Question

A light beam strikes a piece of glass at a 53.00 ∘incident angle. The beam contains two wavelengths, 450.0 nm and 700.0 nm, for which the index of refraction of the glass is 1.4831 and 1.4754, respectively.

Part A

What is the angle between the two refracted beams?

Answer 1

Solution:

The angle of incidence of light beam, θ = 53.00^o

For the light of wavelength 450.0 nm, glass has index of refraction n₁ = 1.4831

For the light of wavelength 700.0 nm, glass has index of refraction n₂ = 1.4754

From the Snell’s law, the angle of refraction θ₁ of the light of wavelength 450.0 nm is given by,

n₁*sinθ₁ = n*sinθ                  (n = 1 for air if not mentioned)

sinθ₁ = (n/n₁)*sinθ

θ₁ = sin^-1[(n/n₁)*sinθ]

θ₁ = sin^-1[(1/1.4831)*sin53.00^o]

θ₁ = 32.58^o

From the Snell’s law, the angle of refraction θ₂ of the light of wavelength 700.0 nm is given by,

n₂*sinθ₂ = n*sinθ            

sinθ₂ = (n/n₂)*sinθ

θ₂ = sin^-1[(n/n₂)*sinθ]

θ₂ = sin^-1[(1/1.4754)*sin53.00^o]

θ₂ = 32.77^o

Hence the angle between two refracted beams is

θ₂ - θ₁ = 32.77^o - 32.58^o = 0.19^o     (zero point nineteen degrees)

Hence the answer is 0.19^o