Question

A split lens has its two parts separated by distance $a$ and its focal length is $f$. An object is placed at a distance $\frac{3f}{2}$ on the axis of the undivided lens as shown. The distance between the images formed is

• A. $a$
• B. $2a$
• C. $3a$
• D. $4a$

Answer 1

The correct option is C $3a$


Focal lens of each part will be $f$.
Consider $u=-\frac{3f}{2},f=f$
Using lens formula,
$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\Rightarrow \frac{1}{v}-\frac{1}{\frac{-3f}{2}}=\frac{1}{f}$
$\Rightarrow \frac{1}{v}=\frac{1}{f}-\frac{2}{3f}=\frac{3-2}{3f}$
$\Rightarrow v=3f$

So each part will form a real image of the point object $O$ at distance $3f$ as shown.

For similar triangles $O{L}_1{L}_2$ and $O{I}_1{I}_2$,
$\frac{L_1{L}_2}{\frac{3f}{2}}=\frac{I_1{I}_2}{\frac{3f}{2}+3f}\Rightarrow \frac{a}{\frac{3f}{2}}=\frac{I_1{I}_2}{\frac{9f}{2}}$

$\Rightarrow I_1{I}_2=3a$

Alternate solution :

From the above figure we can say that the distance of the object from the principle axis of both of the parts is $\frac{a}{2}.$
Therefore, we can consider the object as an extended object of height $\frac{a}{2}$ for both of the lenses.

Now, magnification due to upper part of the lens is given by :
$m=\frac{h_i}{h_o}=\frac{v}{u}$
From the data :
$\Rightarrow \frac{h_1}{-a/2}=\frac{3f}{-3f/2}$
$\Rightarrow h_1=a$
From the symmetry we can say that :
$h_1=h_2=a$
Therefore, the distance between the images is :
$=a+h_1+h_2=a+a+a$
$=3a$

$\begin{array}{c}\text{Why this Question}?\\ \text{Understanding the similar triangles through ray diagrams}\end{array}$