Diffraction Grating wavelengths

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SUMMARY

The discussion focuses on calculating the angles for first and second-order maxima of light wavelengths 520nm and 630nm passing through a diffraction grating with 6000 lines/cm. The formula used is \( \sin(\theta) = \frac{n \lambda}{d} \), where \( d \) is the grating spacing. The lowest value of \( m \) for which the 520nm line no longer exists is determined by the condition that \( \sin(\theta) \) cannot exceed one. Participants emphasize the importance of independently solving the calculations while providing guidance on the process.

PREREQUISITES
  • Understanding of diffraction grating principles
  • Familiarity with the formula \( \sin(\theta) = \frac{n \lambda}{d} \)
  • Knowledge of light wavelengths and their measurements in nanometers
  • Basic trigonometry for angle calculations
NEXT STEPS
  • Study the derivation of the diffraction grating formula
  • Learn about the impact of grating density on diffraction patterns
  • Explore the concept of maxima and minima in wave interference
  • Practice calculating angles for various wavelengths using different grating spacings
USEFUL FOR

Students preparing for exams in optics, physics educators, and anyone interested in the practical applications of diffraction gratings in experimental setups.

rojasharma
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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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