Question

A converging lens forms a five fold magnified image of an object. The screen is moved towards the object by a distance $d=0.5\text{m}$ and the lens is shifted so that the image has the same size as the object. Find the power of lens and the initial distance between the object and the screen respectively.

• A. $6.4\text{D},11.25\text{m}$
• B. $3.2\text{D},9.25\text{m}$
• C. $6.4\text{D},22.5\text{m}$
• D. None of the above

Answer 1

The correct option is A $6.4\text{D},11.25\text{m}$

In the first case image is five times magnified. Hence, $|v|=5|u|$.

In the second case, image and object are of equal size. Hence, $|v|=|u|$.

From the two figures shown below.


$6x=2y+d$ or $6x-2y=0.5.....\left(i\right)$

Using the lens formula for both the cases,

$\frac{1}{5x}-\frac{1}{-x}=\frac{1}{f}$ or $\frac{6}{5x}=\frac{1}{f}....\left(ii\right)$

$\frac{1}{y}-\frac{1}{-y}=\frac{1}{f}$ or $\frac{2}{y}=\frac{1}{f}.....\left(iii\right)$

Solving these three equations we get,

$x=0.1875\text{m}$ and $f=0.15625\text{m}$

Therefore, initial distance between the object and the screen , $6x=1.125\text{m}$

Power of the lens, $P=\frac{1}{f}$
$\Rightarrow P=\frac{1}{0.15625}=6.4\text{D}$

Hence, option (a) is the correct answer.