A certain glass has a refractive index of 1.50 when in contact with air. Calculate i) the angle of refraction in glass when a light ray is incident at 30° with the normal.

ii) at what angle of incidence would one need to shine a beam of light so as to emerge in glass at 36°.

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i) The angle of refraction is approximately 19.47°

ii) The angle of the incident light should be approximately 61.85°

Explanation:

Snell's law for the relationship between the angle of incident  and the angle of refraction of light is given mathematically as follows;

n₁·sin(θ₁) = n₂·sin(θ₂)

i) The characteristics of the glass are;

The refractive index of the glass, n₂ = 1.50

The angle of incident of the light ray in air, θ₁ = 30°

The refractive index of air, n₁ = 1

Let 'θ₂' represent the angle of the light refracted by the glass, by Snell's law we have;

1 × sin(30°) = 1.5 × sin(θ₂)

∴ sin(θ₂) = 1 × sin(30°)/1.5 = 1 × 0.5/1.5 = 1/3

θ₂ = arcsin(1/3) ≈ 19.47°

∴  The angle of refraction, θ₂ ≈ 19.47°

ii) When the angle at which the light incident from air emerges in the glass is 36°, we have;

The angle of refraction of the light through the glass, θ₂ = 36°

The refractive index of the glass remains, n₂ = 1.50

The refractive index of air, n₁ = 1

Let 'θ₁' represent the angle of the incident light, we get;

1 × sin(θ₁) = 1.5 × sin(36°)

∴ sin(θ₁) = 1.5 × sin(36°) = 1.5 × √(10 - 2·√5)/4

θ₁ = arcsin( 1.5 × √(10 - 2·√5)/4 ) ≈ 61.85°

The angle of the incident light should be, θ₁ ≈ 61.85°.

waves moving from a fast medium to a slow medium will experience a reflection as light travels more slowly than red light. total internal reflection occurs when the critical angle is.