Monochromator and changing grating question.

[hlee]

Homework Statement

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

Homework Equations

nλ=d (sini + sinj)
sin(a+b)=sin(a)cos(b)+sin(b)cos(a)

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 won't produce -1.6 degree?

any help would be greatful. Thank you.

 

[mark726]

Dear forum post,

Thank you for your question. I am happy to assist you with finding the angle of rotation for the grating monochromator. First, let's review the information given in the problem. We know that the incident angle (Ain) and the output angle (Aout) are both 50 degrees. We also know that the blaze density of the grating is 1200 lines/mm. Finally, we are looking for the angle of rotation for the grating to allow the first order diffraction peak for light of wavelength 615 nm to pass through the exit slit.

To solve this problem, we will need to use the sum-angle trigonometric identity, as mentioned in the hint. This identity states that sin(a+b)=sin(a)cos(b)+sin(b)cos(a). In this case, we can use it to find the angle of rotation, let's call it x, by setting a to be the incident angle (50 degrees) and b to be the output angle (also 50 degrees). We also know that the diffraction order (n) is 1 and the wavelength (λ) is 615 nm. Therefore, our equation becomes nλ=d (sini + sinj), which we can rewrite as 1(615 nm)=1200 lines/mm (sin50 + sinx).

Now, we can solve for x by rearranging the equation to get sinx = (615 nm)/(1200 lines/mm) - sin50. Using a calculator, we find that sinx = 0.0005. To find the angle x, we need to take the inverse sine of 0.0005, which gives us x = 0.03 degrees. Therefore, the grating needs to be rotated by approximately 0.03 degrees for the first order diffraction peak for light of wavelength 615 nm to pass through the exit slit.

I hope this helps you solve the problem. Keep in mind that this solution assumes that the grating is rotated around the entrance slit, and not the exit slit. If the grating is rotated around the exit slit, the solution will be different. Also, make sure to double check your calculations and units to ensure accuracy. Let me know if you have any further questions. Good luck with your research!



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