Question

Refractive index of a rectangular glass slab is $\mu =\sqrt{3}.$ A light ray incident at an angle ${60}^{\circ }$ is displaced laterally through $2.5cm.$ Distance travelled by light in the slab is

• A. $4cm$
• B. $5cm$
• C. $2.5\sqrt{3}cm$
• D. $3cm$

Answer 1

The correct option is B $5cm$

Given, R.I. of the glass slab $\mu =\sqrt{3}$,

Angle of incidence at the first surface of the glass, ${\theta }_i={60}^{\circ }$.

We know, lateral shift, $d=t\frac{sin({\theta }_i-{\theta }_e)}{cos{\theta }_e}=2.5cm$ (given).

Again, from geometry, $t=lcos{\theta }_e$.

Here, $l$ be the distance travelled by light in the slab,

And, ${\theta }_e$ and $t$ be the angle of refraction at the first surface and depth of the slab.

So, we have, $d=lsin({\theta }_i-{\theta }_e)$......... (1)

Now, using Snell's law,

$sin{\theta }_i=\mu sin{\theta }_e$

Or, $sin{60}^{\circ }=\sqrt{3}sin{\theta }_e$

Or, $sin{\theta }_e=\frac{1}{2}$

Or, ${\theta }_e={30}^{\circ }$

Now, from the equation (1),

$2.5=l\times sin({60}^{\circ }-{30}^{\circ })=l\times sin{30}^{\circ }$

Or, $l=5cm$

So, the correct option is (B).