Oona has a glass block (n = 1.50) sitting in an unknown clear liquid. A ray leaves the glass at 40.0° from the normal and refracts into the liquid at 45.0°. She wants the liquid's index.
Snell's law has four quantities. Give it any three, and the routine digs out the fourth. This lesson solves for an index, and for an incidence angle.
In Oona's crossing, three of the four quantities are known: the glass's index, 1.50; the 40.0° angle inside the glass; and the 45.0° angle in the liquid. The liquid's index is the unknown.
Snell's law ties them: n₁ sin θ₁ = n₂ sin θ₂. Rearrange for the unknown by dividing both sides by sin θ₂: n₂ = n₁ sin θ₁ ÷ sin θ₂.
Substituting Oona's numbers gives n₂ = 1.36. The liquid's index is about 1.36.
Sense-check it: the ray bent AWAY from the normal (45.0° is larger than 40.0°), so the liquid must be less optically dense than the glass. 1.36 is below 1.50. It passes.
When the unknown is an ANGLE on the incoming side instead, rearrange for its sine: sin θ₁ = n₂ sin θ₂ ÷ n₁, then undo the sine with sin⁻¹, as in the last lesson.
Either way, the method is the same four moves: write the values, rearrange the one law, substitute, and sense-check the result against the bend's direction.
Light leaves a glass block (n = 1.50) at 40.0° from the normal and refracts into an unknown liquid at 45.0° from the normal. Calculate the liquid's index of refraction, to two decimal places.
Write down the values:
n₁ = 1.50, θ₁ = 40.0°, θ₂ = 45.0°, n₂ = ?
Write down the equation, and rearrange for n₂:
n₁ sin θ₁ = n₂ sin θ₂, so n₂ = n₁ sin θ₁ ÷ sin θ₂
Substitute in the values, and calculate:
sin 40.0° = 0.643, and 1.50 × 0.643 = 0.965. sin 45.0° = 0.707. So n₂ = 0.965 ÷ 0.707 = 1.36.
The liquid's index of refraction is about 1.36.
Light passes from air (n = 1.00) into ice (n = 1.31) and refracts at 32.0° from the normal. Calculate the angle of incidence, to the nearest degree.
Write down the values:
n₁ = 1.00, n₂ = 1.31, θ₂ = 32.0°, θ₁ = ?
Write down the equation, and rearrange for sin θ₁:
n₁ sin θ₁ = n₂ sin θ₂, so sin θ₁ = n₂ sin θ₂ ÷ n₁
Substitute in the values, and calculate:
sin 32.0° = 0.530, and 1.31 × 0.530 = 0.694, so sin θ₁ = 0.694 ÷ 1.00 = 0.694
Undo the sine with sin⁻¹:
θ₁ = sin⁻¹(0.694) = 43.9°
The angle of incidence is about 44° from the normal.
Light passes from air (n = 1.00) at 60.0° from the normal into a clear liquid, refracting at 40.0° from the normal. Calculate the liquid's index of refraction, to two decimal places. Answers within 0.02 are accepted.
Light passes from air (n = 1.00) into olive oil (n = 1.47), refracting at 30.0° from the normal. Calculate the angle of incidence, to the nearest degree. Answers within 1° are accepted.
Light leaves a window-glass pane (n = 1.52) at 35.0° from the normal and refracts into an unknown liquid at 43.0° from the normal. Calculate the liquid's index of refraction, to two decimal places. Answers within 0.02 are accepted.
Light passes from water (n = 1.33) at 48.0° from the normal into a clear crystal, refracting at 33.0° from the normal. Calculate the crystal's index of refraction, to two decimal places. Answers within 0.02 are accepted.
Light inside a glass block (n = 1.50) reaches the surface and refracts out into the air (n = 1.00) at 70.0° from the normal. Calculate the angle of incidence inside the glass, to the nearest degree. Answers within 1° are accepted.
A ray crosses from a liquid into glass. The two angles are measured, and the glass's index is known. To find the liquid's index with Snell's law, which arrangement is right?