and (b) if it is placed in water (µ=1.33).
ANSWER: R₁ = 20 cm, R₂ =30 cm, µ₂=1.6
(a) For air µ₁ = 1. If f = focal length of the lens, then from the lens maker's formula
1/f =(µ₂/µ₁ - 1)(1/R₁-1/R₂)
→1/f =(1.6-1)(1/20-1/30) =0.6*1/60
→f = 60/0.6 =100 cm
(b) For water µ₁ =1.33 ≈4/3
From, 1/f =(µ₂/µ₁ - 1)(1/R₁-1/R₂)
1/f = (1.6*3/4 -1)(1/20-1/30) =0.2*1/60
→f =60/0.2 =300 cm
50. Lenses are constructed by a material of refractive index 1.50. The magnitudes of the radii of curvature are 20 cm and 30 cm. Find the focal lengths of the possible lenses with the above specifications.
ANSWER: Given, µ₂ = 1.5, µ₁ = 1, R₁ = 士20 cm,
R₂ =士30 cm.
1/f =(µ₂/µ₁ - 1)(1/R₁-1/R₂)
→1/f =(1.5-1)(士1/20 士30)
→1/f = 0.5*(士3士2)/60 =(士3士2)/120
→f = 120/(士3士2)
Four focal lengths are possible.
1. f = 120/(3+2) = 120/5 =24 cm
2. f = 120/(-3-2) =-24 cm
3. f = 120/(3-2) =120 cm
4. f = 120/(-3+2) =120 cm
Hence the focal lengths of the possible lens are 士24 cm, 士120 cm.
41. A biconvex thick lens is constructed with glass (µ=1.50). Each of the surfaces has a radius of 10 cm and the thickness at the middle is 5 cm. Locate the image of an object placed far away from the lens.
ANSWER: For the refraction at the first surface, the origin is at the intersection of the axis with the surface. So,
u = -∞, R = 10 cm, µ₁ = 1, µ₂ = 1.5, v = v' =?
From, µ₂/v - µ₁/u = (µ₂ - µ₁)/R
1.5/v' - 1/(-∞) = (1.5 - 1)/10 =0.5/10 =1/20
→1.5/v' = 1/20
→v' = 20*1.5 =30 cm
For the refraction at the farther surface, the origin is at the intersection of the axis with this surface. So, here object is the image for the first surface which distance from the new origin is lesser by the thickness of the lens. It is the virtual object.
u = v' = 30 - 5 = 25 cm. R = -10 cm, µ₁ = 1.5, µ₂ = 1.0, v =?
From, µ₂/v - µ₁/u = (µ₂ - µ₁)/R
1/v - 1.5/25 = -(1-1.5)/10 = 1/20
→1/v = 1/20 + 3/50 = (5+6)/100 =11/100
→v = 100/11 =9.1 cm.
It is on the positive side. Hence the final image is 9.1 cm from the farther surface on the other side
42. A narrow pencil of parallel light is incident normally on a solid transparent sphere of radius r. What should be the refractive index if the pencil is to be focussed (a) at the surface of the sphere, (b) at the center of the sphere?
ANSWER: (a) Let the radius of the sphere = r. Here u = -∞, v = 2r, R = r, µ₁ = 1, µ₂ = ? From,
µ₂/v - µ₁/u = (µ₂ - µ₁)/R
→µ₂/2r - 1/(-∞) = (µ₂-1)/r
→µ₂/2 = µ₂-1
→µ₂ = 2µ₂-2
→µ₂ = 2
(b) For the image at the center of the sphere,
v = r. Now the equation is
µ₂/r -1/(-∞) = (µ₂ -1)/r
→µ₂ = µ₂ - 1
It is not possible.
But, if µ₂ is large so that µ₂ ≈ µ₂-1, then the pencil will be focused close to the center.
43. One end of a cylindrical glass rod (µ=1.5) of radius 1.0 cm is rounded in the shape of a hemisphere. The rod is immersed in water (µ=4/3) and an object is placed in the water along the axis of the rod at a distance of 8.0 cm from the rounded edge. Locate the image of the object.
ANSWER: Let us see the rod horizontally. Assume the object to the left of the rod. Here, u = -8.0 cm, R=1.0 cm, µ₁ = 4/3, µ₂ = 1.5, v = ?
µ₂/v + µ₁/8 =(1.5-1.33)/1
→1.5/v +4/24 =0.17
→1.5/v = 0.17-1/6 =0.17 - 0.17 = 0
→v = ∞
So the image of the object is at the infinity.
44. A paperweight in the form of a hemisphere of radius 3.0 cm is used to hold down a printed page. An observer looks at the page vertically through the paperweight. At what height above the page will the printed letters near the center appear to the observer?
ANSWER: The first refraction will be on the plane surface for which, u = 0, R =∞, µ₁ =1, µ₂ = 1.5, v = v' =?
From µ₂/v - µ₁/u = (µ₂ - µ₁)/R
→1.5/v' - 1/0 =(1.5-1)/3
→1.5/v'-∞=1/6
→1.5/v' = ∞
→v' =0
So the image is on the flat surface near the object.
For the refraction at the curved surface, this image is the object. Taking the origin at the intersection of the vertical line through the object and the curved surface, u = - 3 cm, R = -3 cm, µ₁ = 1.5, µ₂ = 1, v =?. From µ₂/v - µ₁/u = (µ₂ - µ₁)/R
→1/v + 1.5/3 = -(1-1.5)/3
→1/v = 0.5/3 - 0.5 = (0.5 - 1.5)/3 = -1/3
→v = - 3 cm = u.
Hence no shift is observed.
45. Solve the previous problem if the paperweight is inverted at its place so that the spherical surface touches the paper.
ANSWER: For the first refraction at the spherical surface, u = 0, R = 3 cm, µ₁ = 1, µ₂ = 1.5, v =v' =?
From the µ₂/v - µ₁/u = (µ₂ - µ₁)/R
1.5/v' -1/0 =(1.5-1)/3
→1.5/v' = ∞
→v' = 0
So again the image is at the curved surface near the object. For the refraction through the flat surface, this image position is the object distance which is here 3 cm. Now the problem is like an object under the water surface. Let the image be at h distance below the water, then
h/3 = 1/µ
→h = 3/1.5 = 2 cm
Thus the letters will appear 3 cm-2 cm = 1 cm above the page.
46. A hemispherical portion of the surface of a solid glass sphere (µ=1.5)of radius r is silvered to make the inner side reflecting. An object is placed on the axis of the hemisphere at a distance 3r from the center of the sphere. The light from the object is refracted at the unsilvered part, then reflected from the silvered part and again refracted at the unsilvered part. Locate the final image formed.
ANSWER: We take the point of intersection of the axis and the unsilvered surface as the origin. For the refraction at this surface,
u = -(3r-r) = -2r, R = r, µ₁ = 1, µ₂ = 1.5, v = ?
From the µ₂/v - µ₁/u = (µ₂ - µ₁)/R
1.5/v + 1/2r = (1.5-1)/r
→1.5/v = 0.5/r -0.5/r = 0
→v = ∞
For the reflection at the mirror, this is the object distance. So u = ∞, R = -r, v =? From the mirror formula
1/v + 1/u = 2/R
1/v + 1/∞ =-2/r
→v = -r/2
So the image will be r/2 in front of the center of the silvered surface, which is the focal length of the mirror.
For the refraction again from the unsilvered surface, let us take the direction of the rays as along positive x-axis. So, u = -(r+r/2) = -3r/2
R = -r, µ₁ = 1.5, µ₂ = 1,
So from µ₂/v - µ₁/u = (µ₂ - µ₁)/R
1/v - 1.5/(-3r/2) = (1-1.5)/(-r)
→1/v + 1/r = 0.5/r
→1/v = 0.5/r - 1/r =-0.5/r
→v = -2r.
The negative sign says that the image is on side of the object which is inside the sphere. 2r is the diameter of the sphere. The image is diametrically opposite to the origin at the unsilvered surface i.e. at the reflecting silvered surface of the sphere.
47. The convex surface of a thin concavo-convex lens of the glass of refractive index 1.5 has a radius of curvature 20 cm. The concave surface has a radius of curvature 60 cm. The convex side is silvered and placed on a horizontal surface as shown in figure (18-E13), (a) where should a pin be placed on the axis so that its image is formed at the same place? (b) If the concave part is filled with water (µ=4/3), find the distance through which the pin should be moved so that the image of the pin again coincides with the pin.
ANSWER: (a) If the object is placed at the center of curvature of a mirror, the image is also formed at that place. Due to the glass lens the center of curvature of the composite system changes. Let the equivalent focal length of the system = F.
The focal length of the concavo-convex lens (f) can be found out with the help of the lens maker's formula,
1/f =(µ₂/µ₁ - 1)(1/R₁-1/R₂)
Here µ₁ = 1, µ₂ = 1.5, R₁ = -60 cm, R₂ =-20 cm
→1/f = (1.5-1)(-1/60 + 1/20)
→1/f = 0.5*2/60 =1/60
→f = 60 cm
To form the final image the rays from the object passes through the lens then reflects from the mirror and again passes through the lens and come out. Thus the equivalent focal length F can be calculated as
1/F = 1/f + 1/fₘ + 1/f
(fₘ = focal length of the mirror =R/2 = 20/2 =10 cm)
→1/F = 2/f + 1/fₘ
→1/F = 2/60 + 1/10 =1/30 + 1/10
→1/F = (1+3)/30 =4/30 =2/15
→F = 15/2 =7.5 cm
So the equivalent radius of curvature of the system =2F =2*7.5 =15 cm
So the pin should be placed at 15 cm from the lens on the axis.
(b) When the concave part of the mirror is filled with water it acts as a lens. Let the focal length of this lens =fᵢ. For it R₁=∞, R₂ = -60 cm, µ₂ = 4/3, µ₁ = 1.
1/fᵢ = (4/3 -1)(1/∞ + 1/60) =(1/3)*1/60 =1/180
→fᵢ =180 cm
Now the equivalent focal length F is
1/F =2/fᵢ + 2/f +1/fₘ
→1/F =2/180 + 2/60 + 1/10
→1/F = (2+6+18)/180 =26/180
→F = 180/26 =6.93 cm
Equivalent radius of curvature R = 2F =2*6.93 =13.86 cm.
If the pin is kept at 13.86 cm from the lens the image will coincide with it. Thus the pin has to be moved 15 - 13.86 =1.14 cm towards the lens.
48. A double convex lens has a focal length of 25 cm. The radius of curvature of one of the surface is double of the other. Find the radii, if the refractive index of the material of the lens is 1.5.
ANSWER: Given, f =25 cm, R₁ = r (say), R₂ =-2r, µ₂=1.5, µ₁ = . From the lens maker's formula
1/f =(µ₂/µ₁ - 1)(1/R₁-1/R₂)
→1/25 =(1.5-1)(1/r+1/2r)
→1/25 =0.5*3/2r =3/4r
→4r = 75
→r = 75/4 =18.75 cm =R₁
And R₂ = 2*18.75 =37.5 cm
49. The radii of the curvature of the lens are +20 cm and +30 cm. The material of the lens has a refracting index of 1.6. Find the focal length of the lens (a) if it is placed in the air, and (b) if it is placed in water (µ=1.33).
ANSWER: R₁ = 20 cm, R₂ =30 cm, µ₂=1.6
(a) For air µ₁ = 1. If f = focal length of the lens, then from the lens maker's formula
1/f =(µ₂/µ₁ - 1)(1/R₁-1/R₂)
→1/f =(1.6-1)(1/20-1/30) =0.6*1/60
→f = 60/0.6 =100 cm
(b) For water µ₁ =1.33 ≈4/3
From, 1/f =(µ₂/µ₁ - 1)(1/R₁-1/R₂)
1/f = (1.6*3/4 -1)(1/20-1/30) =0.2*1/60
→f =60/0.2 =300 cm
50. Lenses are constructed by a material of refractive index 1.50. The magnitudes of the radii of curvature are 20 cm and 30 cm. Find the focal lengths of the possible lenses with the above specifications.
ANSWER: Given, µ₂ = 1.5, µ₁ = 1, R₁ = 士20 cm,
R₂ =士30 cm.
1/f =(µ₂/µ₁ - 1)(1/R₁-1/R₂)
→1/f =(1.5-1)(士1/20 士30)
→1/f = 0.5*(士3士2)/60 =(士3士2)/120
→f = 120/(士3士2)
Four focal lengths are possible.
1. f = 120/(3+2) = 120/5 =24 cm
2. f = 120/(-3-2) =-24 cm
3. f = 120/(3-2) =120 cm
4. f = 120/(-3+2) =120 cm
Hence the focal lengths of the possible lens are 士24 cm, 士120 cm.
ANSWER: For the refraction at the first surface, the origin is at the intersection of the axis with the surface. So,
u = -∞, R = 10 cm, µ₁ = 1, µ₂ = 1.5, v = v' =?
From, µ₂/v - µ₁/u = (µ₂ - µ₁)/R
1.5/v' - 1/(-∞) = (1.5 - 1)/10 =0.5/10 =1/20
→1.5/v' = 1/20
→v' = 20*1.5 =30 cm
 |
| Diagram for Q-41 |
For the refraction at the farther surface, the origin is at the intersection of the axis with this surface. So, here object is the image for the first surface which distance from the new origin is lesser by the thickness of the lens. It is the virtual object.
u = v' = 30 - 5 = 25 cm. R = -10 cm, µ₁ = 1.5, µ₂ = 1.0, v =?
From, µ₂/v - µ₁/u = (µ₂ - µ₁)/R
1/v - 1.5/25 = -(1-1.5)/10 = 1/20
→1/v = 1/20 + 3/50 = (5+6)/100 =11/100
→v = 100/11 =9.1 cm.
It is on the positive side. Hence the final image is 9.1 cm from the farther surface on the other side
42. A narrow pencil of parallel light is incident normally on a solid transparent sphere of radius r. What should be the refractive index if the pencil is to be focussed (a) at the surface of the sphere, (b) at the center of the sphere?
ANSWER: (a) Let the radius of the sphere = r. Here u = -∞, v = 2r, R = r, µ₁ = 1, µ₂ = ? From,
µ₂/v - µ₁/u = (µ₂ - µ₁)/R
→µ₂/2r - 1/(-∞) = (µ₂-1)/r
→µ₂/2 = µ₂-1
→µ₂ = 2µ₂-2
→µ₂ = 2
(b) For the image at the center of the sphere,
v = r. Now the equation is
µ₂/r -1/(-∞) = (µ₂ -1)/r
→µ₂ = µ₂ - 1
It is not possible.
But, if µ₂ is large so that µ₂ ≈ µ₂-1, then the pencil will be focused close to the center.
43. One end of a cylindrical glass rod (µ=1.5) of radius 1.0 cm is rounded in the shape of a hemisphere. The rod is immersed in water (µ=4/3) and an object is placed in the water along the axis of the rod at a distance of 8.0 cm from the rounded edge. Locate the image of the object.
ANSWER: Let us see the rod horizontally. Assume the object to the left of the rod. Here, u = -8.0 cm, R=1.0 cm, µ₁ = 4/3, µ₂ = 1.5, v = ?
µ₂/v + µ₁/8 =(1.5-1.33)/1
→1.5/v +4/24 =0.17
→1.5/v = 0.17-1/6 =0.17 - 0.17 = 0
→v = ∞
So the image of the object is at the infinity.
44. A paperweight in the form of a hemisphere of radius 3.0 cm is used to hold down a printed page. An observer looks at the page vertically through the paperweight. At what height above the page will the printed letters near the center appear to the observer?
ANSWER: The first refraction will be on the plane surface for which, u = 0, R =∞, µ₁ =1, µ₂ = 1.5, v = v' =?
From µ₂/v - µ₁/u = (µ₂ - µ₁)/R
→1.5/v' - 1/0 =(1.5-1)/3
→1.5/v'-∞=1/6
→1.5/v' = ∞
→v' =0
So the image is on the flat surface near the object.
For the refraction at the curved surface, this image is the object. Taking the origin at the intersection of the vertical line through the object and the curved surface, u = - 3 cm, R = -3 cm, µ₁ = 1.5, µ₂ = 1, v =?. From µ₂/v - µ₁/u = (µ₂ - µ₁)/R
→1/v + 1.5/3 = -(1-1.5)/3
→1/v = 0.5/3 - 0.5 = (0.5 - 1.5)/3 = -1/3
→v = - 3 cm = u.
Hence no shift is observed.
45. Solve the previous problem if the paperweight is inverted at its place so that the spherical surface touches the paper.
ANSWER: For the first refraction at the spherical surface, u = 0, R = 3 cm, µ₁ = 1, µ₂ = 1.5, v =v' =?
From the µ₂/v - µ₁/u = (µ₂ - µ₁)/R
1.5/v' -1/0 =(1.5-1)/3
→1.5/v' = ∞
→v' = 0
So again the image is at the curved surface near the object. For the refraction through the flat surface, this image position is the object distance which is here 3 cm. Now the problem is like an object under the water surface. Let the image be at h distance below the water, then
h/3 = 1/µ
→h = 3/1.5 = 2 cm
Thus the letters will appear 3 cm-2 cm = 1 cm above the page.
46. A hemispherical portion of the surface of a solid glass sphere (µ=1.5)of radius r is silvered to make the inner side reflecting. An object is placed on the axis of the hemisphere at a distance 3r from the center of the sphere. The light from the object is refracted at the unsilvered part, then reflected from the silvered part and again refracted at the unsilvered part. Locate the final image formed.
ANSWER: We take the point of intersection of the axis and the unsilvered surface as the origin. For the refraction at this surface,
u = -(3r-r) = -2r, R = r, µ₁ = 1, µ₂ = 1.5, v = ?
From the µ₂/v - µ₁/u = (µ₂ - µ₁)/R
1.5/v + 1/2r = (1.5-1)/r
→1.5/v = 0.5/r -0.5/r = 0
→v = ∞
For the reflection at the mirror, this is the object distance. So u = ∞, R = -r, v =? From the mirror formula
1/v + 1/u = 2/R
1/v + 1/∞ =-2/r
→v = -r/2
So the image will be r/2 in front of the center of the silvered surface, which is the focal length of the mirror.
 |
| Diagram for Q-46 |
For the refraction again from the unsilvered surface, let us take the direction of the rays as along positive x-axis. So, u = -(r+r/2) = -3r/2
R = -r, µ₁ = 1.5, µ₂ = 1,
So from µ₂/v - µ₁/u = (µ₂ - µ₁)/R
1/v - 1.5/(-3r/2) = (1-1.5)/(-r)
→1/v + 1/r = 0.5/r
→1/v = 0.5/r - 1/r =-0.5/r
→v = -2r.
The negative sign says that the image is on side of the object which is inside the sphere. 2r is the diameter of the sphere. The image is diametrically opposite to the origin at the unsilvered surface i.e. at the reflecting silvered surface of the sphere.
47. The convex surface of a thin concavo-convex lens of the glass of refractive index 1.5 has a radius of curvature 20 cm. The concave surface has a radius of curvature 60 cm. The convex side is silvered and placed on a horizontal surface as shown in figure (18-E13), (a) where should a pin be placed on the axis so that its image is formed at the same place? (b) If the concave part is filled with water (µ=4/3), find the distance through which the pin should be moved so that the image of the pin again coincides with the pin.
 |
| The figure for Q-47 |
ANSWER: (a) If the object is placed at the center of curvature of a mirror, the image is also formed at that place. Due to the glass lens the center of curvature of the composite system changes. Let the equivalent focal length of the system = F.
The focal length of the concavo-convex lens (f) can be found out with the help of the lens maker's formula,
1/f =(µ₂/µ₁ - 1)(1/R₁-1/R₂)
Here µ₁ = 1, µ₂ = 1.5, R₁ = -60 cm, R₂ =-20 cm
→1/f = (1.5-1)(-1/60 + 1/20)
→1/f = 0.5*2/60 =1/60
→f = 60 cm
To form the final image the rays from the object passes through the lens then reflects from the mirror and again passes through the lens and come out. Thus the equivalent focal length F can be calculated as
1/F = 1/f + 1/fₘ + 1/f
(fₘ = focal length of the mirror =R/2 = 20/2 =10 cm)
→1/F = 2/f + 1/fₘ
→1/F = 2/60 + 1/10 =1/30 + 1/10
→1/F = (1+3)/30 =4/30 =2/15
→F = 15/2 =7.5 cm
So the equivalent radius of curvature of the system =2F =2*7.5 =15 cm
So the pin should be placed at 15 cm from the lens on the axis.
(b) When the concave part of the mirror is filled with water it acts as a lens. Let the focal length of this lens =fᵢ. For it R₁=∞, R₂ = -60 cm, µ₂ = 4/3, µ₁ = 1.
1/fᵢ = (4/3 -1)(1/∞ + 1/60) =(1/3)*1/60 =1/180
→fᵢ =180 cm
Now the equivalent focal length F is
1/F =2/fᵢ + 2/f +1/fₘ
→1/F =2/180 + 2/60 + 1/10
→1/F = (2+6+18)/180 =26/180
→F = 180/26 =6.93 cm
Equivalent radius of curvature R = 2F =2*6.93 =13.86 cm.
If the pin is kept at 13.86 cm from the lens the image will coincide with it. Thus the pin has to be moved 15 - 13.86 =1.14 cm towards the lens.
48. A double convex lens has a focal length of 25 cm. The radius of curvature of one of the surface is double of the other. Find the radii, if the refractive index of the material of the lens is 1.5.
ANSWER: Given, f =25 cm, R₁ = r (say), R₂ =-2r, µ₂=1.5, µ₁ = . From the lens maker's formula
1/f =(µ₂/µ₁ - 1)(1/R₁-1/R₂)
→1/25 =(1.5-1)(1/r+1/2r)
→1/25 =0.5*3/2r =3/4r
→4r = 75
→r = 75/4 =18.75 cm =R₁
And R₂ = 2*18.75 =37.5 cm
49. The radii of the curvature of the lens are +20 cm and +30 cm. The material of the lens has a refracting index of 1.6. Find the focal length of the lens (a) if it is placed in the air, and (b) if it is placed in water (µ=1.33).
ANSWER: R₁ = 20 cm, R₂ =30 cm, µ₂=1.6
(a) For air µ₁ = 1. If f = focal length of the lens, then from the lens maker's formula
1/f =(µ₂/µ₁ - 1)(1/R₁-1/R₂)
→1/f =(1.6-1)(1/20-1/30) =0.6*1/60
→f = 60/0.6 =100 cm
(b) For water µ₁ =1.33 ≈4/3
From, 1/f =(µ₂/µ₁ - 1)(1/R₁-1/R₂)
1/f = (1.6*3/4 -1)(1/20-1/30) =0.2*1/60
→f =60/0.2 =300 cm
50. Lenses are constructed by a material of refractive index 1.50. The magnitudes of the radii of curvature are 20 cm and 30 cm. Find the focal lengths of the possible lenses with the above specifications.
ANSWER: Given, µ₂ = 1.5, µ₁ = 1, R₁ = 士20 cm,
R₂ =士30 cm.
1/f =(µ₂/µ₁ - 1)(1/R₁-1/R₂)
→1/f =(1.5-1)(士1/20 士30)
→1/f = 0.5*(士3士2)/60 =(士3士2)/120
→f = 120/(士3士2)
Four focal lengths are possible.
1. f = 120/(3+2) = 120/5 =24 cm
2. f = 120/(-3-2) =-24 cm
3. f = 120/(3-2) =120 cm
4. f = 120/(-3+2) =120 cm
Hence the focal lengths of the possible lens are 士24 cm, 士120 cm.
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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
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CHAPTER- 7 - Circular Motion
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CHAPTER- 6 - Friction
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CHAPTER- 5 - Newton's Laws of Motion
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CHAPTER- 4 - The Forces
The Forces-
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CHAPTER- 3 - Kinematics - Rest and Motion
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CHAPTER- 2 - "Vector related Problems"
CHAPTER- 2 - "Vector related Problems"
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