Question

A point object O is placed at a distance $30cm$ from a convex lens (focal length $20cm$) cut into two halves each of which is displaced by $0.05cm,$ perpendicular to the principal axis. Distance between the two images formed is

• A. $0.1cm$
• B. $0.2cm$
• C. $0.3cm$
• D. $0.4cm$

Answer 1

Answer: (C)

$0.3cm$

Using lens formula,

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

We have,

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

Given,

$u=-30cm$

$f=20cm$

The focal length of both the lens after getting halved along the principal axis remains the same as before and is equal to $20cm$.

Now,

$\frac{1}{v}=\frac{1}{20}+\frac{1}{-30}=\frac{1}{20}-\frac{1}{30}=\frac{3-2}{60}$

$\Rightarrow \frac{1}{v}=\frac{1}{60}$

$\Rightarrow v=60cm$

We know,

$m=\frac{v}{u}=\frac{60}{-30}=-2$

Also, $m=\frac{h_i}{h_o}=\frac{h_i}{-0.05}$

$\Rightarrow \frac{h_i}{-0.05}=-2$

$\Rightarrow h_i=0.1cm$

Image $I_1$ will be $0.1cm$ above Lens $L{1}^{\prime }s$ optic axis

Image $I_2$ will be $0.1cm$ below Lens $L{2}^{\prime }s$ optic axis

From the figure,

Distance between the two images = $0.1+0.1+Separationbetweenthetwolenses=0.1+0.1+0.1=0.3cm$

Hence, the correct answer is OPTION C.