Question

A convex lens of focal length f produces a real image of size is $\frac{1}{n}$ times the size of the object. Find the position of the object.

Answer 1

Given,

Focal length $=f$

magnification, $m=\frac{1}{n}$

We know,

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

$-\frac{1}{n}=\frac{v}{n}$

$u=-nv.....\left(1\right)$

from lens formula

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

$=\frac{1}{v}+\frac{1}{nv}$

$=\frac{nv+v}{v.nv}$

$=\frac{v(n+1)}{v\left(nv\right)}$

$\Rightarrow$ $\frac{1}{f}=\frac{f(n+1)}{n}.....\left(2\right)$

from (1) and (2)

$u=-n\frac{f(n+1)}{n}$

$u=-f(n+1)$