Question

A glass slab of thickness $3\text{cm}$ and refractive index $\left({\mu }_g\right)=\frac{3}{2}$ is placed in front of a concave mirror as shown in figure.



At what distance the final image will be formed from the mirror?

• A. $18\text{cm}$
• B. $21\text{cm}$
• C. $20\text{cm}$
• D. $15\text{cm}$

Answer 1

The correct option is B $21\text{cm}$


Due to presence of glass slab, the position of object from mirror gets shifted by,

$S=t(1-\frac{1}{\mu })$

$\Rightarrow S=3(1-\frac{1}{\left(\frac{3}{2}\right)})=1\text{cm}$

The new object distance for mirror is,
$u=21-1=20\text{cm}$

Thus, object appears to be at centre of curvature $(R=20\text{cm})$ of mirror.

$\Rightarrow$ Light ray will retrace its path and final image will be formed at object.
$\therefore v=21\text{cm}$

Hence, option (b) is the correct answer.

Why this question? Concepts: apply shift produced by a glass slab in association with reflection from mirror.