Question

A virtual erect image by a diverging lens is represented by
($u,v,f$ are coordinates)

• A.
• B.
• C.
• D.

Answer 1

The correct option is D

By the lens formula
$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$
$\frac{f}{v}-\frac{f}{u}=1$
$\frac{1}{\frac{v}{f}}=\frac{1}{\frac{u}{f}}+1$
Now comparing equation with
$\frac{1}{y}=\frac{1}{x}+1$
$\therefore y=\frac{v}{f},x=\frac{u}{f}$
So, the graph will be of the form
$\frac{1}{y}=\frac{1}{x}+1$

$y=\frac{x}{1+x}$