Question

A ray of light is incident on a glass slab. The light ray travels distance of 5 cm inside the glass slab before emerging out of the slab. If the incident ray suffers a deviation of ${30}^{\circ }$ after the first referaction, the perpendicular distance between incident and the emergent ray is :

• A. 5 cm
• B. 2.5 cm
• C. 7.5 cm
• D. 10 cm

Answer 1

The correct option is B 2.5 cm


The deviation suffered by incident ray after first referaction is,
$\delta =i-r$
$\Rightarrow (i-r)={30}^{\circ }$
'MN' is distance between incident and emergent ray and PM is the distance travelled by refracted ray in slab.

From $\Delta PMN,$
$\angle MPN=i-r={30}^{\circ }$
$\Rightarrow sin{30}^{\circ }=\frac{MN}{PM}$
or $\frac{1}{2}=\frac{MN}{5}$ ($\because PM=5cm$)
$\therefore MN=\frac{5}{2}=2.5cm$

Why this question? It intends to test your understanding of geometry of ray diagram and concept of lateral shift. Tip: Distance travelled by light ray is measured from point of incidence to point of emergence.