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 $v=1{\text{ms}}^{–1}$ 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 double the size of the image.

• A. $3{\text{ms}}^{–1}$ away from screen
• B. $3{\text{ms}}^{–1}$ towards screen
• C. $5{\text{ms}}^{–1}$ towards screen
• D. $5{\text{ms}}^{–1}$ away from screen

Answer 1

The correct option is B $3{\text{ms}}^{–1}$ towards screen

Given, velocity of the lens $v=1{\text{ms}}^{–1}$

Let us assume that, lens to be stationary and screen is moving with velocity $v$ away from the lens.

Using lens formula,

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

On differentiating both sides, we get,

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

$\frac{du}{dt}=\frac{u^2}{v^2}\frac{dv}{dt}$
$u=\frac{1}{m^2}v$

Thus, the object is moving with velocity $v=1{\text{ms}}^{–1}$ with respect to the lens and towards it (i.e., towards the screen).

Velocity of the object with respect to the screen,

$v_{os}=v-u=v-\frac{v}{m^2}$

$\text{As,}m=\frac{1}{2},$

$\Rightarrow v_{os}=[1-\frac{1}{{\left(\frac{1}{2}\right)}^2}]=-3v$

$\therefore v_{os}=3{\text{ms}}^{-1}$ towards the screen.

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(B\right)$ is the correct answer.