Question

The drawing shows a rectangular block of glass (*n* =
1.55) surrounded by liquid carbon disulfide (*n* = 1.68). A
ray of light is incident on the glass at point A with a
θ_{1} = 43.0° angle of incidence. At what angle of
refraction does the ray leave the glass at point B?

Answer #1

Given , μ1 = 1.68 , μ2 = 1.55 ,θ1 = 43°

Accordingto Snell law, we have

μ_{1}sinθ_{1} =μ_{2}sinθ_{2}

1.68*sin(43°) = 1.55*sin(θ2)

sin(θ2) = 1.68sin(43°)/1.55 = 0.74

θ2 = sin^-1(0.74) = 47.663°

Therefore the refraction angle at A is θ2 = 47.663°

The angle of incidence at B = 90 - θ_{2} = 90 - 47.663°
= 42.336°

Again using the snells law,

1.55*sin(42.336°) = 1.68*sin(θ_{3})

sin θ_{3} =1.55*sin(42.336°)/1.68

sin θ_{3} = 0.621

θ3 = sin^-1(0.621) = 38.416°

Thus , the angl of refraction is 38.416°

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