Question:

An incident light ray traveling in air strikes the surface of a glass block with an incident angle of 30°. The refractive index of glass is 1.5.

a) Calculate the angle of refraction at the air-glass interface.

b) If the incident ray enters the glass block and travels at an angle of 45° with the normal, calculate the angle at which the refracted ray leaves the glass block and enters the air.

c) Determine the critical angle for total internal reflection at the air-glass interface.

Answer:

a) To calculate the angle of refraction at the air-glass interface, we can use Snell's law, which states that the ratio of the sine of the angle of incidence to the sine of the angle of refraction is equal to the ratio of the velocities of the wave in the two media or the ratio of the refractive indices:

n₁ * sin(θ₁) = n₂ * sin(θ₂)

where n₁ is the refractive index of the first medium, n₂ is the refractive index of the second medium, θ₁ is the angle of incidence, and θ₂ is the angle of refraction.

Given:

• n₁ (refractive index of air) = 1.0003
• n₂ (refractive index of glass) = 1.5
• θ₁ (angle of incidence) = 30°

Using Snell's law, we can rearrange the equation to solve for θ₂:

sin(θ₂) = (n₁/n₂) * sin(θ₁)

θ₂ = arcsin((n₁/n₂) * sin(θ₁))

Substituting the given values:

θ₂ = arcsin((1.0003/1.5) * sin(30°))

θ₂ ≈ arcsin(0.666866)

θ₂ ≈ 41.8°

Therefore, the angle of refraction at the air-glass interface is approximately 41.8°.

b) To calculate the angle at which the refracted ray leaves the glass block and enters the air, we can apply Snell's law again. This time, the refractive indices are reversed, and we know the angle of refraction (θ₂) is 45°.

Using Snell's law:

n₂ * sin(θ₂) = n₁ * sin(θ₃)

Rearranging the equation:

sin(θ₃) = (n₂/n₁) * sin(θ₂)

θ₃ = arcsin((n₂/n₁) * sin(θ₂))

Substituting the given values:

θ₃ = arcsin((1.5/1.0003) * sin(45°))

θ₃ ≈ arcsin(1.49925)

θ₃ ≈ 61.8°

Therefore, the angle at which the refracted ray leaves the glass block and enters the air is approximately 61.8°.

c) The critical angle for total internal reflection is the angle of incidence that results in an angle of refraction of 90°. In other words, it is the maximum angle of incidence at which light can pass through the surface and refract into the second medium.

Using Snell's law:

n₁ * sin(θ_c) = n₂ * sin(90°)

Since sin(90°) = 1, we can simplify the equation:

sin(θ_c) = n₂/n₁

θ_c = arcsin(n₂/n₁)

Substituting the given values:

θ_c = arcsin(1.5/1.0003)

θ_c ≈ arcsin(1.4998)

θ_c ≈ 48.8°

Therefore, the critical angle for total internal reflection at the air-glass interface is approximately 48.8°.