Question

A thin equiconvex lens has focal length $10cm$ 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

Answer: 5.00

For unsilvered lens.

$\frac{1}{f_L}=(\mu -1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$
$\therefore \frac{1}{10}=(1.5-1)\left(\frac{1}{R}-\frac{1}{-R}\right)$
$\therefore \frac{1}{10}=\frac{1}{2}\times \frac{2}{R}$
$\therefore R=10cm$

For silvered lens.
$\frac{1}{F}=-\frac{2}{f_L}+\frac{2}{R_2}$
$\therefore \frac{1}{F}=-\frac{2}{10}+\frac{2}{-10}$
$\Rightarrow F=-2.5cm$
It will behave as concave mirror of focal length $2.5cm$.
If object is placed at $2F$ then image forms at object itself.
$\therefore u=2f=2\times 2.5=5cm$