How do you get phi 1 max? I have been able to solve the rest of the parts of the

Question

How do you get phi 1 max?

I have been able to solve the rest of the parts of the question, but I can't seem to figure this one out.

Red light is incident in air on a 300, 600, 90° prism as shown. The incident beam is directed at an angle of 1 = 43.49 with respect to the horizontal and enters the prism at a height h 19 cm above the base. The beam leaves the prism to the air at a distance d-49.7 cm along the baeas shown. 91. 300 2 3 1) what is 2, the angle the beam in the prism makes with the horizontal axis? 48.53 degrees Submit

Explanation / Answer

using snell's law we have

ngsin thetha1=na sin thetha2

sin thetha1=1*sin 90/1.4

thetha1=arcsin(1/1.4)

thetha1=43.79 dgerees

thetha2 =90-43.79

=46.2 degrees

let 's take it as a triangle where light is incident we can say that

180=30+46.2+A

A=103.79 degrees

now it is making 90 degrees with the vertucal so angle between them is

103.79-90

=13.79 degrees

using snell's law again

ngsin thetha1=nasin thetha2

(1.4)sin 13.79=1*sinthetha2

sin-1((1.4)sin 13.79)=thetha2

thetha2=201.5 degrees

so now taking all the angles from the incident one we get

90=30-20.15+thethamax

thethamax=39.29 degrees--answer

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