Monochromatic light through a triangular prism

Classical Mechanics Level 3

Monochromatic light of wavelength 600 nm 600 \text{ nm} strike an equilateral triangular prism with an angle of incidence θ i = 3 0 ∘ . \theta_i = 30^\circ. The index of refraction for the prism is n = 1.5. n=1.5. What is the angle of refraction of the light as it leaves the prism rounded to the nearest degree?


The answer is 77.

July Thomas by July Thomas , 30, USA

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1 solution

July Thomas
May 11, 2016

n 1 sin ⁡ θ 1 = n 2 sin ⁡ θ 2 n_1 \sin\theta_1 = n_2\sin\theta_2

θ 2 = sin ⁡ − 1 ( n 1 sin ⁡ θ 1 n 2 ) = 19.4 7 ∘ \theta_2 = \sin^{-1} (\frac{n_1 \sin\theta_1}{n_2}) = 19.47^\circ

180 = Σ angles = 60 + ( 90 − 19.47 ) + ( 90 − θ 3 ) 180 = \Sigma\text{angles} = 60 + (90-19.47) + (90-\theta_3)

θ 3 = 40.5 3 ∘ \theta_3 = 40.53^\circ

n 2 sin ⁡ θ 3 = n 1 sin ⁡ θ 4 n_2\sin\theta_3 = n_1\sin\theta_4

1.5 sin ⁡ ( 40.53 ) = 1 sin ⁡ θ 4 1.5\sin(40.53) = 1\sin\theta_4

θ 4 = 77.1 0 ∘ \theta_4 = 77.10^\circ

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