Question

When unpolarized light is incident at an angle of 60° on a transparent medium from air. The reflected ray is completely polarized. The angle of refraction in the medium is

• A. 30°
• B. 60°
• C. 90°
• D. 45°

Answer 1

The Correct Option is A

To solve the problem of determining the angle of refraction when unpolarized light is incident at an angle of 60° on a transparent medium and the reflected ray is completely polarized, we can use Brewster's Law. Brewster's Law states that the tangent of the polarizing angle (also called Brewster's angle) is equal to the refractive index of the medium, or mathematically:

\(\tan(\theta_B) = n\)

Here, \(\theta_B\) is the angle of incidence which is 60°, and \(n\) is the refractive index of the transparent medium.

According to Brewster’s Law, when light is incident at the Brewster angle, the reflected ray is completely polarized perpendicular to the plane of incidence, and the angle of incidence is the polarizing angle.

From the relationship given by Snell's Law:

\(\sin(\theta_i) = n \cdot \sin(\theta_r)\)

where \(\theta_i\) is the angle of incidence, and \(\theta_r\) is the angle of refraction.

Since the angle of incidence \(\theta_B = 60°\) and in such a case where the reflected light is fully polarized, the angle of refraction \(\theta_r\) and the incident angle add up to 90 degrees:

\(\theta_B + \theta_r = 90°\)

This implies:

\(\theta_r = 90° - \theta_B\)

Substituting \(\theta_B = 60°\):

\(\theta_r = 90° - 60° = 30°\)

Therefore, the angle of refraction in the medium is 30°.

Hence, the correct answer is 30°.

Answer 2

According to Brewster’s law, the reflected ray is completely polarized when the angle of incidence \( \theta_p \) satisfies:

\[ \tan \theta_p = n, \]

where \( n \) is the refractive index of the medium relative to air.

Given \( \theta_p = 60^\circ \), we have:

\[ n = \tan 60^\circ = \sqrt{3}. \]

Using Snell’s law \( n_1 \sin \theta_1 = n_2 \sin \theta_2 \), with \( n_1 = 1 \) (for air) and \( \theta_1 = 60^\circ \):

\[ \sin \theta_2 = \frac{\sin 60^\circ}{\sqrt{3}} = \frac{\sqrt{3}/2}{\sqrt{3}} = \frac{1}{2}. \]

Therefore, \( \theta_2 = 30^\circ \).