Nilo sends a ray from water (n = 1.33) into a glass block (n = 1.50) at 40.0° from the normal. He wants the angle of refraction before he measures it.
Snell's law holds the answer. This lesson is the routine that digs the angle out.
Nilo's crossing has three knowns: the water's index, 1.33; the glass's, 1.50; and the 40.0° angle of incidence. The unknown is the angle of refraction.
Snell's law ties the four together: rearrange n₁ sin θ₁ = n₂ sin θ₂ into sin θ₂ = n₁ sin θ₁ ÷ n₂.
Substituting Nilo's numbers gives sin θ₂ = 0.570. That is a sine, not yet an angle.
To finish, you need the step that turns a sine back into its angle.
You have used this kind of step before, for the tangent: tan⁻¹ takes a tangent value and returns the angle that made it.
So sin⁻¹(0.570) = 34.7°, and the refraction angle is about 35°.
Sense-check the result against the direction rule: the glass is denser than the water, so the ray should bend toward the normal, and 35° is indeed smaller than 40°.
One warning sign: if a substitution ever hands you a sine above 1, stop. No angle has a sine above 1, and a right answer never needs one.
Light passes from water (n = 1.33) into glass (n = 1.50) at 40.0° from the normal. Calculate the angle of refraction, to the nearest degree.
Write down the values:
n₁ = 1.33, θ₁ = 40.0°, n₂ = 1.50, θ₂ = ?
Write down the equation, and rearrange for sin θ₂:
n₁ sin θ₁ = n₂ sin θ₂, so sin θ₂ = n₁ sin θ₁ ÷ n₂
Substitute in the values, and calculate:
sin 40.0° = 0.643, and 1.33 × 0.643 = 0.855, so sin θ₂ = 0.855 ÷ 1.50 = 0.570
Undo the sine with sin⁻¹:
θ₂ = sin⁻¹(0.570) = 34.7°
The angle of refraction is about 35° from the normal.
Light passes from air (n = 1.00) into water (n = 1.33) at 55.0° from the normal. Calculate the angle of refraction, to the nearest degree.
Write down the values:
n₁ = 1.00, θ₁ = 55.0°, n₂ = 1.33, θ₂ = ?
Write down the equation, and rearrange for sin θ₂:
sin θ₂ = n₁ sin θ₁ ÷ n₂
Substitute in the values, and calculate:
sin 55.0° = 0.819, and 1.00 × 0.819 = 0.819, so sin θ₂ = 0.819 ÷ 1.33 = 0.616
Undo the sine with sin⁻¹:
θ₂ = sin⁻¹(0.616) = 38.0°
The angle of refraction is about 38° from the normal.
Light passes from air (n = 1.00) into glycerin (n = 1.47) at 30.0° from the normal. Calculate sin θ₂, the sine of the refraction angle, to three decimal places. Answers within 0.005 are accepted.
Light passes from air into glycerin, and the sine of the refraction angle comes out as 0.340. Calculate the refraction angle itself, to the nearest degree. Answers within 1° are accepted.
Light passes from air (n = 1.00) into window glass (n = 1.52) at 45.0° from the normal. Calculate the angle of refraction, to the nearest degree. Answers within 1° are accepted.
Light passes from water (n = 1.33) out into air (n = 1.00) at 35.0° from the normal. Calculate the angle of refraction in the air, to the nearest degree. Answers within 1° are accepted.
Light passes from a glass block (n = 1.50) into water (n = 1.33) at 22.0° from the normal. Calculate the angle of refraction in the water, to the nearest degree. Answers within 1° are accepted.
Light is about to pass from air (n = 1.00) into a diamond (n = 2.42) at 40.0° from the normal. Before any calculation, what must be true of the refraction angle?