(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

A simple grating monochromator requires the incident light to arrive from the entrance slit at angle "A incident" relative to the "neutral" position, and only light reflected at angle "Aout" will travel through the exit slit. For this monochromator, Ain = Aout = 50 degrees.

If the grating has a blaze density of 1200 lines/mm, at what angle would the grating need to be rotated to to allow the first order diffraction peak for light of wavelength 615 nm to pass through the exit slit?

Give your answer in degrees.

Hint: You will need to use the sum-angle trigonometric identity to solve this question

2. Relevant equations

nλ=d (sini + sinj)

sin(a+b)=sin(a)cos(b)+sin(b)cos(a)

3. The attempt at a solution

using using nλ=d (sini + sinj) I found that when grating angle is not changed, the 1st order reflection would be at -1.6 degrees. I am not quite sure how to use sum angle to find angle of rotation of grating so that light will exit thru the exit slit.

I would assume grating normal changes thus incident light changes and thus output wont produce -1.6 degree?

any help would be greatful. Thank you.

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# Monochromator and changing grating question.

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