53 SSM www ILW In Fig. 33-53, a ray is incident on one face of a triangular glass prism in air. The angle of incidence e is chosen so that the emerging ray also makes the same angle e with the nor- mal to the other face. Show that the index of refraction n of the glass prism is given by sin ( + 6) sin o where o is the vertex angle of the prism and is the deviation angle, the total angle through which the beam is turned in passing through the prism. (Under these conditions the deviation angle u has the smallest possible value, which is called the angle of mini- mum deviation.) Figure 33-53 Problems 53 and 64.

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53 SSM www ILW In Fig. 33-53, a ray is incident on one face
of a triangular glass prism in air. The angle of incidence e is chosen
so that the emerging ray also makes the same angle e with the nor-
mal to the other face. Show that the index of refraction n of the
glass prism is given by
sin ( + 6)
sin o
where o is the vertex angle of the prism and is the deviation
angle, the total angle through which the beam is turned in passing
through the prism. (Under these conditions the deviation angle u
has the smallest possible value, which is called the angle of mini-
mum deviation.)
Figure 33-53 Problems 53 and 64.
Transcribed Image Text:53 SSM www ILW In Fig. 33-53, a ray is incident on one face of a triangular glass prism in air. The angle of incidence e is chosen so that the emerging ray also makes the same angle e with the nor- mal to the other face. Show that the index of refraction n of the glass prism is given by sin ( + 6) sin o where o is the vertex angle of the prism and is the deviation angle, the total angle through which the beam is turned in passing through the prism. (Under these conditions the deviation angle u has the smallest possible value, which is called the angle of mini- mum deviation.) Figure 33-53 Problems 53 and 64.
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