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Diffraction gratings and intensity

  • Thread starter Deneb Cyg
  • Start date
  • #1
10
0

Homework Statement



The interference pattern in this image:
http://session.masteringphysics.com/problemAsset/1074248/3/22.P45.jpg
is seen on a screen 1.0m behind an 950 lines/mm diffraction grating.

What is the wavelength of the light? (expressed in nanometers to 2 sig figs)

Homework Equations



I know that the equation relating the variables is: d sin(theta)=m(lambda) where theta is the angle from the center maximum to the mth maxima. d being the distance between slits in the grating.

And using the small angle approximation I think you can replace sin(theta) with y/L where y is the distance between the central maximum and the mth one and L is the distance of the screen from the grating

The Attempt at a Solution



I tried plugging the variables into the equation but I can't seem to get the right answer.

ie:
d(y/L)=m(lambda)
(1/950000)(.436/1)=1lambda
lambda=460nm which is wrong

What am I doing wrong?
 

Answers and Replies

  • #2
rl.bhat
Homework Helper
4,433
5
0.436/1 is tanθ. Find sinθ
 
  • #3
10
0
Thank you. That worked. I can't believe I missed that.
 

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