Jiso P.

asked • 07/28/20# Physics Energy 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.

## 2 Answers By Expert Tutors

Frank A. answered • 07/29/20

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Sidney P. answered • 07/28/20

Minored in physics in college, 2 years of recent teaching experience

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 ϑ_{1} be incoming angle to the normal which is 50^{o}, and ϑ_{2} be refracted angle relative to normal inside prism. Index of refraction in air is taken to be n_{1} = 1.00 and in prism is given as n_{2} = 1.50.

Snell’s Law gives n_{2} sin ϑ_{2} = n_{1} sin ϑ_{1} from which sin ϑ_{2} = (1.00 sin 50) / 1.50 = 0.5107, so ϑ_{2} = 30.71^{o}. Let α be_{ }the angle from refracted ray to upper left side of prism, so that α = 90 - ϑ_{2} = 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 ϑ_{3} = 90 – β = 39.29^{o}.

Letting ϑ_{4} be the angle “?” to be determined, n_{1} sin ϑ_{4} = n_{2} sin ϑ_{3} and sin ϑ_{4} = (1.50 sin 39.29)/ 1.00 = 0.9499. Finally ϑ_{4} = 71.8^{o}.

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Sidney P.

07/28/20