Question

A light ray is incident at an angle of 45° with the normal to a √2 cm thick plate (μ = 2.0). Find the shift in the path of the light as it emerges out from the plate.

Answer 1

Given,
Angle of incidence = 45˚
Thickness of the plate = \[\sqrt{2} \text{ cm }\] 
Refractive index (μ) of the plate = 2.0

Applying Snell's law,
\[\frac{\sin i}{\sin r} = \frac{1}{\mu}\]
\[ \Rightarrow \frac{\sin i}{\sin r} = \frac{2}{1}\]
\[ \Rightarrow \sin r = \frac{\sin 45^\circ }{2}\]
\[ \Rightarrow \sin r = \frac{1}{2\sqrt{2}}\]
\[ \Rightarrow r = 21^\circ\]
Therefore,
\[\frac{\sin i}{\sin r} = \frac{1}{\mu}\]
\[ \Rightarrow \frac{\sin i}{\sin r} = \frac{2}{1}\]
\[ \Rightarrow \sin r = \frac{\sin 45^\circ }{2}\]
\[ \Rightarrow \sin r = \frac{1}{2\sqrt{2}}\]
\[ \Rightarrow r = 21^\circ\]
From figure, we know that BD is the shift in path which is equal to (AB)sin 24˚
\[= 0 . 406 \times AB = \frac{AE}{\cos 21^\circ } \times 0 . 406\]
\[ = 0 . 62 cm\]