Question

A ray of light passes from 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 \( n \) is

• A. \( 25 \sin^{-1} \left( \frac{1}{n} \right) \)
• B. \( \cos^{-1} \left( \frac{1}{n} \right) \)
• C. \( \sin^{-1} \left( \frac{n}{2} \right) \)
• D. \( 2 \cos^{-1} \left( \frac{1}{2n} \right) \)

Hint:

When using Snell's law, remember that the angle of incidence and refraction are related by the refractive index of the medium.

Answer 1

The Correct Option is C

The angle of incidence \( i \) is twice the angle of refraction \( r \). This means: \[ i = 2r \] Using Snell’s law: \[ n_1 \sin(i) = n_2 \sin(r) \] For this case, \( n_1 = 1 \) (vacuum), so: \[ \sin(i) = n \sin(r) \] Substitute \( i = 2r \): \[ \sin(2r) = n \sin(r) \] Using the identity \( \sin(2r) = 2 \sin(r) \cos(r) \): \[ 2 \sin(r) \cos(r) = n \sin(r) \] Dividing both sides by \( \sin(r) \) (assuming \( \sin(r) \neq 0 \)): \[ 2 \cos(r) = n \] Thus: \[ \cos(r) = \frac{n}{2} \] Now, the angle of refraction is: \[ r = \cos^{-1} \left( \frac{n}{2} \right) \] And since \( i = 2r \), the angle of incidence is: \[ i = 2 \cos^{-1} \left( \frac{n}{2} \right) \] Thus, the correct answer is option (4): \( 2 \cos^{-1} \left( \frac{1}{2n} \right) \).