Question

An object is placed $20cm$ to the left of a converging lens having focal length $16cm$. A second, identical lens is placed to the right of the first lens, such that the image formed by the combination is of the same size and orientation as the object is. Find the separation between the lenses.

Answer 1

Given,

Object distance, $u=-20cm$

Focal length, $f=16cm$

Using lens formula,

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

We have,

$\Rightarrow \frac{1}{16}=\frac{1}{v}-\frac{1}{-20}$

$\Rightarrow \frac{1}{v}=\frac{1}{16}-\frac{1}{20}$

$\Rightarrow \frac{1}{v}=\frac{20-16}{20\times 16}=\frac{4}{320}$

$\Rightarrow v=80cm$

Now, this image acts as object for the second lens.

Let the separation between the lens be $dcm$.

Object distance, $u=80-dcm$

Focal length, $f=16cm$

$\Rightarrow \frac{1}{16}=\frac{1}{v}-\frac{1}{80-d}$

$\Rightarrow \frac{1}{v}=\frac{1}{16}+\frac{1}{80-d}$

$\Rightarrow \frac{1}{v}=\frac{96-d}{16(80-d)}$

$\Rightarrow v=\frac{16(80-d)}{96-d}$

We know,

$m=\frac{v}{u}$

Magnification of first image, $m_1=\frac{80}{-20}=-4$

Magnification of second image, $m_2=\frac{(\frac{16(80-d)}{96-d})}{80-d}=\frac{16}{96-d}$

Given, image formed by the combination is of the same size and orientation as the original object.

$\Rightarrow m_1\times m_2=1$

$\Rightarrow -4\times \frac{16}{96-d}=1$

$\Rightarrow -64=96-d$

$\Rightarrow d=96+64=160cm$

Hence, the correct answer is $160cm$