A light ray enters water (n ≈ 1.33) from air (n ≈ 1.00) at 30°. Using Snell's law, what is the refracted angle approximately?

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Multiple Choice

A light ray enters water (n ≈ 1.33) from air (n ≈ 1.00) at 30°. Using Snell's law, what is the refracted angle approximately?

Snell's law relates the incident angle, the refractive indices, and the refracted angle: n1 sin(theta1) = n2 sin(theta2). Here light goes from air (n1 ≈ 1.00) into water (n2 ≈ 1.33) with theta1 = 30°. Solve for theta2: sin(theta2) = (n1/n2) sin(theta1) = (1.00/1.33) × sin(30°) ≈ 0.376. Therefore theta2 ≈ arcsin(0.376) ≈ 22°. Since the ray enters a denser medium, it bends toward the normal, giving a refracted angle smaller than the incident angle. That’s why the angle is about 22 degrees.

Snell's law relates the incident angle, the refractive indices, and the refracted angle: n1 sin(theta1) = n2 sin(theta2). Here light goes from air (n1 ≈ 1.00) into water (n2 ≈ 1.33) with theta1 = 30°. Solve for theta2: sin(theta2) = (n1/n2) sin(theta1) = (1.00/1.33) × sin(30°) ≈ 0.376. Therefore theta2 ≈ arcsin(0.376) ≈ 22°. Since the ray enters a denser medium, it bends toward the normal, giving a refracted angle smaller than the incident angle. That’s why the angle is about 22 degrees.

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