Question

A double convex lens forms a real image of an object on a screen which is fixed. Now the lens is given a constant velocity $v_0$ along its axis and away from the screen. For the purpose of forming the image always on the screen, the object is also required to be given an appropriate velocity. Find the velocity of the object at the instant its size is n times the size of image.(Take $n=1/2$ and $v_0=1$m/s).

Answer 1

Given: A double convex lens forms a real image of an object on a screen which is fixed. Now the lens is given a constant velocity $v_0$ along its axis and away from the screen. For the purpose of forming the image always on the screen, the object is also required to be given an appropriate velocity.

To find the velocity of the object at the instant its size is $n$ times the size of image

Solution:

Let us take the lens to be stationary and screen moving with velocity $v$ away from the lens.

Then,

$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$

Differentiating with respect to t,we get

$0=-\frac{1}{v^2}.\frac{dv}{dt}+\frac{1}{u^2}.\frac{du}{dt}⟹\frac{du}{dt}=\frac{u^2}{v^2}.\frac{dv}{dt}⟹u^{\prime }=\frac{1}{m^2}.v^{\prime }$

Thus, the object is moving with velocity of $\frac{1}{m^2}$V with respect to the lens and towards it (i.e., towards the screen).

given, object size is n times the size of the image, $⟹m=\frac{1}{2}$

Velocity of the object with respect to the screen,

$v_{os}=v_0-\frac{v_0}{m^2}=1-\frac{1}{{\left(\frac{1}{2}\right)}^2}⟹v_{os}=1-4=-3m/s$

the negative sign indicates the object is moving towards the screen with a velocity of 3m/s.