Physics Energy Question

Question

A ray of light strikes a side of lucite ( = 1.50) prism at 40° as shown below. Find the angle that the light leaves the prism.

The prism is a isosceles with a degree of 70 at the top.

Answer 1

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Answer 2

A very similar Ask entry had link to a diagram, which is critical to the solution because the 40^o angle is not from the normal but from the lower left side of the prism, below where the ray enters the prism. If this is true for the un-shown diagram of this Ask, then the following is the multi-step solution --

At left entrance into prism, let ϑ₁ be incoming angle to the normal which is 50^o, and ϑ₂ be refracted angle relative to normal inside prism. Index of refraction in air is taken to be n₁ = 1.00 and in prism is given as n₂ = 1.50.

Snell’s Law gives n₂ sin ϑ₂ = n₁ sin ϑ₁ from which sin ϑ₂ = (1.00 sin 50) / 1.50 = 0.5107, so ϑ₂ = 30.71^o. Let α be the angle from refracted ray to upper left side of prism, so that α = 90 - ϑ₂ = 59.29^o. Let β be the angle at right side between refracted ray and inside upper right side of prism; β = 180 – 70 – 59.29 = 50.71^o. Now angle of ray to normal on inside right side ϑ₃ = 90 – β = 39.29^o.

Letting ϑ₄ be the angle “?” to be determined, n₁ sin ϑ₄ = n₂ sin ϑ₃ and sin ϑ₄ = (1.50 sin 39.29)/ 1.00 = 0.9499. Finally ϑ₄ = 71.8^o.