Question

A light ray passes from air (refractive index \( n_1 = 1 \)) into water (refractive index \( n_2 = 1.33 \)). If the angle of incidence is \( 30^\circ \), what is the angle of refraction in the water?

• A. \( 22^\circ \)
• B. \( 30^\circ \)
• C. \( 23.6^\circ \)
• D. \( 40^\circ \)

Hint:

Snell's law relates the angles of incidence and refraction through the refractive indices. Always use the sine of the angles in the law.

Answer 1

The Correct Option is C

Step 1: Apply Snell's Law of refraction \[n_1 \sin \theta_1 = n_2 \sin \theta_2\]Definitions: - \( n_1 \): refractive index of air - \( n_2 \): refractive index of water - \( \theta_1 \): angle of incidence - \( \theta_2 \): angle of refractionGiven values: - \( n_1 = 1 \) - \( n_2 = 1.33 \) - \( \theta_1 = 30^\circ \)Substitute known values into Snell's Law:\[1 \times \sin(30^\circ) = 1.33 \times \sin(\theta_2)\]Calculate \( \sin(\theta_2) \):\[\sin(\theta_2) = \frac{\sin(30^\circ)}{1.33} = \frac{0.5}{1.33} \approx 0.3759\]Determine \( \theta_2 \):\[\theta_2 = \sin^{-1}(0.3759) \approx 23.6^\circ\]Result: The angle of refraction in water is approximately \( 23.6^\circ \). This corresponds to option (3).