If a light ray strikes the interface from air into water at an incidence of 45°, what is the approximate refracted angle? (Assume n air = 1, n water = 1.33)

If a light ray strikes the interface from air into water at an incidence of 45°, what is the approximate refracted angle? (Assume n air = 1, n water = 1.33)

• A. About 8°
• B. About 60°
• C. About 45°
• D. About 32.1°

Answer: (D)

When light passes from one medium to another, its path bends in a way described by Snell's law: n1 sin(theta1) = n2 sin(theta2). Here, air has n1 ≈ 1.00 and the ray hits the boundary at 45°, entering water with n2 ≈ 1.33. Solve for the refracted angle: sin(theta2) = (n1/n2) sin(theta1) = (1/1.33) × sin(45°) ≈ 0.531. Taking the inverse sine gives theta2 ≈ 32.1°. The refracted ray is closer to the normal than the incident ray because it slows down in the denser medium, causing the bend toward the normal. So the approximate refracted angle is about 32°.

When light passes from one medium to another, its path bends in a way described by Snell's law: n1 sin(theta1) = n2 sin(theta2). Here, air has n1 ≈ 1.00 and the ray hits the boundary at 45°, entering water with n2 ≈ 1.33. Solve for the refracted angle: sin(theta2) = (n1/n2) sin(theta1) = (1/1.33) × sin(45°) ≈ 0.531. Taking the inverse sine gives theta2 ≈ 32.1°. The refracted ray is closer to the normal than the incident ray because it slows down in the denser medium, causing the bend toward the normal. So the approximate refracted angle is about 32°.