Question

Consider a coaxial system of two thin convex lenses of focal length $f$ each separated by a distance $d$. The left column of the table below gives different possible conditions between $d$ and $f$, and the right column represents the nature of the lens combination. Choose the correct alternative.

$\text{Conditions}$	$\text{Nature}$
$\left(1\right)$ $d<f$	$\left(\text{P}\right)$ $\text{Convex}$
$\left(2\right)$ $d=2f$	$\left(\text{Q}\right)$ $\text{Plane glass slab}$
$\left(3\right)$ $d>2f$	$\left(\text{R}\right)$ $\text{Concave}$

• A. $\text{1-P, 2-R, 3-Q}$
• B. $\text{1-R, 2-P, 3-Q}$
  
• C. $\text{1-P, 2-Q, 3-R}$
• D. $\text{1-Q, 2-P, 3-R}$

Answer 1

The correct option is C $\text{1-P, 2-Q, 3-R}$

For the two lenses seperated by a distance $d$, we have
$$\frac{1}{f_{eq}}=\frac{1}{f_1}+\frac{1}{f_2}-\frac{d}{f_1{f}_2}...\left(1\right)$$
Given,
$f_1=f_2=f\left(\text{say}\right)$

Substituting the values in $\left(1\right)$, we get

$\frac{1}{f_{eq}}=\frac{1}{f}+\frac{1}{f}-\frac{d}{f\times f}$

$\Rightarrow f_{eq}=\frac{f^2}{2f-d}$

This is a quadratic equation of $f$.

$\text{Condition 1}$:

When , $d<f$

So, we can conclude that , $f_{eq}<f$ and positive

So, the system will behave as a convex lens but of focal length $<f$.

$\text{Condition 2}$:

When , $d=2f$

$\Rightarrow f_{eq}$ will be infinite, the system will behave as a plane glass plate of infinite focal length.

$\text{Condition 3}$:

When , $d>2f$

$\Rightarrow f_{eq}$ will be negative. the system will behave as a concave lens.

Hence, option (c) is the correct answer.