micole

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White light strikes a diffraction grating (930 lines/mm) at normal incidence. What is the longest wavelength that forms a second-order maximum?
I used the equation [(1/d)sin(90)]/m = wavelength: [(1/0.93m)(1)] /2 = 0.5376m. To convert to nm I multiplied by 10^-9. This is incorrect. Can anyone help? Thanks

Isn't d the reciprocal of 930/mm? or

or 1/d= .001m/930=1.08E-8 m or .108nm?

check my thinking. In your equation, unit wise, if you are to have length on the right, you have to have length units on the left.

I remember the grading equation as

d sinTheta /n= lambda, where d is in meters.
Perhaps that is your error.

  • micole -

    The answer should follow such that
    Lines/mm --> d --> (1/930mm)/1000mm/m
    d = 1.075e-6 meters
    set sin theta = 90o = (1)
    (dsin(Theta))m=mlambda
    (1.075e-6 (1))/2)= 5.375e-7 or 537nm.

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