Question

A convex lens forms a real image three times larger than the size object on a screen. Object and screen are moved until the image becomes twice the size of the object. If the shift of the object is $6\text{cm}$ (away from the lens), the shift of screen is :

Answer 1

Given that,
Case I: When the object is at $O$ and image is formed at $I$.
Initial magnification of lens $m=-3$
Let object is at, $u=-x$


then,
$m=\frac{v}{u}=\frac{v}{-x}=-3$
$\Rightarrow v=3x$
By using lens formula,
$\frac{1}{3x}-\frac{1}{-x}=\frac{1}{f}$
$\frac{4}{3x}=\frac{1}{f}......\left(1\right)$

Case II: When object is at $O^{\prime }$ and image is formed at $I^{\prime }$
Now when object is shifted by $6\text{cm}$ :
Object distance, $u=-(6+x)\text{cm}$
New magnification, $m^{\prime }=\frac{v}{u}=-2$
Therefore, image distance, $v=2(6+x)\text{cm}$
Applying lens formula,
$\frac{1}{2(6+x)}-\frac{1}{-(6+x)}=\frac{1}{f}$
$\frac{3}{2(6+x)}=\frac{1}{f}......\left(2\right)$

Equation $\left(1\right)$ and $\left(2\right)$ gives
$\frac{4}{3x}=\frac{3}{2(6+x)}$
$\therefore 9x=48+8x$
$x=48\text{cm}$

Shift of screen$=3x-2(6+x)$
$=3\times 48-2(6+48)$
$=36\text{cm}$