Question

A point object $O$ is placed in front of a concave mirror of focal length $10cm$. A glass slab of refractive index $\mu =3/2$ and thickness $6cm$ is inserted between the object and mirror.
Find the position and nature of the final image when the distance $x$ shown in figure, is $20cm$.

• A. $17cm$, virtual
• B. $17cm$, real
• C. $12cm$, virtual
• D. $15cm$, virtual

Answer 1

Answer: (A)

$17cm$, virtual

The normal shift produced by a glass slab is,
$△x=(1-\frac{1}{\mu })t=(1-\frac{2}{3})\left(6\right)=2cm$
i.e., for the mirror, the object is placed at a distanced $(32-△x)=30cm$ from it.
Applying mirror formula,
$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$
$\frac{1}{v}+\frac{1}{30}=-\frac{1}{10}$
or, $v=-15cm$
a. When $x=5cm$: The light falls on the slab on its return journey as shown. But the slab will again shift it by a distance.
$△x=2cm$. Hence, the final real image is formed at a distance $(15+2)=17cm$ from the mirror.
b. When $x=20cm$: This time also the final image is at a distance of $17cm$ from the mirror but it is virtual as shown.