Question

A thin equiconvex lens has focal length $10\, cm$ and refractive index $1.5$. One of its faces is now silvered and for an object placed at a distance $u$ in front of it, the image coincides with the object. The value of $u$ is (in $cm$)

Answer 1

For unsilvered lens.

$\frac{1}{ f _{ L }}=(\mu-1)\left(\frac{1}{ R _{1}}-\frac{1}{ R _{2}}\right)$
$\therefore \frac{1}{10}=(1.5-1)\left(\frac{1}{ R }-\frac{1}{- R }\right)$
$\therefore \frac{1}{10}=\frac{1}{2} \times \frac{2}{ R }$
$\therefore R =10\, cm$

For silvered lens.
$\frac{1}{ F }=-\frac{2}{ f _{ L }}+\frac{2}{ R _{2}} $
$\therefore \frac{1}{ F }=-\frac{2}{10}+\frac{2}{-10} $
$\Rightarrow F =-2.5\, cm$
It will behave as concave mirror of focal length $2.5 \,cm$.
If object is placed at $2 \,F$ then image forms at object itself.
$\therefore u =2 f =2 \times 2.5=5\, cm$