Question

Light enters from air into a given medium at an angle of 45° with interface of the air-medium surface. After refraction, the light ray is deviated through an angle of 15° from its original direction. The refractive index of the medium is

• A. 1.732
• B. 1.333
• C. 1.414
• D. 2.732

Answer 1

The Correct Option is C

To solve this question, we will use Snell's law, which relates the angle of incidence and refraction for light travelling between two media of different refractive indices. Snell's law is given by the equation:

n_1 \sin \theta_1 = n_2 \sin \theta_2

where:

• n_1 is the refractive index of the first medium (air, in this case).
• n_2 is the refractive index of the second medium.
• \theta_1 is the angle of incidence.
• \theta_2 is the angle of refraction.

For this problem:

• The angle of incidence, \theta_1 , is 45°.
• The deviation of the light ray after refraction is given as 15° from its original path.

The deviation can be expressed as the difference between the angle of incidence ( \theta_1 ) and the angle of refraction ( \theta_2 ):

\text{Deviation} = \theta_1 - \theta_2 = 15^\circ

From this, we can find:

\theta_2 = \theta_1 - 15^\circ = 45^\circ - 15^\circ = 30^\circ

Using the values in Snell's law:

n_1 \sin 45^\circ = n_2 \sin 30^\circ

Given n_1 = 1 for air (since we assume the refractive index of air to be approximately 1):

\sin 45^\circ = \frac{\sqrt{2}}{2}

\sin 30^\circ = \frac{1}{2}

Therefore, Snell's law becomes:

1 \times \frac{\sqrt{2}}{2} = n_2 \times \frac{1}{2}

Simplifying gives:

n_2 = \sqrt{2} \approx 1.414

Thus, the refractive index of the medium is 1.414. Therefore, the correct answer is:

Option 1.414.