Diffraction Grating wavelengths

AI Thread Summary
The discussion focuses on calculating the angles for the first and second-order maxima of light wavelengths 520nm and 630nm passing through a diffraction grating with 6000 lines/cm. Participants clarify that different angles are needed for each wavelength when determining the maxima. The lowest value of m for which the 520nm line no longer exists is also discussed, emphasizing that sine values cannot exceed one. One participant expresses frustration with previous incorrect answers and seeks guidance for exam preparation. The conversation highlights the importance of understanding the calculations rather than simply obtaining answers.
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light wavelengths of 520nm and 630nm passes through a diffraction grating that contains 6000lines/cm. a) sketch a diagram of the image produced from m=0 to m=2. label the order of each fringe.b) calculate the angles for the first and second-order maxima that would appear on the screen. c) What is the lowest value of m for which the 520-nm line no longer exists?

For part b...what wavlength do i use to find the angles for the first and second oreder maxima?...:S
 
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can anyone help me ...please..
 
Both.
They have different angles in both the first and second maxima.
c)
(10^-2/6000)sintheta =n(520*10^-9)
sintheta cannot exceed one
so calculate the n your own

you should calculate and do homework yourself.
I am to guide you, not to help u finish your homework
 
this is not my homework..It was my homework...but i got it wrong, unfortunately my teacher did not write the correct answer...as my exam is approching...i am preparing for it. I am doing this course through ilc, ilc tutors are not available right now because of the break, that's why i wanted to know the answer to this question.
So for part b) i am supposed to find the first and sec order maxima for both wavelengths?...i see now that's why i got it wrong in first place.
 

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