GEOMETRICAL OPTICS
EXERCISES- Q51 to Q60
51. A thin lens made of a material of refractive index µ₂ has a medium of refractive index µ₁ on one side and a medium of refractive index µ₃ on the other side. The lens is biconvex and the two radii of curvature have equal magnitude R. A beam of light traveling parallel to the principal axis is incident on the lens. Where will the image be formed if the beam is incident from (a) the medium µ₁ and (b) from the medium µ₃?
ANSWER: (a) For the refraction from the first surface,
u = -∞, R = +R, Hence from
µ₂/v - µ₁/u = (µ₂ - µ₁)/R
→µ₂/v -µ₁/(-∞) = (µ₂ - µ₁)/R
→v =µ₂R/(µ₂-µ₁)
Diagram for Q-51
For the refraction from the second surface, it will be the object distance. i.e. u = µ₂R/(µ₂-µ₁), R→ -R, we shall replace µ₂ by µ₃ and µ₁ by µ₂. Hence now
µ₃/v -µ₂/u = -(µ₃-µ₂)/R
→µ₃/v =µ₂(µ₂-µ₁)/µ₂R -(µ₃-µ₂)/R
=(µ₂-µ₁-µ₃+µ₂)/R
→µ₃/v =(2µ₂-µ₁-µ₃)/R
→v = µ₃R/(2µ₂-µ₁-µ₃)
So when the beam is incident from medium µ₁, the image is formed at a distance of µ₃R/(2µ₂-µ₁-µ₃) from the lens on the other side.
(b) Similarly, it can be shown that when the beam is incident from medium µ₃, the image is formed at a distance of µ₁R/(2µ₂-µ₁-µ₃) from the lens on the other side.
52. A convex lens has a focal length of 10 cm. Find the location and nature of the image if a point object is placed on the principal axis at a distance of (a) 9.8 cm, (b) 10.2 cm from the lens.
ANSWER: From the lens formula
1/v - 1/ u = 1/f
(a) Here, u = -9.8 cm, f = 10 cm. v =?
1/v +1/9.8 =1/10
→1/v =1/10 -1/9.8 =(9.8-10)/98
→1/v = -0.2/98 =-1/490
→v = -490 cm
Negative sign shows that the image is formed at 490 cm from the lens on the side of the object.
Magnification m= v/u
=-490/(-9.8) =50.
Positive magnification indicates the image is erect. Since the rays go to the other side of the lens after refraction but the image is on the side of the object, it means the rays cross each other when produced backward. Thus the image is virtual.
(b) For u = -10.2 cm
1/v -1/(-10.2) =1/10
→1/v = 1/10 - 1/10.2 = (10.2-10)/102
→1/v = 0.2/102 = 1/510
→v = 510 cm
Positive sign shows that the image is on the other side of the object at 510 cm from the lens.
Magnification m = v/u = 510/(-10.2) = -50
Negative sign shows that the image is inverted. Since the image is on the other side of the object where refracted rays pass through, the image is real.
53. A slide projector has to project a 35 mm slide (35 mm x 23 mm) on a 2 m x 2 m screen at a distance of 10 m from the lens. What should be the focal length of the lens in the projector?
ANSWER: Here, v = 10 m,
m=v/u = h₂/h₁ =200/3.5
→1000/u = 200/3.5
(If the larger side fits on the screen then full slide will be projected, hence m = 200/3.5 )
→u = 3.5*1000/200 =17.5 cm
For the lence formula,
u = -17.5 cm, v = 1000 cm, hence
1/f =1/1000 -1/(-17.5)
→1/f = 1/1000 +10/175 = 0.001+0.057 =0.058
→f = 17.24 cm
54. A particle executes a simple harmonic motion of amplitude 1.0 cm along the principal axis of a convex lens of focal length 12 cm. The mean position of oscillation is at 20 cm from the lens. Find the amplitude of oscillation of the image of the particle.
ANSWER: For the lens, f = 12 cm. Since the mean position is at 20 cm from the lens. The extreme points will be at 19 cm and 21 cm from the lens. If the object distance be x cm then u = -x cm. Hence
1/v - 1/(-x) =1/12
→1/v =1/12 - 1/x =(x-12)/12x
→v = 12x/(x-12)
For x = 19 cm, v = 12*19/(19-12)
=12*19/7 ≈32.6 cm
For x = 21 cm, v = 12*21/(21-12) =12*21/9
=28 cm
So the image will vibrate between 28 cm to 32.57 cm. Hence the amplitude of vibration of the image
=(32.6-28)/2
=4.6/2
=2.3 cm
55. An extended object is placed at a distance of 5.0 cm from a convex lens of focal length 8.0 cm. (a) Draw the ray diagram (to the scale) to locate the image and from this, measure the distance of the image from the lens. (b) Find the position of the image from the lens formula and see how close the drawing is to the correct result.
ANSWER: (a) Let us draw the ray diagram on a graph paper. One division of the graph represents 1 cm. Locate the object position AB at 5 cm and focal length OF at 8 cm. We take two rays, one AC parallel to the principal axis which passes through the lens and goes along the focus, ray CF. Another ray AO is taken through the optical center O which goes undeviated through the lens. As we see that these two rays CF and AO diverge out and will not meet in reality. If we trace them back they cross at A'. Thus the virtual and erect image A'B' is located at B'. The measured distance of the image from the lens B'O ≈ 13.3 cm.
Diagram for Q-55(a)
(b) From the lens formula
1/v -1/u = 1/f
Here u = -5 cm, f = 8 cm. Thus
1/v + 1/5 = 1/8
→1/v =1/8-1/5 = (5-8)/40 =-3/40
→v = -40/3 =-13.33 cm
It also shows that the virtual image is formed on the side of the object at 13.33 cm from the lens which matches with the traced and measured distance on the graph paper in part (a).
56. A pin of length 2.0 cm is placed perpendicular to the principal axis of a converging lens. An inverted image of size 1.00 cm is formed at a distance of 40.0 cm from the pin. Find the focal length of the lens and its distance from the pin.
ANSWER: Let the distance of the pin from the lens = x cm. Since an inverted image is formed on the other side of the lens, the distance of the image, v = 40-x cm, u =-x cm. Also, h₁ = 2 cm, h₂ =-1.0 cm, magnification m = v/u =h₂/h₁
→(40-x)/(-x) =-1/2
→80-2x =x
→3x = 80 cm
→x = 80/3 cm =26.67 cm
So the object is at 26.67 cm from the lens.
Now, v = 40-26.67 =13.33 cm
Hence from the lens formula
1/13.33 + 1/26.67 = 1/f
→1/f =(26.67+13.33)/(13.33*26.67)
→f = 13.33*26.67/40 =8.89 cm
57. A convex lens produces a double size real image when an object is placed at a distance of 18 cm from it. Where should the object be placed to produce a triple size real image?
ANSWER: In a convex lens the real image is formed inverted and on the other side of the lens.
h₁ = h (say), h₂ = -2h
magnification, m = v/u =h₂/h₁ =-2h/h =-2
→v = -2u
Given, u = -18 cm. Hence v = -2*(-18) =36 cm
From the lens formula,
1/v -1/u =1/f
→1/36 + 1/18 = 1/f
→1/f = (2+1)/36 =3/36 =1/12
→f = 12 cm
For a triple size image,
v/u = h₂/h₁ =-3h/h =-3
→v = -3u
From the lens formula,
1/v - 1/u = 1/f
→1/(-3u) -1/u = 1/12
→u/12 = -(1+1/3) =-4/3
→u =-48/3 =-16 cmHence the object should be placed at 16 cm from the lens to get a triple size of image.
58. A pin of length 2.0 cm lies along the principal axis of a converging lens, the center being at a distance of 11 cm from the lens. The focal length of the lens is 6 cm. Find the size of the image.
ANSWER: The ends of the pin are at 11-1 =10 cm and 11+1 =12 cm. Focal length f = 6 cm. For one end of the object, u =-10 cm. Hence
1/v -1/(-10) = 1/6
→1/v =1/6 - 1/10 =(5-3)/30 =2/30 =1/15
→v = 15 cm
For the other end u = -12 cm
1/v' +1/12 =1/6
→1/v' =1/6-1/12 =1/12
→v' = 12 cm
Hence the size of the image =v-v' =15-12 =3 cm
59. The diameter of the sun is 1.4 x 10⁹ m and its distance from the earth is 1.5 x10¹¹ m. Find the radius of the image of the sun formed by a lens of focal length 20 cm.
ANSWER: For the sun image distance u =-1.5 x10¹¹ m. Its image will be formed at the focus of a converging lens, hence v = f =20 cm = 0.20 m
Magnification, m = v/u =h₂/h₁
Given h₁ =1.4 x 10⁹ m
Hence, 0.20/(-1.5 x10¹¹ ) =h₂/1.4 x 10⁹
→h₂ = -1.4 x 10⁹*0.20/1.5 x10¹¹
=-1.86x10⁻³ m =-1.86 mm
The negative sign says that the image is inverted and the diameter of the sun's image is 1.86 mm. So the radius of the image is = 1.86/2 = 0.93 cm.
60. A 5.0 diopter lens forms a virtual image which is four times the object placed perpendicularly on the principal axis of the lens. Find the distance of the object from the lens.
ANSWER: The power of the lens is given as 5.0 diopter, if the focal length is f then
→1/f = 5.0
→f =1/5 =0.20 m =20 cm
Since the power is positive, the lens is convex and the virtual image forms on the side of the object.
Also, h₁ = h, h₂ = 4h
magnification, m = v/u =h₂/h₁ =4h/h =4
→v =4u
From the lens formula,
1/v - 1/u =1/f
→1/4u - 1/u =1/20
→u/20 =(1/4 - 1) =-3/4
→u =-20*3/4 = -15 cm
So the object is at 15 cm from the lens.
51. A thin lens made of a material of refractive index µ₂ has a medium of refractive index µ₁ on one side and a medium of refractive index µ₃ on the other side. The lens is biconvex and the two radii of curvature have equal magnitude R. A beam of light traveling parallel to the principal axis is incident on the lens. Where will the image be formed if the beam is incident from (a) the medium µ₁ and (b) from the medium µ₃?
ANSWER: (a) For the refraction from the first surface,
u = -∞, R = +R, Hence from
µ₂/v - µ₁/u = (µ₂ - µ₁)/R
→µ₂/v -µ₁/(-∞) = (µ₂ - µ₁)/R
→v =µ₂R/(µ₂-µ₁)
For the refraction from the second surface, it will be the object distance. i.e. u = µ₂R/(µ₂-µ₁), R→ -R, we shall replace µ₂ by µ₃ and µ₁ by µ₂. Hence now
µ₃/v -µ₂/u = -(µ₃-µ₂)/R
→µ₃/v =µ₂(µ₂-µ₁)/µ₂R -(µ₃-µ₂)/R
=(µ₂-µ₁-µ₃+µ₂)/R
→µ₃/v =(2µ₂-µ₁-µ₃)/R
→v = µ₃R/(2µ₂-µ₁-µ₃)
So when the beam is incident from medium µ₁, the image is formed at a distance of µ₃R/(2µ₂-µ₁-µ₃) from the lens on the other side.
(b) Similarly, it can be shown that when the beam is incident from medium µ₃, the image is formed at a distance of µ₁R/(2µ₂-µ₁-µ₃) from the lens on the other side.
52. A convex lens has a focal length of 10 cm. Find the location and nature of the image if a point object is placed on the principal axis at a distance of (a) 9.8 cm, (b) 10.2 cm from the lens.
ANSWER: From the lens formula
1/v - 1/ u = 1/f
(a) Here, u = -9.8 cm, f = 10 cm. v =?
1/v +1/9.8 =1/10
→1/v =1/10 -1/9.8 =(9.8-10)/98
→1/v = -0.2/98 =-1/490
→v = -490 cm
Negative sign shows that the image is formed at 490 cm from the lens on the side of the object.
Magnification m= v/u
=-490/(-9.8) =50.
Positive magnification indicates the image is erect. Since the rays go to the other side of the lens after refraction but the image is on the side of the object, it means the rays cross each other when produced backward. Thus the image is virtual.
(b) For u = -10.2 cm
1/v -1/(-10.2) =1/10
→1/v = 1/10 - 1/10.2 = (10.2-10)/102
→1/v = 0.2/102 = 1/510
→v = 510 cm
Positive sign shows that the image is on the other side of the object at 510 cm from the lens.
Magnification m = v/u = 510/(-10.2) = -50
Negative sign shows that the image is inverted. Since the image is on the other side of the object where refracted rays pass through, the image is real.
53. A slide projector has to project a 35 mm slide (35 mm x 23 mm) on a 2 m x 2 m screen at a distance of 10 m from the lens. What should be the focal length of the lens in the projector?
ANSWER: Here, v = 10 m,
m=v/u = h₂/h₁ =200/3.5
→1000/u = 200/3.5
(If the larger side fits on the screen then full slide will be projected, hence m = 200/3.5 )
→u = 3.5*1000/200 =17.5 cm
For the lence formula,
u = -17.5 cm, v = 1000 cm, hence
1/f =1/1000 -1/(-17.5)
→1/f = 1/1000 +10/175 = 0.001+0.057 =0.058
→f = 17.24 cm
54. A particle executes a simple harmonic motion of amplitude 1.0 cm along the principal axis of a convex lens of focal length 12 cm. The mean position of oscillation is at 20 cm from the lens. Find the amplitude of oscillation of the image of the particle.
ANSWER: For the lens, f = 12 cm. Since the mean position is at 20 cm from the lens. The extreme points will be at 19 cm and 21 cm from the lens. If the object distance be x cm then u = -x cm. Hence
1/v - 1/(-x) =1/12
→1/v =1/12 - 1/x =(x-12)/12x
→v = 12x/(x-12)
For x = 19 cm, v = 12*19/(19-12)
=12*19/7 ≈32.6 cm
For x = 21 cm, v = 12*21/(21-12) =12*21/9
=28 cm
So the image will vibrate between 28 cm to 32.57 cm. Hence the amplitude of vibration of the image
=(32.6-28)/2
=4.6/2
=2.3 cm
55. An extended object is placed at a distance of 5.0 cm from a convex lens of focal length 8.0 cm. (a) Draw the ray diagram (to the scale) to locate the image and from this, measure the distance of the image from the lens. (b) Find the position of the image from the lens formula and see how close the drawing is to the correct result.
ANSWER: (a) Let us draw the ray diagram on a graph paper. One division of the graph represents 1 cm. Locate the object position AB at 5 cm and focal length OF at 8 cm. We take two rays, one AC parallel to the principal axis which passes through the lens and goes along the focus, ray CF. Another ray AO is taken through the optical center O which goes undeviated through the lens. As we see that these two rays CF and AO diverge out and will not meet in reality. If we trace them back they cross at A'. Thus the virtual and erect image A'B' is located at B'. The measured distance of the image from the lens B'O ≈ 13.3 cm.
(b) From the lens formula
1/v -1/u = 1/f
Here u = -5 cm, f = 8 cm. Thus
1/v + 1/5 = 1/8
→1/v =1/8-1/5 = (5-8)/40 =-3/40
→v = -40/3 =-13.33 cm
It also shows that the virtual image is formed on the side of the object at 13.33 cm from the lens which matches with the traced and measured distance on the graph paper in part (a).
56. A pin of length 2.0 cm is placed perpendicular to the principal axis of a converging lens. An inverted image of size 1.00 cm is formed at a distance of 40.0 cm from the pin. Find the focal length of the lens and its distance from the pin.
ANSWER: Let the distance of the pin from the lens = x cm. Since an inverted image is formed on the other side of the lens, the distance of the image, v = 40-x cm, u =-x cm. Also, h₁ = 2 cm, h₂ =-1.0 cm, magnification m = v/u =h₂/h₁
→(40-x)/(-x) =-1/2
→80-2x =x
→3x = 80 cm
→x = 80/3 cm =26.67 cm
So the object is at 26.67 cm from the lens.
Now, v = 40-26.67 =13.33 cm
Hence from the lens formula
1/13.33 + 1/26.67 = 1/f
→1/f =(26.67+13.33)/(13.33*26.67)
→f = 13.33*26.67/40 =8.89 cm
57. A convex lens produces a double size real image when an object is placed at a distance of 18 cm from it. Where should the object be placed to produce a triple size real image?
ANSWER: In a convex lens the real image is formed inverted and on the other side of the lens.
h₁ = h (say), h₂ = -2h
magnification, m = v/u =h₂/h₁ =-2h/h =-2
→v = -2u
Given, u = -18 cm. Hence v = -2*(-18) =36 cm
From the lens formula,
1/v -1/u =1/f
→1/36 + 1/18 = 1/f
→1/f = (2+1)/36 =3/36 =1/12
→f = 12 cm
For a triple size image,
v/u = h₂/h₁ =-3h/h =-3
→v = -3u
From the lens formula,
1/v - 1/u = 1/f
→1/(-3u) -1/u = 1/12
→u/12 = -(1+1/3) =-4/3
→u =-48/3 =-16 cmHence the object should be placed at 16 cm from the lens to get a triple size of image.
58. A pin of length 2.0 cm lies along the principal axis of a converging lens, the center being at a distance of 11 cm from the lens. The focal length of the lens is 6 cm. Find the size of the image.
ANSWER: The ends of the pin are at 11-1 =10 cm and 11+1 =12 cm. Focal length f = 6 cm. For one end of the object, u =-10 cm. Hence
1/v -1/(-10) = 1/6
→1/v =1/6 - 1/10 =(5-3)/30 =2/30 =1/15
→v = 15 cm
For the other end u = -12 cm
1/v' +1/12 =1/6
→1/v' =1/6-1/12 =1/12
→v' = 12 cm
Hence the size of the image =v-v' =15-12 =3 cm
59. The diameter of the sun is 1.4 x 10⁹ m and its distance from the earth is 1.5 x10¹¹ m. Find the radius of the image of the sun formed by a lens of focal length 20 cm.
ANSWER: For the sun image distance u =-1.5 x10¹¹ m. Its image will be formed at the focus of a converging lens, hence v = f =20 cm = 0.20 m
Magnification, m = v/u =h₂/h₁
Given h₁ =1.4 x 10⁹ m
Hence, 0.20/(-1.5 x10¹¹ ) =h₂/1.4 x 10⁹
→h₂ = -1.4 x 10⁹*0.20/1.5 x10¹¹
=-1.86x10⁻³ m =-1.86 mm
The negative sign says that the image is inverted and the diameter of the sun's image is 1.86 mm. So the radius of the image is = 1.86/2 = 0.93 cm.
60. A 5.0 diopter lens forms a virtual image which is four times the object placed perpendicularly on the principal axis of the lens. Find the distance of the object from the lens.
ANSWER: The power of the lens is given as 5.0 diopter, if the focal length is f then
→1/f = 5.0
→f =1/5 =0.20 m =20 cm
Since the power is positive, the lens is convex and the virtual image forms on the side of the object.
Also, h₁ = h, h₂ = 4h
magnification, m = v/u =h₂/h₁ =4h/h =4
→v =4u
From the lens formula,
1/v - 1/u =1/f
→1/4u - 1/u =1/20
→u/20 =(1/4 - 1) =-3/4
→u =-20*3/4 = -15 cm
So the object is at 15 cm from the lens.
ANSWER: (a) For the refraction from the first surface,
u = -∞, R = +R, Hence from
µ₂/v - µ₁/u = (µ₂ - µ₁)/R
→µ₂/v -µ₁/(-∞) = (µ₂ - µ₁)/R
→v =µ₂R/(µ₂-µ₁)
 |
| Diagram for Q-51 |
For the refraction from the second surface, it will be the object distance. i.e. u = µ₂R/(µ₂-µ₁), R→ -R, we shall replace µ₂ by µ₃ and µ₁ by µ₂. Hence now
µ₃/v -µ₂/u = -(µ₃-µ₂)/R
→µ₃/v =
=(µ₂-µ₁-µ₃+µ₂)/R
→µ₃/v =(2µ₂-µ₁-µ₃)/R
→v = µ₃R/(2µ₂-µ₁-µ₃)
So when the beam is incident from medium µ₁, the image is formed at a distance of µ₃R/(2µ₂-µ₁-µ₃) from the lens on the other side.
(b) Similarly, it can be shown that when the beam is incident from medium µ₃, the image is formed at a distance of µ₁R/(2µ₂-µ₁-µ₃) from the lens on the other side.
52. A convex lens has a focal length of 10 cm. Find the location and nature of the image if a point object is placed on the principal axis at a distance of (a) 9.8 cm, (b) 10.2 cm from the lens.
ANSWER: From the lens formula
1/v - 1/ u = 1/f
(a) Here, u = -9.8 cm, f = 10 cm. v =?
1/v +1/9.8 =1/10
→1/v =1/10 -1/9.8 =(9.8-10)/98
→1/v = -0.2/98 =-1/490
→v = -490 cm
Negative sign shows that the image is formed at 490 cm from the lens on the side of the object.
Magnification m= v/u
=-490/(-9.8) =50.
Positive magnification indicates the image is erect. Since the rays go to the other side of the lens after refraction but the image is on the side of the object, it means the rays cross each other when produced backward. Thus the image is virtual.
(b) For u = -10.2 cm
1/v -1/(-10.2) =1/10
→1/v = 1/10 - 1/10.2 = (10.2-10)/102
→1/v = 0.2/102 = 1/510
→v = 510 cm
Positive sign shows that the image is on the other side of the object at 510 cm from the lens.
Magnification m = v/u = 510/(-10.2) = -50
Negative sign shows that the image is inverted. Since the image is on the other side of the object where refracted rays pass through, the image is real.
53. A slide projector has to project a 35 mm slide (35 mm x 23 mm) on a 2 m x 2 m screen at a distance of 10 m from the lens. What should be the focal length of the lens in the projector?
ANSWER: Here, v = 10 m,
m=v/u = h₂/h₁ =200/3.5
→1000/u = 200/3.5
(If the larger side fits on the screen then full slide will be projected, hence m = 200/3.5 )
→u = 3.5*1000/200 =17.5 cm
For the lence formula,
u = -17.5 cm, v = 1000 cm, hence
1/f =1/1000 -1/(-17.5)
→1/f = 1/1000 +10/175 = 0.001+0.057 =0.058
→f = 17.24 cm
54. A particle executes a simple harmonic motion of amplitude 1.0 cm along the principal axis of a convex lens of focal length 12 cm. The mean position of oscillation is at 20 cm from the lens. Find the amplitude of oscillation of the image of the particle.
ANSWER: For the lens, f = 12 cm. Since the mean position is at 20 cm from the lens. The extreme points will be at 19 cm and 21 cm from the lens. If the object distance be x cm then u = -x cm. Hence
1/v - 1/(-x) =1/12
→1/v =1/12 - 1/x =(x-12)/12x
→v = 12x/(x-12)
For x = 19 cm, v = 12*19/(19-12)
=12*19/7 ≈32.6 cm
For x = 21 cm, v = 12*21/(21-12) =12*21/9
=28 cm
So the image will vibrate between 28 cm to 32.57 cm. Hence the amplitude of vibration of the image
=(32.6-28)/2
=4.6/2
=2.3 cm
55. An extended object is placed at a distance of 5.0 cm from a convex lens of focal length 8.0 cm. (a) Draw the ray diagram (to the scale) to locate the image and from this, measure the distance of the image from the lens. (b) Find the position of the image from the lens formula and see how close the drawing is to the correct result.
ANSWER: (a) Let us draw the ray diagram on a graph paper. One division of the graph represents 1 cm. Locate the object position AB at 5 cm and focal length OF at 8 cm. We take two rays, one AC parallel to the principal axis which passes through the lens and goes along the focus, ray CF. Another ray AO is taken through the optical center O which goes undeviated through the lens. As we see that these two rays CF and AO diverge out and will not meet in reality. If we trace them back they cross at A'. Thus the virtual and erect image A'B' is located at B'. The measured distance of the image from the lens B'O ≈ 13.3 cm.
 |
| Diagram for Q-55(a) |
(b) From the lens formula
1/v -1/u = 1/f
Here u = -5 cm, f = 8 cm. Thus
1/v + 1/5 = 1/8
→1/v =1/8-1/5 = (5-8)/40 =-3/40
→v = -40/3 =-13.33 cm
It also shows that the virtual image is formed on the side of the object at 13.33 cm from the lens which matches with the traced and measured distance on the graph paper in part (a).
56. A pin of length 2.0 cm is placed perpendicular to the principal axis of a converging lens. An inverted image of size 1.00 cm is formed at a distance of 40.0 cm from the pin. Find the focal length of the lens and its distance from the pin.
ANSWER: Let the distance of the pin from the lens = x cm. Since an inverted image is formed on the other side of the lens, the distance of the image, v = 40-x cm, u =-x cm. Also, h₁ = 2 cm, h₂ =-1.0 cm, magnification m = v/u =h₂/h₁
→(40-x)/(-x) =-1/2
→80-2x =x
→3x = 80 cm
→x = 80/3 cm =26.67 cm
So the object is at 26.67 cm from the lens.
Now, v = 40-26.67 =13.33 cm
Hence from the lens formula
1/13.33 + 1/26.67 = 1/f
→1/f =(26.67+13.33)/(13.33*26.67)
→f = 13.33*26.67/40 =8.89 cm
57. A convex lens produces a double size real image when an object is placed at a distance of 18 cm from it. Where should the object be placed to produce a triple size real image?
ANSWER: In a convex lens the real image is formed inverted and on the other side of the lens.
h₁ = h (say), h₂ = -2h
magnification, m = v/u =h₂/h₁ =-2h/h =-2
→v = -2u
Given, u = -18 cm. Hence v = -2*(-18) =36 cm
From the lens formula,
1/v -1/u =1/f
→1/36 + 1/18 = 1/f
→1/f = (2+1)/36 =3/36 =1/12
→f = 12 cm
For a triple size image,
v/u = h₂/h₁ =-3h/h =-3
→v = -3u
From the lens formula,
1/v - 1/u = 1/f
→1/(-3u) -1/u = 1/12
→u/12 = -(1+1/3) =-4/3
→u =-48/3 =-16 cmHence the object should be placed at 16 cm from the lens to get a triple size of image.
58. A pin of length 2.0 cm lies along the principal axis of a converging lens, the center being at a distance of 11 cm from the lens. The focal length of the lens is 6 cm. Find the size of the image.
ANSWER: The ends of the pin are at 11-1 =10 cm and 11+1 =12 cm. Focal length f = 6 cm. For one end of the object, u =-10 cm. Hence
1/v -1/(-10) = 1/6
→1/v =1/6 - 1/10 =(5-3)/30 =2/30 =1/15
→v = 15 cm
For the other end u = -12 cm
1/v' +1/12 =1/6
→1/v' =1/6-1/12 =1/12
→v' = 12 cm
Hence the size of the image =v-v' =15-12 =3 cm
59. The diameter of the sun is 1.4 x 10⁹ m and its distance from the earth is 1.5 x10¹¹ m. Find the radius of the image of the sun formed by a lens of focal length 20 cm.
ANSWER: For the sun image distance u =-1.5 x10¹¹ m. Its image will be formed at the focus of a converging lens, hence v = f =20 cm = 0.20 m
Magnification, m = v/u =h₂/h₁
Given h₁ =1.4 x 10⁹ m
Hence, 0.20/(-1.5 x10¹¹ ) =h₂/1.4 x 10⁹
→h₂ = -1.4 x 10⁹*0.20/1.5 x10¹¹
=-1.86x10⁻³ m =-1.86 mm
The negative sign says that the image is inverted and the diameter of the sun's image is 1.86 mm. So the radius of the image is = 1.86/2 = 0.93 cm.
60. A 5.0 diopter lens forms a virtual image which is four times the object placed perpendicularly on the principal axis of the lens. Find the distance of the object from the lens.
ANSWER: The power of the lens is given as 5.0 diopter, if the focal length is f then
→1/f = 5.0
→f =1/5 =0.20 m =20 cm
Since the power is positive, the lens is convex and the virtual image forms on the side of the object.
Also, h₁ = h, h₂ = 4h
magnification, m = v/u =h₂/h₁ =4h/h =4
→v =4u
From the lens formula,
1/v - 1/u =1/f
→1/4u - 1/u =1/20
→u/20 =(1/4 - 1) =-3/4
→u =-20*3/4 = -15 cm
So the object is at 15 cm from the lens.
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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
Click here for → OBJECTIVE-I
Click here for → OBJECTIVE-II
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Click here for → Question for Short Answers
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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
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CHAPTER- 6 - Friction
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CHAPTER- 6 - Friction
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Click here for → Questions for Short Answer
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Click here for → Friction - OBJECTIVE-II
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For more practice on problems on friction solve these- "New Questions on Friction".
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Click here for → EXERCISES (21-31)
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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)
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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
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CHAPTER- 3 - Kinematics - Rest and Motion
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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"
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