Question

A flat glass slab of thickness 6 cm and refractive index 1.5 is placed in front of a plane mirror. An observer is standing behind the glass slab and looking at the mirror. The actual distance of the observer from the mirror is 50 cm. The distance of his image from himself, as seen by the observer is

• A. 94 cm
• B. 96 cm
• C. 98 cm
• D. 100 cm

Answer 1

The correct option is B 96 cm

Given, thickness of slab (t) = 6 cm
${\mu }_g=1.5$

Since the ray incident from 'O' will be going towards plane mirror, shift produced by slab is:
$S=t(1-\frac{1}{{\mu }_g})$
$\Rightarrow S=6(1-\frac{2}{3})=2cm$
Thus object will appear at O' i.e. at $(u=50-2=48cm)$ behind mirror for observer.
Distance of image from 'O' is,
$d=50cm+(48-2)cm$
$\therefore d=96cm$

Why this question? It intends to test the application of normal shift produced by glass slab along with fundamental plane mirror.