Question

A point object $\text{O}$ is placed in front of a concave mirror of focal length $10\text{cm}$ as shown in the figure. A glass slab of refractive index $\mu =3/2$ and thickness $6\text{cm}$ is inserted between object and mirror. Find the position of final image when the distance between mirror and slab is $5\text{cm}$.

• A. $15\text{cm}$ from the mirror
• B. $17\text{cm}$ from the mirror
• C. $12\text{cm}$ from the mirror
• D. $16\text{cm}$ from the mirror

Answer 1

The correct option is B $17\text{cm}$ from the mirror

The normal shift (s) produced by the glass slab in the direction of incident ray is,
$s=t(1-\frac{1}{\mu })$

$\Rightarrow s=6(1-\frac{1}{3/2})$

$\Rightarrow s=6(1-\frac{2}{3})=2\text{cm}$

Thus, for the mirror, object placed at distance of $32\text{cm}$ will appear at,
$u^{\prime }=32-s=32-2=30\text{cm}$
Applying mirror formula,

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

Substituting the given values we get,

$\frac{1}{v}+\frac{1}{(-30)}=\frac{1}{(-10)}$

or,

$\frac{1}{v}=\frac{-1}{10}+\frac{1}{30}=\frac{-3+1}{30}$

$\therefore v=-15\text{cm}$

when the light falls on slab after getting reflected from mirror, the slab will again cause a shift of $s=2\text{cm}$ in direction of light ray.

$\Rightarrow$ Thus, final real image is formed at,
$v^{\prime }=-(15+2)=-17\text{cm}$
i.e. $17\text{cm}$ from mirror.

Hence, option (a) is the correct answer.

Why this question? Tip: Always remember that slab will cause the shift during journey of incident ray as well as reflected ray.