Diffraction gratings and intensity

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The discussion revolves around calculating the wavelength of light using a diffraction grating with 950 lines/mm and an interference pattern observed on a screen 1.0m away. The relevant equation is d sin(theta) = m(lambda), where d is the distance between slits, and the small angle approximation allows sin(theta) to be replaced with y/L. The user initially miscalculated the wavelength by incorrectly using tan(theta) instead of sin(theta). After correcting this mistake, they successfully found the right wavelength, demonstrating the importance of accurate trigonometric functions in such calculations. The conversation highlights common pitfalls in physics problem-solving related to diffraction gratings.
Deneb Cyg
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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?
 
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0.436/1 is tanθ. Find sinθ
 
Thank you. That worked. I can't believe I missed that.
 
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