Question
The refractive index of the material of a prism is $$\sqrt 2 $$ and its refracting angle is $${30^ \circ }.$$ One of the refracting surfaces of the prism is made a mirror inwards. A beam of monochromatic light entering the prism from the other face will retrace its path after reflection from the mirrored surface, if its angle of incidence on the prism is
A.
$${45^ \circ }$$
B.
$${60^ \circ }$$
C.
$${0^ \circ }$$
D.
$${30^ \circ }$$
Answer :
$${45^ \circ }$$
Solution :
According to the given condition, the beam of light will retrace its path after reflection from $$BC.$$
So, $$\angle CPQ = {90^ \circ }$$
Thus, angle of refraction at surface $$AC$$
$$\angle PQN = \angle r = {90^ \circ } - {60^ \circ } = {30^ \circ }$$
By Snell's law
$$\eqalign{ & \mu = \frac{{\sin i}}{{\sin r}} \Rightarrow \sqrt 2 = \frac{{\sin i}}{{\sin {{30}^ \circ }}} \cr & \therefore \sqrt 2 \times \sin {30^ \circ } = \sin i \cr & \Rightarrow \sqrt 2 \times \frac{1}{2} = \sin i \cr & \Rightarrow \sin i = \frac{1}{{\sqrt 2 }} = \sin {45^ \circ } \cr & \therefore i = {45^ \circ } \cr} $$
According to the given condition, the beam of light will retrace its path after reflection from $$BC.$$
So, $$\angle CPQ = {90^ \circ }$$
Thus, angle of refraction at surface $$AC$$
$$\angle PQN = \angle r = {90^ \circ } - {60^ \circ } = {30^ \circ }$$
By Snell's law
$$\eqalign{ & \mu = \frac{{\sin i}}{{\sin r}} \Rightarrow \sqrt 2 = \frac{{\sin i}}{{\sin {{30}^ \circ }}} \cr & \therefore \sqrt 2 \times \sin {30^ \circ } = \sin i \cr & \Rightarrow \sqrt 2 \times \frac{1}{2} = \sin i \cr & \Rightarrow \sin i = \frac{1}{{\sqrt 2 }} = \sin {45^ \circ } \cr & \therefore i = {45^ \circ } \cr} $$
