GEOMETRICAL OPTICS
EXERCISES- Q71 to Q79
71. A point object is placed at a distance of 15 cm from a convex lens. The image is formed on the other side at a distance of 30 cm from the lens. When a concave lens is placed in contact with the convex lens, the image shifts away further by 30 cm. Calculate the focal lengths of the two lenses.
ANSWER: Let the focal length of the convex lens = f.
u = -15 cm, v = 30 cm. From the lens formula
1/v - 1/u = 1/f
→1/f = 1/30 +1/15 =(1+2)/30 =3/30 =1/10
→f = 10 cm.
Let the combined focal length of both lenses = F. In this case, u = -15 cm, v = 30+30 = 60 cm, hence
1/F = 1/60 +1/15 =(1+4)/60 =5/60 =1/12
→F = 12 cm.
If f' is the focal length of the concave lens, then
1/f + 1/f' =1/F
→1/10 +1/f' =1/12
→1/f' =1/12-1/10 =(5-6)/60 =-1/60
→f' =-60 cm
So the focal length of the concave lens is 60 cm.
72. Two convex lenses, each of focal length 10 cm, are placed at a separation of 15 cm with their principal axes coinciding. (a) Show that a light beam coming parallel to the principal axis diverges as it comes out of the lens system. (b) Find the location of the virtual image formed by the lens system of an object placed far away. (c) Find the focal length of the equivalent lens. (Note that the sign of the focal length is positive although the lens system actually diverges a parallel beam incident on it).
ANSWER: (a) Since the parallel beams converge at the focus of the convex lens, the light beam coming parallel to the principal axis on the first lens will concentrate at a distance of 10 cm (focal length) on the other side. It will be the object for the second lens for which, u =-(15-10) =-5 cm, f =10 cm. from the lens formula
1/v - 1/(-5) = 1/10
→1/v =1/10 - 1/5 =(1-2)/10 =-1/10
→v = -10 cm
Diagram for Q-72
The Negative sign shows that the image is towards the object side at 10 cm from the lens and the image is virtual. The virtual image is formed because the rays diverge when comes out of the second lens.
(b) The virtual image is formed at 10 cm behind the second lens or 15-10 = 5 cm from the first lens towards the second lens.
(c) Let the focal length of the system = F, then
1/F =1/f + 1/f' -d/ff'
→1/F =1/10 +1/10 -15/(10*10)
→1/F =2/10 -15/100 =(20-15)/100 =5/100 =1/20
→F = 20 cm
73. A ball is kept at a height h above the surface of a heavy transparent sphere made of a material of refractive index µ. The radius of the sphere is R. At t=0, the ball is dropped to fall normally on the sphere. Find the speed of the image formed as a function of time for t<√(2h/g). Consider only the image by single refraction.
ANSWER: At time t = 0, the ball is dropped. For the given time t <√(2h/g) means when the ball height is less than h.
At any instant t, the object distance from the sphere
u = -(h-½gt²), µ₁ = 1, µ₂ = µ, for the first refraction at the spherical surface
µ₂/v - µ₁/u = (µ₂-µ₁)/R
→µ/v +1/(h-½gt²) =(µ-1)/R
→µ/v =(µ-1)/R - 1/(h-½gt²)
→µ/v ={(µ-1)(h-½gt²) - R}/R(h-½gt²)
→µ/v =(µh-½gµt²-h+½gt²)/R(h-½gt²)
→v/µ =R(h-½gt²)/{(µ-1)(h-½gt²) - R}
→v = µR(h-½gt²)/{(µ-1)(h-½gt²) - R}
This is the image distance at time t. Hence the speed of the image,
s = dv/dt
= {(-µRgt){(µ-1)(h-½gt²) - R}+(µ-1)gt*µR(h-½gt²)}/{(µ-1)(h-½gt²) - R}²
={-(µRgt)(µ-1)(h-½gt²)+µR²gt+(µRgt)(µ-1)(h-½gt²)}/{(µ-1)(h-½gt²)-R}²
=µR²gt/{(µ-1)(h-½gt²)-R}²
74. A particle is moving at a constant speed V from a large distance towards a concave mirror of radius R along the principal axis. Find the speed of the image formed by the mirror as a function of the distance x of the particle from the mirror.
ANSWER: Let the distance of the object from the mirror = x. Hence its speed V =dx/dt. For the reflection from the mirror,
u = -x, f =-R/2, v =? From the mirror formula,
1/v + 1/u = 1/f
→1/v -1/x =-2/R
→1/v =1/x - 2/R =(R-2x)/xR
→v = xR/(R-2x)
Hence the speed of the image formed, S =dv/dt
→S = d{xR/(R-2x)}/dt
→S ={(R-2x)R*dx/dt-xR*(-2dx/dt)}/(R-2x)²
→S ={R(R-2x)*V+2xRV}/(R-2x)²
→S ={R²V-2xRV+2xRV}/(R-2x)²
→S =R²V/(2x-R)²
75. A small block of mass m and a concave mirror of radius R fitted with a stand, lie on a smooth horizontal table with a separation d between them. the mirror together with its stand has a mass m. The block is pushed at t=0 towards the mirror so that it starts moving towards the mirror at a constant speed V and collides with it. The collision is perfectly elastic. Find the velocity of the image (a) at a time t< d/V, (b) at a time t>d/V.
ANSWER: (a) Let the object distance from the mirror =x. At any time t, the distance traveled = Vt. hence
x = d-Vt.
For the time t < d/V, the block is approaching the mirror before impact. As in the previous problem, the speed of the image S =R²V/(2x-R)²
→S =R²V/{2(d-Vt)-R}²
(b) For a time t>d/V, the movements are after the collision. Since both, the block and the mirror have the same mass and the collision is perfectly elastic the block will come to rest and the mirror will move with a constant speed V.
Now the distance of the object with respect to the mirror, u = -(Vt-d) =d-Vt, f = -R/2
From the mirror formula
1/v +1/(d-Vt) = -2/R
→1/v =-1/(d-Vt)-2/R
→1/v = -{R+2(d-Vt)}/R(d-Vt)
→v =-R(d-Vt)/{R+2(d-Vt)}
So the image distance from the mirror is
y = R(d-Vt)/{R+2(d-Vt)}.
Hence the image velocity
S =dy/dt
S=d[R(d-Vt)/{R+2(d-Vt)}]/dt
=[{R+2(d-Vt)}*(-VR)-R(d-Vt)*(-2V)}/{R+2(d-Vt)}²]
→S=[{-VR²-2VR(d-Vt)+2VR(d-Vt)}/{R+2(d-Vt)}²]
→S = -VR²/{R-2(Vt-d)}²
But this speed of the image is with respect to the mirror. Since the mirror is moving with a speed V, the speed of the image with respect to the table
=V+S
=V-VR²/{R-2(Vt-d)}²
=V[1-R²/{R-2(Vt-d)}²]
=V[1-R²/{2(Vt-d)-R}²]
76. A gun of mass M fires a bullet of mass m with a horizontal speed V. The gun is fitted with a concave mirror of focal length f facing towards the receding bullet. Find the speed of the separation of the bullet and the image just after the gun was fired.
ANSWER: Let the speed of the gun after the recoil = v.
From the law of conservation of momentum,
Mv+mV = 0, because the initial momentum of the combined gun and bullet was zero.
→v =-mV/M
So the gun and hence the mirror moves with a speed of mV/M in the opposite direction of the bullet. At any instant t after the bullet is fired, the distance of the bullet from the mirror = (V+mV/M)t=V(1+m/M)t
For the mirror with sign convention,
u =-V(1+m/M)t
f = -f
v =? From the mirror formula,
1/v -1/V(1+m/M)t =-1/f
→1/v =1/V(1+m/M)t -1/f
→1/v ={f-V(1+m/M)t}/fV(1+m/M)t
→v = fV(1+m/M)t/{f-V(1+m/M)t}
Hence at any instant t, the speed of separation of the image with respect to the mirror is
S =dv/dt
=[fV(1+m/M)*{f-V(1+m/M)t}-fV(1+m/M)t*V(1+m/M)]/{f-V(1+m/M)t}²
Since the mirror and the bullet are themselves separating with a speed V+mV/M=V(1+m/M), hence the separation of the image at any instant t,
=[fV(1+m/M)*{f-V(1+m/M)t}-fV(1+m/M)t*V(1+m/M)]/{f-V(1+m/M)t}² +V(1+m/M)
We need S just when the bullet is fired. Hence for t=0,
S₀ =f²V(1+m/M)/f² +V(1+m/M) =2V(1+m/M)
77. A mass m=50 g is dropped on a vertical spring of spring constant 500 N/m from a height h=10 cm as shown in figure (18-E14). The mass sticks to the spring and executes simple harmonic oscillations after that. A concave mirror of focal length 12 cm facing the mass is fixed with its principal axis coinciding the line of motion of the mass, its pole being at a distance of 30 cm from the free end of the spring. Find the length in which the image of the mass oscillates.
The figure for Q-77
ANSWER: Given that m = 50 g =50/1000 kg =0.05 kg, k = 500 N/m. The mean position of the simple harmonic motion will be the position at which the mass just rests on the spring. In this position the compression of the spring =z =mg/k =0.05*10/500 =0.5/500 =0.001 m =0.1 cm
Hence the distance of the mean position from the mirror = 30+0.1 =30.1 cm
Let the maximum compression of the spring due to fallen weight = x. The potential energy stored in the spring =½kx²
The initial potential energy of the mass with respect to the maximum compressed position
=mg(10/100+x) =mg(x+0.1)
The two will be equal,
½kx² =mg(x+0.1)
→250x² =0.05*10(x+0.1) =0.5x+0.05
→250x²-0.5x-0.05=0
→50x²-0.1x -0.01=0
x =[0.1土√(0.1²+4*50*0.01)}/2*50
=[0.1土1.42]/100
=1.52/100 m or -1.32/100 m
=1.52 cm or -1.32 cm
We will take the greater x =1.52 cm ≈1.5 cm
Hence the farthest position of the mass in oscillation with respect to the mirror will be 30+1.5 =31.5 cm. The maximum amplitude of the vibration =31.5-30.1
= 1.4 cm.
Diagram for Q-77
The nearest point of the oscillating mass from the mirror =31.5-2*1.4 =28.7 cm
For u = -31.5 cm and f = -12 cm, from the mirror formula,
1/v -1/31.5 =-1/12
→1/v = 1/31.5 - 1/12 =(12 -31.5)/(31.5*12)
→v =-31.5*12/19.5 =-19.38 cm
For u = -28.7 cm
1/v -1/28.7 = -1/12
→1/v =1/28.7 -1/12 =(12-28.7)/(28.7*12)
→v = -28.7*12/16.7 =-20.62 cm
Hence the length in which the image of the mass oscillates =20.62-19.38 =1.24 cm
78. Two concave mirrors of equal radii of curvature R are fixed on a stand facing opposite directions. The whole system has a mass m and is kept on a frictionless horizontal table (figure 18-E15).
Two blocks A and B, each of mass m, are placed on the two sides of the stand. At t=0, the separation between A and the mirrors is 2R and also the separation between B and the mirrors is 2R. Block B moves towards the mirror at a speed v. All collisions that take place are elastic. Taking the original position of the mirror-stand system to be x=0 and x-axis along AB, find the position of the images of A and B at t=
(a) R/v (b) 3R/v (c) 5R/v.
The figure for Q-78
ANSWER: (a) At time t =R/v, the position of the block B =v*R/v = R. It is at the center of curvature of the mirror. In a concave mirror, the image of an object at the center of curvature is formed at the same position. It can also be found out by the mirror formula. u =-R, f =-R/2
1/v-1/R =-2/R
→1/v =1/R-2/R =-1/R
→v = -R. So the image of B is at the center of curvature of the mirror.
At time t =R/v, block A is at 2R distance from the mirror. Here u =-2R. Hence
1/v -1/2R =-2/R
→1/v = 1/2R -2/R =(1-4)/2R =-3/2R
→v =-2R/3.
So the image of block A is at x= -2R/3 distance from the mirror and that of the block B at x =R
(b) For the t =3R/v, the distance of the block B should be =v*3R/v =3R. But it is at 2R distance from the mirror initially. Hence B collides with the mirror. Since the collision is elastic and the masses of the block and the mirror are the same, hence after the collision B stops and the mirror moves with a constant speed v. So at t =3R/v, the blocks are at 2R distance apart and the mirror is exactly at midway. i.e. Block A is at x =-2R, mirror at x =-R and block B at x=0. So for each block, u = -R. Hence their images form at the respective blocks itself. The image of block A is at x = -2R and the image of the block B at x=0.
(c) For the time t = 5R/v. Since block B hits the mirror at time t =2R/v and stands at rest at that position i.e. at x = 0. Now the mirror moves with a speed v and at t=4R/v hits the block A. Equal mass and elastic collision makes the mirror stand at x=-2R and moves the block A with speed v. In the next 5R/v-4R/v =R/v time the block A moves v*R/v =R distance from the mirror and its position is x =-3R.
Diagram for Q-78
Image of block A
Block A is at R distance from the mirror, hence again its image will be at itself i.e. at x = -3R.
Image of the block B
The block B is at 2R distance from the mirror. Hence u=-2R, f = -R/2. Hence
1/v -1/2R =-2/R
→1/v =1/2R - 2/R =(1-4)/2R =-3/2R
→v =-2R/3
So the image of the block B is at 2R/3 distance from the mirror on the side of the block i.e. at
x =-2R-(-2R/3) =-2R+2R/3 =(-6R+2R)/3 =-4R/3.
79. Consider the situation shown in the figure (18-E16). The elevator is going up with an acceleration of 2.00 m/s² and the focal length of the mirror is 12.0 cm. All the surfaces are smooth and the pulley is light. The mass-pulley system is released from rest (with respect to the elevator) at t= 0 when the distance of B from the mirror is 42.0 cm. Find the distance between the image of the block B and the mirror at t = 0.200 s. Take g = 10 m/s².
The figure for Q-79
ANSWER: The acceleration of the elevator will have no effect on the horizontal movement of block A. The acceleration of the elevator will be added to the acceleration due to gravity for block B. If T is tension in the string, then for A
T = ma'. {a' = acceleration of the blocks.}
For block B,
m(g+a) -T =ma'
→m(g+a) =T+ma' =ma'+ma' =2ma'
→a' =m(g+a)/2m =(g+a)/2 =(10+2)/2 =6 m/s²
In time t = 0.200 s, the block B will fall by
=½*6*(0.20)² =3*0.04 m =0.12 m =12 cm.
So at t =0.200 s, the distance of the block B from the mirror is =42-12 =30 cm. It is a convex lens, hence
f =12 cm, u =-30 cm, hence from the mirror formula,
1/v - 1/30 =1/12
→1/v =1/30 + 1/12 =(2+5)/60 =7/60
→v =60/7 = 8.57 cm
The positive sign shows that the image is on the other side of the object.
Hence at t =0.200 s, the image of the block B is at 8.57 cm from the mirror.
71. A point object is placed at a distance of 15 cm from a convex lens. The image is formed on the other side at a distance of 30 cm from the lens. When a concave lens is placed in contact with the convex lens, the image shifts away further by 30 cm. Calculate the focal lengths of the two lenses.
ANSWER: Let the focal length of the convex lens = f.
u = -15 cm, v = 30 cm. From the lens formula
1/v - 1/u = 1/f
→1/f = 1/30 +1/15 =(1+2)/30 =3/30 =1/10
→f = 10 cm.
Let the combined focal length of both lenses = F. In this case, u = -15 cm, v = 30+30 = 60 cm, hence
1/F = 1/60 +1/15 =(1+4)/60 =5/60 =1/12
→F = 12 cm.
If f' is the focal length of the concave lens, then
1/f + 1/f' =1/F
→1/10 +1/f' =1/12
→1/f' =1/12-1/10 =(5-6)/60 =-1/60
→f' =-60 cm
So the focal length of the concave lens is 60 cm.
72. Two convex lenses, each of focal length 10 cm, are placed at a separation of 15 cm with their principal axes coinciding. (a) Show that a light beam coming parallel to the principal axis diverges as it comes out of the lens system. (b) Find the location of the virtual image formed by the lens system of an object placed far away. (c) Find the focal length of the equivalent lens. (Note that the sign of the focal length is positive although the lens system actually diverges a parallel beam incident on it).
ANSWER: (a) Since the parallel beams converge at the focus of the convex lens, the light beam coming parallel to the principal axis on the first lens will concentrate at a distance of 10 cm (focal length) on the other side. It will be the object for the second lens for which, u =-(15-10) =-5 cm, f =10 cm. from the lens formula
1/v - 1/(-5) = 1/10
→1/v =1/10 - 1/5 =(1-2)/10 =-1/10
→v = -10 cm
The Negative sign shows that the image is towards the object side at 10 cm from the lens and the image is virtual. The virtual image is formed because the rays diverge when comes out of the second lens.
(b) The virtual image is formed at 10 cm behind the second lens or 15-10 = 5 cm from the first lens towards the second lens.
(c) Let the focal length of the system = F, then
1/F =1/f + 1/f' -d/ff'
→1/F =1/10 +1/10 -15/(10*10)
→1/F =2/10 -15/100 =(20-15)/100 =5/100 =1/20
→F = 20 cm
73. A ball is kept at a height h above the surface of a heavy transparent sphere made of a material of refractive index µ. The radius of the sphere is R. At t=0, the ball is dropped to fall normally on the sphere. Find the speed of the image formed as a function of time for t<√(2h/g). Consider only the image by single refraction.
ANSWER: At time t = 0, the ball is dropped. For the given time t <√(2h/g) means when the ball height is less than h.
At any instant t, the object distance from the sphere
u = -(h-½gt²), µ₁ = 1, µ₂ = µ, for the first refraction at the spherical surface
µ₂/v - µ₁/u = (µ₂-µ₁)/R
→µ/v +1/(h-½gt²) =(µ-1)/R
→µ/v =(µ-1)/R - 1/(h-½gt²)
→µ/v ={(µ-1)(h-½gt²) - R}/R(h-½gt²)
→µ/v =(µh-½gµt²-h+½gt²)/R(h-½gt²)
→v/µ =R(h-½gt²)/{(µ-1)(h-½gt²) - R}
→v = µR(h-½gt²)/{(µ-1)(h-½gt²) - R}
This is the image distance at time t. Hence the speed of the image,
s = dv/dt
= {(-µRgt){(µ-1)(h-½gt²) - R}+(µ-1)gt*µR(h-½gt²)}/{(µ-1)(h-½gt²) - R}²
={-(µRgt)(µ-1)(h-½gt²)+µR²gt+(µRgt)(µ-1)(h-½gt²)}/{(µ-1)(h-½gt²)-R}²
=µR²gt/{(µ-1)(h-½gt²)-R}²
74. A particle is moving at a constant speed V from a large distance towards a concave mirror of radius R along the principal axis. Find the speed of the image formed by the mirror as a function of the distance x of the particle from the mirror.
ANSWER: Let the distance of the object from the mirror = x. Hence its speed V =dx/dt. For the reflection from the mirror,
u = -x, f =-R/2, v =? From the mirror formula,
1/v + 1/u = 1/f
→1/v -1/x =-2/R
→1/v =1/x - 2/R =(R-2x)/xR
→v = xR/(R-2x)
Hence the speed of the image formed, S =dv/dt
→S = d{xR/(R-2x)}/dt
→S ={(R-2x)R*dx/dt-xR*(-2dx/dt)}/(R-2x)²
→S ={R(R-2x)*V+2xRV}/(R-2x)²
→S ={R²V-2xRV+2xRV}/(R-2x)²
→S =R²V/(2x-R)²
75. A small block of mass m and a concave mirror of radius R fitted with a stand, lie on a smooth horizontal table with a separation d between them. the mirror together with its stand has a mass m. The block is pushed at t=0 towards the mirror so that it starts moving towards the mirror at a constant speed V and collides with it. The collision is perfectly elastic. Find the velocity of the image (a) at a time t< d/V, (b) at a time t>d/V.
ANSWER: (a) Let the object distance from the mirror =x. At any time t, the distance traveled = Vt. hence
x = d-Vt.
For the time t < d/V, the block is approaching the mirror before impact. As in the previous problem, the speed of the image S =R²V/(2x-R)²
→S =R²V/{2(d-Vt)-R}²
(b) For a time t>d/V, the movements are after the collision. Since both, the block and the mirror have the same mass and the collision is perfectly elastic the block will come to rest and the mirror will move with a constant speed V.
Now the distance of the object with respect to the mirror, u = -(Vt-d) =d-Vt, f = -R/2
From the mirror formula
1/v +1/(d-Vt) = -2/R
→1/v =-1/(d-Vt)-2/R
→1/v = -{R+2(d-Vt)}/R(d-Vt)
→v =-R(d-Vt)/{R+2(d-Vt)}
So the image distance from the mirror is
y = R(d-Vt)/{R+2(d-Vt)}.
Hence the image velocity
S =dy/dt
S=d[R(d-Vt)/{R+2(d-Vt)}]/dt
=[{R+2(d-Vt)}*(-VR)-R(d-Vt)*(-2V)}/{R+2(d-Vt)}²]
→S=[{-VR²-2VR(d-Vt)+2VR(d-Vt)}/{R+2(d-Vt)}²]
→S = -VR²/{R-2(Vt-d)}²
But this speed of the image is with respect to the mirror. Since the mirror is moving with a speed V, the speed of the image with respect to the table
=V+S
=V-VR²/{R-2(Vt-d)}²
=V[1-R²/{R-2(Vt-d)}²]
=V[1-R²/{2(Vt-d)-R}²]
76. A gun of mass M fires a bullet of mass m with a horizontal speed V. The gun is fitted with a concave mirror of focal length f facing towards the receding bullet. Find the speed of the separation of the bullet and the image just after the gun was fired.
ANSWER: Let the speed of the gun after the recoil = v.
From the law of conservation of momentum,
Mv+mV = 0, because the initial momentum of the combined gun and bullet was zero.
→v =-mV/M
So the gun and hence the mirror moves with a speed of mV/M in the opposite direction of the bullet. At any instant t after the bullet is fired, the distance of the bullet from the mirror = (V+mV/M)t=V(1+m/M)t
For the mirror with sign convention,
u =-V(1+m/M)t
f = -f
v =? From the mirror formula,
1/v -1/V(1+m/M)t =-1/f
→1/v =1/V(1+m/M)t -1/f
→1/v ={f-V(1+m/M)t}/fV(1+m/M)t
→v = fV(1+m/M)t/{f-V(1+m/M)t}
Hence at any instant t, the speed of separation of the image with respect to the mirror is
S =dv/dt
=[fV(1+m/M)*{f-V(1+m/M)t}-fV(1+m/M)t*V(1+m/M)]/{f-V(1+m/M)t}²
Since the mirror and the bullet are themselves separating with a speed V+mV/M=V(1+m/M), hence the separation of the image at any instant t,
=[fV(1+m/M)*{f-V(1+m/M)t}-fV(1+m/M)t*V(1+m/M)]/{f-V(1+m/M)t}² +V(1+m/M)
We need S just when the bullet is fired. Hence for t=0,
S₀ =f²V(1+m/M)/f² +V(1+m/M) =2V(1+m/M)
77. A mass m=50 g is dropped on a vertical spring of spring constant 500 N/m from a height h=10 cm as shown in figure (18-E14). The mass sticks to the spring and executes simple harmonic oscillations after that. A concave mirror of focal length 12 cm facing the mass is fixed with its principal axis coinciding the line of motion of the mass, its pole being at a distance of 30 cm from the free end of the spring. Find the length in which the image of the mass oscillates.
ANSWER: Given that m = 50 g =50/1000 kg =0.05 kg, k = 500 N/m. The mean position of the simple harmonic motion will be the position at which the mass just rests on the spring. In this position the compression of the spring =z =mg/k =0.05*10/500 =0.5/500 =0.001 m =0.1 cm
Hence the distance of the mean position from the mirror = 30+0.1 =30.1 cm
Let the maximum compression of the spring due to fallen weight = x. The potential energy stored in the spring =½kx²
The initial potential energy of the mass with respect to the maximum compressed position
=mg(10/100+x) =mg(x+0.1)
The two will be equal,
½kx² =mg(x+0.1)
→250x² =0.05*10(x+0.1) =0.5x+0.05
→250x²-0.5x-0.05=0
→50x²-0.1x -0.01=0
x =[0.1土√(0.1²+4*50*0.01)}/2*50
=[0.1土1.42]/100
=1.52/100 m or -1.32/100 m
=1.52 cm or -1.32 cm
We will take the greater x =1.52 cm ≈1.5 cm
Hence the farthest position of the mass in oscillation with respect to the mirror will be 30+1.5 =31.5 cm. The maximum amplitude of the vibration =31.5-30.1
= 1.4 cm.
The nearest point of the oscillating mass from the mirror =31.5-2*1.4 =28.7 cm
For u = -31.5 cm and f = -12 cm, from the mirror formula,
1/v -1/31.5 =-1/12
→1/v = 1/31.5 - 1/12 =(12 -31.5)/(31.5*12)
→v =-31.5*12/19.5 =-19.38 cm
For u = -28.7 cm
1/v -1/28.7 = -1/12
→1/v =1/28.7 -1/12 =(12-28.7)/(28.7*12)
→v = -28.7*12/16.7 =-20.62 cm
Hence the length in which the image of the mass oscillates =20.62-19.38 =1.24 cm
78. Two concave mirrors of equal radii of curvature R are fixed on a stand facing opposite directions. The whole system has a mass m and is kept on a frictionless horizontal table (figure 18-E15).
Two blocks A and B, each of mass m, are placed on the two sides of the stand. At t=0, the separation between A and the mirrors is 2R and also the separation between B and the mirrors is 2R. Block B moves towards the mirror at a speed v. All collisions that take place are elastic. Taking the original position of the mirror-stand system to be x=0 and x-axis along AB, find the position of the images of A and B at t=
(a) R/v (b) 3R/v (c) 5R/v.
ANSWER: (a) At time t =R/v, the position of the block B =v*R/v = R. It is at the center of curvature of the mirror. In a concave mirror, the image of an object at the center of curvature is formed at the same position. It can also be found out by the mirror formula. u =-R, f =-R/2
1/v-1/R =-2/R
→1/v =1/R-2/R =-1/R
→v = -R. So the image of B is at the center of curvature of the mirror.
At time t =R/v, block A is at 2R distance from the mirror. Here u =-2R. Hence
1/v -1/2R =-2/R
→1/v = 1/2R -2/R =(1-4)/2R =-3/2R
→v =-2R/3.
So the image of block A is at x= -2R/3 distance from the mirror and that of the block B at x =R
(b) For the t =3R/v, the distance of the block B should be =v*3R/v =3R. But it is at 2R distance from the mirror initially. Hence B collides with the mirror. Since the collision is elastic and the masses of the block and the mirror are the same, hence after the collision B stops and the mirror moves with a constant speed v. So at t =3R/v, the blocks are at 2R distance apart and the mirror is exactly at midway. i.e. Block A is at x =-2R, mirror at x =-R and block B at x=0. So for each block, u = -R. Hence their images form at the respective blocks itself. The image of block A is at x = -2R and the image of the block B at x=0.
(c) For the time t = 5R/v. Since block B hits the mirror at time t =2R/v and stands at rest at that position i.e. at x = 0. Now the mirror moves with a speed v and at t=4R/v hits the block A. Equal mass and elastic collision makes the mirror stand at x=-2R and moves the block A with speed v. In the next 5R/v-4R/v =R/v time the block A moves v*R/v =R distance from the mirror and its position is x =-3R.
Image of block A
Block A is at R distance from the mirror, hence again its image will be at itself i.e. at x = -3R.
Image of the block B
The block B is at 2R distance from the mirror. Hence u=-2R, f = -R/2. Hence
1/v -1/2R =-2/R
→1/v =1/2R - 2/R =(1-4)/2R =-3/2R
→v =-2R/3
So the image of the block B is at 2R/3 distance from the mirror on the side of the block i.e. at
x =-2R-(-2R/3) =-2R+2R/3 =(-6R+2R)/3 =-4R/3.
79. Consider the situation shown in the figure (18-E16). The elevator is going up with an acceleration of 2.00 m/s² and the focal length of the mirror is 12.0 cm. All the surfaces are smooth and the pulley is light. The mass-pulley system is released from rest (with respect to the elevator) at t= 0 when the distance of B from the mirror is 42.0 cm. Find the distance between the image of the block B and the mirror at t = 0.200 s. Take g = 10 m/s².
ANSWER: The acceleration of the elevator will have no effect on the horizontal movement of block A. The acceleration of the elevator will be added to the acceleration due to gravity for block B. If T is tension in the string, then for A
T = ma'. {a' = acceleration of the blocks.}
For block B,
m(g+a) -T =ma'
→m(g+a) =T+ma' =ma'+ma' =2ma'
→a' =m(g+a)/2m =(g+a)/2 =(10+2)/2 =6 m/s²
In time t = 0.200 s, the block B will fall by
=½*6*(0.20)² =3*0.04 m =0.12 m =12 cm.
So at t =0.200 s, the distance of the block B from the mirror is =42-12 =30 cm. It is a convex lens, hence
f =12 cm, u =-30 cm, hence from the mirror formula,
1/v - 1/30 =1/12
→1/v =1/30 + 1/12 =(2+5)/60 =7/60
→v =60/7 = 8.57 cm
The positive sign shows that the image is on the other side of the object.
Hence at t =0.200 s, the image of the block B is at 8.57 cm from the mirror.
ANSWER: Let the focal length of the convex lens = f.
u = -15 cm, v = 30 cm. From the lens formula
1/v - 1/u = 1/f
→1/f = 1/30 +1/15 =(1+2)/30 =3/30 =1/10
→f = 10 cm.
Let the combined focal length of both lenses = F. In this case, u = -15 cm, v = 30+30 = 60 cm, hence
1/F = 1/60 +1/15 =(1+4)/60 =5/60 =1/12
→F = 12 cm.
If f' is the focal length of the concave lens, then
1/f + 1/f' =1/F
→1/10 +1/f' =1/12
→1/f' =1/12-1/10 =(5-6)/60 =-1/60
→f' =-60 cm
So the focal length of the concave lens is 60 cm.
72. Two convex lenses, each of focal length 10 cm, are placed at a separation of 15 cm with their principal axes coinciding. (a) Show that a light beam coming parallel to the principal axis diverges as it comes out of the lens system. (b) Find the location of the virtual image formed by the lens system of an object placed far away. (c) Find the focal length of the equivalent lens. (Note that the sign of the focal length is positive although the lens system actually diverges a parallel beam incident on it).
ANSWER: (a) Since the parallel beams converge at the focus of the convex lens, the light beam coming parallel to the principal axis on the first lens will concentrate at a distance of 10 cm (focal length) on the other side. It will be the object for the second lens for which, u =-(15-10) =-5 cm, f =10 cm. from the lens formula
1/v - 1/(-5) = 1/10
→1/v =1/10 - 1/5 =(1-2)/10 =-1/10
→v = -10 cm
 |
| Diagram for Q-72 |
The Negative sign shows that the image is towards the object side at 10 cm from the lens and the image is virtual. The virtual image is formed because the rays diverge when comes out of the second lens.
(b) The virtual image is formed at 10 cm behind the second lens or 15-10 = 5 cm from the first lens towards the second lens.
(c) Let the focal length of the system = F, then
1/F =1/f + 1/f' -d/ff'
→1/F =1/10 +1/10 -15/(10*10)
→1/F =2/10 -15/100 =(20-15)/100 =5/100 =1/20
→F = 20 cm
73. A ball is kept at a height h above the surface of a heavy transparent sphere made of a material of refractive index µ. The radius of the sphere is R. At t=0, the ball is dropped to fall normally on the sphere. Find the speed of the image formed as a function of time for t<√(2h/g). Consider only the image by single refraction.
ANSWER: At time t = 0, the ball is dropped. For the given time t <√(2h/g) means when the ball height is less than h.
At any instant t, the object distance from the sphere
u = -(h-½gt²), µ₁ = 1, µ₂ = µ, for the first refraction at the spherical surface
µ₂/v - µ₁/u = (µ₂-µ₁)/R
→µ/v +1/(h-½gt²) =(µ-1)/R
→µ/v =(µ-1)/R - 1/(h-½gt²)
→µ/v ={(µ-1)(h-½gt²) - R}/R(h-½gt²)
→µ/v =(µh-½gµt²-h+½gt²)/R(h-½gt²)
→v/µ =R(h-½gt²)/{(µ-1)(h-½gt²) - R}
→v = µR(h-½gt²)/{(µ-1)(h-½gt²) - R}
This is the image distance at time t. Hence the speed of the image,
s = dv/dt
= {(-µRgt){(µ-1)(h-½gt²) - R}+(µ-1)gt*µR(h-½gt²)}/{(µ-1)(h-½gt²) - R}²
={-(µRgt)(µ-1)(h-½gt²)+µR²gt+(µRgt)(µ-1)(h-½gt²)}/{(µ-1)(h-½gt²)-R}²
=µR²gt/{(µ-1)(h-½gt²)-R}²
74. A particle is moving at a constant speed V from a large distance towards a concave mirror of radius R along the principal axis. Find the speed of the image formed by the mirror as a function of the distance x of the particle from the mirror.
ANSWER: Let the distance of the object from the mirror = x. Hence its speed V =dx/dt. For the reflection from the mirror,
u = -x, f =-R/2, v =? From the mirror formula,
1/v + 1/u = 1/f
→1/v -1/x =-2/R
→1/v =1/x - 2/R =(R-2x)/xR
→v = xR/(R-2x)
Hence the speed of the image formed, S =dv/dt
→S = d{xR/(R-2x)}/dt
→S ={(R-2x)R*dx/dt-xR*(-2dx/dt)}/(R-2x)²
→S ={R(R-2x)*V+2xRV}/(R-2x)²
→S ={R²V-2xRV+2xRV}/(R-2x)²
→S =R²V/(2x-R)²
75. A small block of mass m and a concave mirror of radius R fitted with a stand, lie on a smooth horizontal table with a separation d between them. the mirror together with its stand has a mass m. The block is pushed at t=0 towards the mirror so that it starts moving towards the mirror at a constant speed V and collides with it. The collision is perfectly elastic. Find the velocity of the image (a) at a time t< d/V, (b) at a time t>d/V.
ANSWER: (a) Let the object distance from the mirror =x. At any time t, the distance traveled = Vt. hence
x = d-Vt.
For the time t < d/V, the block is approaching the mirror before impact. As in the previous problem, the speed of the image S =R²V/(2x-R)²
→S =R²V/{2(d-Vt)-R}²
(b) For a time t>d/V, the movements are after the collision. Since both, the block and the mirror have the same mass and the collision is perfectly elastic the block will come to rest and the mirror will move with a constant speed V.
Now the distance of the object with respect to the mirror, u = -(Vt-d) =d-Vt, f = -R/2
From the mirror formula
1/v +1/(d-Vt) = -2/R
→1/v =-1/(d-Vt)-2/R
→1/v = -{R+2(d-Vt)}/R(d-Vt)
→v =-R(d-Vt)/{R+2(d-Vt)}
So the image distance from the mirror is
y = R(d-Vt)/{R+2(d-Vt)}.
Hence the image velocity
S =dy/dt
S=d[R(d-Vt)/{R+2(d-Vt)}]/dt
=[{R+2(d-Vt)}*(-VR)-R(d-Vt)*(-2V)}/{R+2(d-Vt)}²]
→S=[{-VR²-2VR(d-Vt)+2VR(d-Vt)}/{R+2(d-Vt)}²]
→S = -VR²/{R-2(Vt-d)}²
But this speed of the image is with respect to the mirror. Since the mirror is moving with a speed V, the speed of the image with respect to the table
=V+S
=V-VR²/{R-2(Vt-d)}²
=V[1-R²/{R-2(Vt-d)}²]
=V[1-R²/{2(Vt-d)-R}²]
76. A gun of mass M fires a bullet of mass m with a horizontal speed V. The gun is fitted with a concave mirror of focal length f facing towards the receding bullet. Find the speed of the separation of the bullet and the image just after the gun was fired.
ANSWER: Let the speed of the gun after the recoil = v.
From the law of conservation of momentum,
Mv+mV = 0, because the initial momentum of the combined gun and bullet was zero.
→v =-mV/M
So the gun and hence the mirror moves with a speed of mV/M in the opposite direction of the bullet. At any instant t after the bullet is fired, the distance of the bullet from the mirror = (V+mV/M)t=V(1+m/M)t
For the mirror with sign convention,
u =-V(1+m/M)t
f = -f
v =? From the mirror formula,
1/v -1/V(1+m/M)t =-1/f
→1/v =1/V(1+m/M)t -1/f
→1/v ={f-V(1+m/M)t}/fV(1+m/M)t
→v = fV(1+m/M)t/{f-V(1+m/M)t}
Hence at any instant t, the speed of separation of the image with respect to the mirror is
S =dv/dt
=[fV(1+m/M)*{f-V(1+m/M)t}-fV(1+m/M)t*V(1+m/M)]/{f-V(1+m/M)t}²
Since the mirror and the bullet are themselves separating with a speed V+mV/M=V(1+m/M), hence the separation of the image at any instant t,
=[fV(1+m/M)*{f-V(1+m/M)t}-fV(1+m/M)t*V(1+m/M)]/{f-V(1+m/M)t}² +V(1+m/M)
We need S just when the bullet is fired. Hence for t=0,
S₀ =f²V(1+m/M)/f² +V(1+m/M) =2V(1+m/M)
77. A mass m=50 g is dropped on a vertical spring of spring constant 500 N/m from a height h=10 cm as shown in figure (18-E14). The mass sticks to the spring and executes simple harmonic oscillations after that. A concave mirror of focal length 12 cm facing the mass is fixed with its principal axis coinciding the line of motion of the mass, its pole being at a distance of 30 cm from the free end of the spring. Find the length in which the image of the mass oscillates.
 |
| The figure for Q-77 |
ANSWER: Given that m = 50 g =50/1000 kg =0.05 kg, k = 500 N/m. The mean position of the simple harmonic motion will be the position at which the mass just rests on the spring. In this position the compression of the spring =z =mg/k =0.05*10/500 =0.5/500 =0.001 m =0.1 cm
Hence the distance of the mean position from the mirror = 30+0.1 =30.1 cm
Let the maximum compression of the spring due to fallen weight = x. The potential energy stored in the spring =½kx²
The initial potential energy of the mass with respect to the maximum compressed position
=mg(10/100+x) =mg(x+0.1)
The two will be equal,
½kx² =mg(x+0.1)
→250x² =0.05*10(x+0.1) =0.5x+0.05
→250x²-0.5x-0.05=0
→50x²-0.1x -0.01=0
x =[0.1土√(0.1²+4*50*0.01)}/2*50
=[0.1土1.42]/100
=1.52/100 m or -1.32/100 m
=1.52 cm or -1.32 cm
We will take the greater x =1.52 cm ≈1.5 cm
Hence the farthest position of the mass in oscillation with respect to the mirror will be 30+1.5 =31.5 cm. The maximum amplitude of the vibration =31.5-30.1
= 1.4 cm.
 |
| Diagram for Q-77 |
The nearest point of the oscillating mass from the mirror =31.5-2*1.4 =28.7 cm
For u = -31.5 cm and f = -12 cm, from the mirror formula,
1/v -1/31.5 =-1/12
→1/v = 1/31.5 - 1/12 =(12 -31.5)/(31.5*12)
→v =-31.5*12/19.5 =-19.38 cm
For u = -28.7 cm
1/v -1/28.7 = -1/12
→1/v =1/28.7 -1/12 =(12-28.7)/(28.7*12)
→v = -28.7*12/16.7 =-20.62 cm
Hence the length in which the image of the mass oscillates =20.62-19.38 =1.24 cm
78. Two concave mirrors of equal radii of curvature R are fixed on a stand facing opposite directions. The whole system has a mass m and is kept on a frictionless horizontal table (figure 18-E15).
Two blocks A and B, each of mass m, are placed on the two sides of the stand. At t=0, the separation between A and the mirrors is 2R and also the separation between B and the mirrors is 2R. Block B moves towards the mirror at a speed v. All collisions that take place are elastic. Taking the original position of the mirror-stand system to be x=0 and x-axis along AB, find the position of the images of A and B at t=
(a) R/v (b) 3R/v (c) 5R/v.
 |
| The figure for Q-78 |
ANSWER: (a) At time t =R/v, the position of the block B =v*R/v = R. It is at the center of curvature of the mirror. In a concave mirror, the image of an object at the center of curvature is formed at the same position. It can also be found out by the mirror formula. u =-R, f =-R/2
1/v-1/R =-2/R
→1/v =1/R-2/R =-1/R
→v = -R. So the image of B is at the center of curvature of the mirror.
At time t =R/v, block A is at 2R distance from the mirror. Here u =-2R. Hence
1/v -1/2R =-2/R
→1/v = 1/2R -2/R =(1-4)/2R =-3/2R
→v =-2R/3.
So the image of block A is at x= -2R/3 distance from the mirror and that of the block B at x =R
(b) For the t =3R/v, the distance of the block B should be =v*3R/v =3R. But it is at 2R distance from the mirror initially. Hence B collides with the mirror. Since the collision is elastic and the masses of the block and the mirror are the same, hence after the collision B stops and the mirror moves with a constant speed v. So at t =3R/v, the blocks are at 2R distance apart and the mirror is exactly at midway. i.e. Block A is at x =-2R, mirror at x =-R and block B at x=0. So for each block, u = -R. Hence their images form at the respective blocks itself. The image of block A is at x = -2R and the image of the block B at x=0.
(c) For the time t = 5R/v. Since block B hits the mirror at time t =2R/v and stands at rest at that position i.e. at x = 0. Now the mirror moves with a speed v and at t=4R/v hits the block A. Equal mass and elastic collision makes the mirror stand at x=-2R and moves the block A with speed v. In the next 5R/v-4R/v =R/v time the block A moves v*R/v =R distance from the mirror and its position is x =-3R.
 |
| Diagram for Q-78 |
Image of block A
Block A is at R distance from the mirror, hence again its image will be at itself i.e. at x = -3R.
Image of the block B
The block B is at 2R distance from the mirror. Hence u=-2R, f = -R/2. Hence
1/v -1/2R =-2/R
→1/v =1/2R - 2/R =(1-4)/2R =-3/2R
→v =-2R/3
So the image of the block B is at 2R/3 distance from the mirror on the side of the block i.e. at
x =-2R-(-2R/3) =-2R+2R/3 =(-6R+2R)/3 =-4R/3.
79. Consider the situation shown in the figure (18-E16). The elevator is going up with an acceleration of 2.00 m/s² and the focal length of the mirror is 12.0 cm. All the surfaces are smooth and the pulley is light. The mass-pulley system is released from rest (with respect to the elevator) at t= 0 when the distance of B from the mirror is 42.0 cm. Find the distance between the image of the block B and the mirror at t = 0.200 s. Take g = 10 m/s².
 |
| The figure for Q-79 |
ANSWER: The acceleration of the elevator will have no effect on the horizontal movement of block A. The acceleration of the elevator will be added to the acceleration due to gravity for block B. If T is tension in the string, then for A
T = ma'. {a' = acceleration of the blocks.}
For block B,
m(g+a) -T =ma'
→m(g+a) =T+ma' =ma'+ma' =2ma'
→a' =m(g+a)/2m =(g+a)/2 =(10+2)/2 =6 m/s²
In time t = 0.200 s, the block B will fall by
=½*6*(0.20)² =3*0.04 m =0.12 m =12 cm.
So at t =0.200 s, the distance of the block B from the mirror is =42-12 =30 cm. It is a convex lens, hence
f =12 cm, u =-30 cm, hence from the mirror formula,
1/v - 1/30 =1/12
→1/v =1/30 + 1/12 =(2+5)/60 =7/60
→v =60/7 = 8.57 cm
The positive sign shows that the image is on the other side of the object.
Hence at t =0.200 s, the image of the block B is at 8.57 cm from the mirror.
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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
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"
