A light ray travels from air into water with an incidence angle of 60°. Using Snell's law (n_air ≈ 1.00, n_water ≈ 1.33), what is the angle in water approximately?

A light ray travels from air into water with an incidence angle of 60°. Using Snell's law (n_air ≈ 1.00, n_water ≈ 1.33), what is the angle in water approximately?

• A. About 60°
• B. About 23°
• C. About 41°
• D. About 70°

Answer: (C)

When light passes from one medium to another, its path bends according to Snell’s law: n1 sin theta1 = n2 sin theta2. Here theta1 is 60°, n1 about 1.00 (air), and n2 about 1.33 (water). So sin theta2 = (n1/n2) sin theta1 ≈ (1.00/1.33) × sin 60° ≈ 0.652. Taking the arcsin gives theta2 ≈ 41°. So the angle in water is about 41 degrees. The ray bends toward the normal because it enters a denser medium (higher refractive index), which makes the angle smaller than the incident angle. That also explains why it isn’t as large as 60° or 70°, and isn’t as small as 23°.

When light passes from one medium to another, its path bends according to Snell’s law: n1 sin theta1 = n2 sin theta2. Here theta1 is 60°, n1 about 1.00 (air), and n2 about 1.33 (water). So sin theta2 = (n1/n2) sin theta1 ≈ (1.00/1.33) × sin 60° ≈ 0.652. Taking the arcsin gives theta2 ≈ 41°. So the angle in water is about 41 degrees. The ray bends toward the normal because it enters a denser medium (higher refractive index), which makes the angle smaller than the incident angle. That also explains why it isn’t as large as 60° or 70°, and isn’t as small as 23°.