Question

A glass slab of referactive index $\mu =1.5$ and thickness of 15 cm is placed in front of a convex mirror as shown in figure.


The distance between the actual position of pole and image of object $O$ as seen by the observer will be:

• A. $9\text{cm}$
• B. $25\text{cm}$
• C. $15\text{cm}$
• D. $18\text{cm}$

Answer 1

The correct option is C $15\text{cm}$

The distance of object from pole of mirror is :

$u=-(20+15+10)=-45\text{cm}$

However, due to presence of glass slab, the position of object will shift in direction of incident ray.

$S=t(1-\frac{1}{\mu })=15(1-\frac{1}{1.5})$

$\Rightarrow S=5\text{cm}$

$\therefore u=-(45-5)=-40\text{cm}$

$f=+40\text{cm}$

Applying mirror formula,

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

$\Rightarrow \frac{1}{v}+\frac{1}{-40}=\frac{1}{40}or\frac{1}{v}=\frac{2}{40}$

$\therefore v=+20\text{cm}$

i.e. image is formed at a distance $20\text{cm}$ behind mirror.

Once again during journey of reflected ray, slab will produce a shift 'S' in direction of reflected ray.

$\Rightarrow$ Image will get shifted by $S=5cm$ towards the observer.

Now image position as observed by observer is :
$v^{\prime }=+(20-5)=+15\text{cm}$
behind the actual pole of the mirror.

$\therefore$ Distance between the actual position of pole of the mirror and image as seen by the observer is $15\text{cm}$.

Hence, option (c) is the correct answer.

Why this question? Caution: When reflected rays pass through slab, shift will be produced in direction of reflected rays. Hence, image will appear to come closer by 'S' for observer.