Respuesta :
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;
[tex]\dfrac{n_1}{n_2} = \dfrac{sin (\theta_2)}{sin (\theta_1)}[/tex]
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;
[tex]\dfrac{1.66}{1.333} = \dfrac{sin (\theta_2)}{sin (27.2^{\circ})}[/tex]
[tex]sin (\theta_2) = sin (27.2^{\circ}) \times \dfrac{1.66}{1.333} \approx 0.5692292265[/tex]
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;
[tex]\dfrac{1.66}{1.333} = \dfrac{sin (90^{\circ})}{sin (\theta_1)}[/tex]
[tex]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[/tex]
θ₁ ≈ arcsin(0.803) ≈ 53.42°
The angle of incident, θ₁, that would give an angle of refraction of 90° is θ₁ ≈ 53.42°
