Monday

March 30, 2015

March 30, 2015

Posted by **Joseph** on Sunday, March 7, 2010 at 11:26pm.

1. The index of refraction for a prism is given by the equation:

n = 1.58 − (6.41 x 10^4)λ

where λ is the wavelength in meters. Light is incident at 28°.

Find the exit angle ß for light of

a) 800 nm

b) 633 nm

c) 540 nm

2. Will lower wavelengths be allowed to exit the prism? Why or why not?

- Physics -
**drwls**, Monday, March 8, 2010 at 8:04amUse the given wavelength to compute the refractive index at that wavelength, then use Snell's law (twice) to get the refraction angles in and out. You have not said what the angle of the prism is. Shall we assume the cross section is an equilateral triangle? You need to know it when computing the exit angle.

At a wavelength of 540*10^-9 m, the refractive index is 1.545. That is not a lot different from the value at 800 nm, but different enough to disperse a good spectrum. Light incident at any angle will be able to enter the prism, and will find a way to get out somehow.

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