Question

When a Ray of light enters a medium of refractive index new it is observed that the angle of refraction is half of the angle of incidence is then angle of incidence is?

Answer 1

The angle of incidence is given by, $i=2cos^-^1(\frac{\mu}{2} )$

use Snell's law of refraction.

$\mu=\frac{sini}{sinr}$

Substitute $r=\frac{i}{2}$ and simplify

$\mu=\frac{sini}{sinr} \\ =\frac{sini}{sin\frac{i}{2}} \\ =\frac{2sin\frac{i}{2}cos\frac{i}{2}}{sin\frac{i}{2}} \\ =2cos\frac{i}{2}$

Simplify for i.

$cos\frac{i}{2} =\frac{\mu}{2} \\ i=2cos^-^1(\frac{\mu}{2} )$

The angle of incidence is given by $i=2cos^-^1(\frac{\mu}{2} )$

Answer 2

Answer: The correct answer is $i=2cos^{-1}(\frac{\mu}{2})$.

Explanation:

The expression from snell's law is as follows;

$\mu =\frac{sini}{sinr}$

Here, i is the angle of incidence and r is the angle of refraction.

It is given in the problem that the angle of refraction is half of the angle of incidence.

Put $r=\frac{i}{2}$ in the above expression.

$\mu =\frac{sini}{sin\frac{i}{2}}$

Use the formula for $sini= 2sin\frac{i}{2}2cos\frac{i}{2}$.

$\mu =\frac{2sin\frac{i}{2}2cos\frac{i}{2}}{sin\frac{i}{2}}$

$\mu =2cos\frac{i}{2}$

Rearrange the expression for angle of incidence.

$i=2cos^{-1}(\frac{\mu}{2})$

Therefore, the angle of incidence is $i=2cos^{-1}(\frac{\mu}{2})$.