Question

An unpolarized light incident on a plane glass surface gets totally polarized on reflection. If the refractive index of glass is $\tan 57^{\circ}$, then the angle of refraction is:

• A. $90^{\circ}$
• B. $33^{\circ}$
• C. $13^{\circ}$
• D. $37^{\circ}$
• E. $45^{\circ}$

Hint:

Angle of incidence + Angle of refraction = $90^{\circ}$ at Brewster's angle.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
This problem involves the polarization of light by reflection, which is described by Brewster's Law. It also requires the use of Snell's Law to find the angle of refraction.
Step 2: Key Formula or Approach:
1. Brewster's Law: When unpolarized light is incident on a surface, the reflected light is completely polarized if the angle of incidence (\( \theta_i \)) is equal to the polarizing angle or Brewster's angle (\( \theta_p \)). This angle is given by the relation: \[ \tan(\theta_p) = n \] where n is the refractive index of the second medium (glass) relative to the first (usually air, with n \(\approx\) 1). 2. Property at Brewster's Angle: When light is incident at Brewster's angle, the reflected ray and the refracted ray are perpendicular to each other. 3. Snell's Law: Relates the angle of incidence (\( \theta_i \)) and the angle of refraction (\( \theta_r \)): \[ n_1 \sin(\theta_i) = n_2 \sin(\theta_r) \] Assuming the light comes from air (\( n_1=1 \)), we have \( \sin(\theta_i) = n \sin(\theta_r) \). Step 3: Detailed Explanation:
We are told the reflected light is totally polarized. This means the light is incident at Brewster's angle, \( \theta_i = \theta_p \). We are also given that the refractive index of the glass is \( n = \tan(57^\circ) \). From Brewster's Law, \( \tan(\theta_p) = n \). Comparing these two pieces of information: \[ \tan(\theta_p) = \tan(57^\circ) \] This means that Brewster's angle is \( \theta_p = 57^\circ \). So, the angle of incidence is \( \theta_i = 57^\circ \). Now we need to find the angle of refraction, \( \theta_r \). We can use Snell's Law: \[ \sin(\theta_i) = n \sin(\theta_r) \] \[ \sin(57^\circ) = (\tan 57^\circ) \sin(\theta_r) \] \[ \sin(57^\circ) = \left(\frac{\sin 57^\circ}{\cos 57^\circ}\right) \sin(\theta_r) \] Assuming \( \sin(57^\circ) \neq 0 \), we can cancel it from both sides: \[ 1 = \frac{1}{\cos 57^\circ} \sin(\theta_r) \] \[ \sin(\theta_r) = \cos(57^\circ) \] Using the complementary angle identity, \( \cos(\theta) = \sin(90^\circ - \theta) \): \[ \sin(\theta_r) = \sin(90^\circ - 57^\circ) \] \[ \sin(\theta_r) = \sin(33^\circ) \] Therefore, the angle of refraction is \( \theta_r = 33^\circ \). Alternatively, using the property that at Brewster's angle, the reflected and refracted rays are perpendicular: \( \theta_p + \theta_r = 90^\circ \). Since \( \theta_p = 57^\circ \), we have \( 57^\circ + \theta_r = 90^\circ \), which gives \( \theta_r = 90^\circ - 57^\circ = 33^\circ \). Step 4: Final Answer:
The angle of refraction is 33°.