A ray of light of pure single colour is incident on the face of a prism having angle of the prism 30° at an angle of incidence 45°. The refracted ray does not change its direction as it crosses the other face and emerges out of the prism. The refractive index of the material of the prism is:

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$\frac{2}{\sqrt{3}}$

$\sqrt{3}$

$\sqrt{2}$

answer is C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

Use $r_{2} = 0 °$ because the refracted beam exits the prism without altering course. Do not forget to utilise the formula. $A = r_{1} + r_{2}$ and select the appropriate choice.
Formula used: $μ_{21} = \frac{\sin i_{1}}{\sin r_{1}}$ and $A = r_{1} + r_{2}$

Considering the facts provided $A = 30 °$ and $i_{1} = 45 °$
We know that for an angle prism because of the information provided that the refracted beam does not change its orientation. $A, r_{1}$ and $r_{2}$ are the refraction angle
Therefore $A = r_{1} + r_{2}$ (equation 1)
Since the refracted beam exits the prism with no direction change, i.e. $r_{2} = 0 °$
the values in equation 1 are replaced
$\Rightarrow 30 ° = r_{1} + 0 °$
$\Rightarrow r_{1} = 30 °$
Therefore the relation $μ_{21} = \frac{\sin i_{1}}{\sin r_{1}} \Rightarrow \frac{\sin 45 °}{\sin 30 °}$