Question

007 (part 1 of 2) 1.0 points

Answer 1

Answer:

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°

Explanation:

According to Snell's law, we have;

$\dfrac{n_1}{n_2} = \dfrac{sin (\theta_2)}{sin (\theta_1)}$

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;

$\dfrac{1.66}{1.333} = \dfrac{sin (\theta_2)}{sin (27.2^{\circ})}$

$sin (\theta_2) = sin (27.2^{\circ}) \times \dfrac{1.66}{1.333} \approx 0.5692292265$

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;

$\dfrac{1.66}{1.333} = \dfrac{sin (90^{\circ})}{sin (\theta_1)}$

$sin (\theta_1) = \dfrac{sin (90^{\circ})}{\left( \dfrac{1.66}{1.333}\right)} = sin (90^{\circ}) \times \dfrac{1.333}{1.66} \approx 0.803$

θ₁ ≈ arcsin(0.803) ≈ 53.42°

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