Question

A ray of light passes from the vacuum into a medium of refractive index 'n'. If the angle of incidence is twice the angle of refraction, then the angle of incidence in terms of refractive index is

• A. \(sin^{-1}(\frac{n}{2})\)
• B. \(2cos^{-1}(\frac{n}{2})\)
• C. \(2sin^{-1}(\frac{n}{2})\)
• D. \(cos^{-1}(\frac{n}{2})\)

Hint:

Use Snell law and the double angle formula for the sine.

Answer 1

The Correct Option is B

Step 1: Snell law
\(\frac{\sin i}{\sin r} = n\) with \(i = 2r\).

Step 2: Double angle
\(\frac{\sin2r}{\sin r} = 2\cos r = n\), so \(\cos r = \frac n2\) and \(r = \cos^{-1}\frac n2\).

Step 3: Incidence angle
\(i = 2r = 2\cos^{-1}\left(\frac n2\right)\). Option (B).

Final Answer:
The angle of incidence is 2 cos^-1 (n/2). \[ \boxed{\text{(B)}\ 2\cos^{-1}\left(\frac n2\right)} \]