A narrow beam of light with wavelengths from 450 n m to 700 n m is incident perp

Question

A narrow beam of light with wavelengths from 450 nm to 700 nm is incident perpendicular to one face of a 32.00?  prism made of crown glass, for which the index of refraction ranges from n=1.533 to n=1.517 for those wavelengths. What is the angular spread of the beam after passing through the prism? Express your answer using four significant figures. A narrow beam of light with wavelengths from 450 nm to 700 nm is incident perpendicular to one face of a 32.00?  prism made of crown glass, for which the index of refraction ranges from n=1.533 to n=1.517 for those wavelengths. A narrow beam of light with wavelengths from 450 nm to 700 nm is incident perpendicular to one face of a 32.00?  prism made of crown glass, for which the index of refraction ranges from n=1.533 to n=1.517 for those wavelengths. A narrow beam of light with wavelengths from 450 nm to 700 nm is incident perpendicular to one face of a 32.00?  prism made of crown glass, for which the index of refraction ranges from n=1.533 to n=1.517 for those wavelengths. What is the angular spread of the beam after passing through the prism? Express your answer using four significant figures.

Explanation / Answer

ngsin(33) = sin(theta) where theta is the outgoing angle (relative to a line normal to the exit face). there is an angular width because the index of refraction changes with the wavelenght of the light. so, for the largest n we have

ng,maxsin(33) = sin(thetamax)

and for the smallest n

ng,minsin(33) = sin(thetamin).

now you know the two values of n, so you can just calculate sin(thetamax) and sin(thetamin) and get the max and min values of theta... the difference will give you the angular spread of the outgoing beam:

thetamax = 56.61

thetamin = 55.71

so the angular spread is 56.61 - 55.71 = 0.9 degrees.

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