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# What is the angular spread of the beam after passing through the prism?

A narrow beam of light with wavelengths from 450 nm to 700 nm is incident perpendicular to one face of a 31.00degrees prism made of crown glass, for which the index of refraction ranges from n=1.533 to n= 1.517 for those wavelengths.

Explain please

### 2 Answers

- kirchweyLv 71 decade agoFavorite Answer
0.763 degrees.

You have to solve the prism equation (ref.) for each of the two values of N, and take the difference of the exit angles.

exit angle = arcsin(N*sin(alpha - arcsin(sin(theta1)/N))), where alpha is the prism apex angle = 31 deg and theta1 is the incidence angle = 0 deg.

Solving,

for N = 1.517, exit angle = 51.3809280745386 deg

for N = 1.533, exit angle = 52.143792690147 deg

The difference = 0.7628646156084 deg

Source(s): eq. 3 in http://ioannis.virtualcomposer2000.com/spectroscop...- Login to reply the answers

- WendyLv 44 years ago
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