Question

007 (part 1 of 2) 1.0 points

Light moves from flint glass (n = 1.66) into
water at an angle of incidence of 27.2º.
a) What is the angle of refraction?
Answer in units of".
008 (part 2 of 2) 1.0 points
b) At what angle would the light have to be
incident to give an angle of refraction of 90.0°
Answer in units of

Answer 1

a) The angle of refraction is approximately 34.7

b) The angle the light have to be incident to give an angle of refraction of 90° is approximately 53.42°

Step-by-step explanation:

According to Snell's law, we have;

`(n_1)/(n_2) = (sin (\theta_2))/(sin (\theta_1))`

The refractive index of the glass, n₁ = 1.66

The angle of incident of the light as it moves into water, θ₁ = 27.2°

a) The refractive index of water, n₂ = 1.333

Let θ₂ represent the angle of refraction of the light in water

By plugging in the values of the variables in Snell's Law equation gives;

`(1.66)/(1.333) = (sin (\theta_2))/(sin (27.2^(\circ)))`

`sin (\theta_2) = sin (27.2^(\circ)) * (1.66)/(1.333) \approx 0.5692292265`

θ₂ = arcsin(0.5692292265) ≈ 34.7°

The angle of refraction of the light in water, θ₂ ≈ 34.7°

b) When the angle of refraction, θ₂ = 90°, we have;

`(1.66)/(1.333) = (sin (90^(\circ)))/(sin (\theta_1))`

`sin (\theta_1) = (sin (90^(\circ)))/(\left( (1.66)/(1.333)\right)) = sin (90^(\circ)) * (1.333)/(1.66) \approx 0.803`

θ₁ ≈ arcsin(0.803) ≈ 53.42°

The angle of incident, θ₁, that would give an angle of refraction of 90° is θ₁ ≈ 53.42°