Question

Problem 10: A broadband light source is composed of wavelength from 400 nm to 800 nm. The beam is incident on a block of BK7 glass at an angle of 30 degree from the normal. What happens to the beam as it refracts though the glass? Answer this quantitatively by describing the refraction of the 400 nm and 800 nm extremes in wavelength.

I believe snells law has to be used but there is a version sin(pheta1)/ sin(pheta2) = n2/n1 = v1/v2

Answer 1

For BK7 glass, the refractive index of 400nm wavelength is 1.5308 and for 800nm is 1.5108. From Snell's law, the angle of refraction for 400 nm will be:

$\theta_r = sin^{-1}\frac{sin30}{1.5308} = 19.064\degree$

while for 800nm wavelength, this is:

$\theta_r = sin^{-1}\frac{sin30}{1.5108} = 19.326\degree$

this difference in angle of refraction leads to dispersion of light where all the colors between 400nm and 800nm are present in the light beam.