Calculating Single Slit Diffraction Width with 630 nm Light

AI Thread Summary
The discussion focuses on calculating the width of a single slit illuminated by 630 nm light, resulting in a diffraction pattern observed 2.8 m away. The linear distance between the first and second dark fringes is given as 12 cm. The initial calculation for the angle theta was found to be incorrect due to a missed factor of 2 in the formula. Participants suggest writing expressions for the positions of the dark fringes to clarify the calculations. The final width of the slit was confirmed to be 29.4 micrometers after correcting the approach.
muffintop
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Homework Statement


A single slit is illuminated with 630 nm light, and the resulting diffraction pattern is viewed on a screen 2.8 m away.
If the linear distance between the first and second dark fringes of the pattern is 12 cm, what is the width of the slit?

Homework Equations


y=Ltan\theta
W = \lambda m / sin \theta

The Attempt at a Solution


theta = arctan (.12/2.8) = 2.45
W = 630 x 10^-9 x 2 / sin 2.45
 
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Looks good.
 
I got 29.4 micrometers. It's wrong though.
 
muffintop said:
W = 630 x 10^-9 x 2 / sin 2.45
Oops... I missed that factor of 2 before. Why is it there?

Hint: Write expressions for the first dark fringe position and the second dark fringe position, then subtract.
 
I got it thank you so much!
I always get my "m" values wrong.
 
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