Please help with finding the angle of incidence!? - index.of photobucket
Link to image: http://i608.photobucket.com/albums/tt165/trojans_cheerchick/liquid_compare-1.png
The equilateral glass prism shown in the link has a refractive index of 1.56. The prism is surrounded by air. A light beam hits the center of the left face of the prism. What is the smallest angle of incidence for the ray of the right side of the prism sheets?
Index.of Photobucket Please Help With Finding The Angle Of Incidence!?
1 comments:
This can be difficult to follow without a picture, but I'll try
From the beginning we know that the angle of departure at the border 90 ° if sin (is 90) = 1
Thus n_b * sin (θ2) = 1.0 ... and θ2 = arcsin (1/1.56) = 39.87o
Thus, the complementary angle to angle from 90 to 39.87 = 50.13o
However, this upper triangle, the angle of an angle of 50.13 and 60. so that the third angle of the upper triangle is 180 - (50.13 60) = 69.87W
Therefore, the angle of refraction for the input beam is 90-69.87 = 20.13o
Now, for the incident beam 1 * sin (θ) = 1.56 * sin (20.13) => θ = arcsin (1.56 * sin (20.13)) = 32.5o
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