How Many Interference Maxima Appear Within the Central Diffraction Peak?

AI Thread Summary
The discussion centers on calculating the number of interference maxima within the central diffraction peak for light with a wavelength of 530 nm incident on two slits spaced 1.0 mm apart. The initial attempt at solving the problem involves calculating the half-width of the central diffraction peak, resulting in a value of 0.03 degrees. However, the correct calculation reveals that the half-width should be approximately 0.304 degrees. The key takeaway is that the number of maxima is determined by the formula m = d/λ, leading to a total of 21 interference maxima within the central diffraction peak. Understanding the relationship between slit width, spacing, and wavelength is crucial for solving such diffraction problems.
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Homework Statement


Light with a wavelength of 530 nm is incident on 2 slits that are spaced 1.0mm apart. If each slit has a width of .1mm, how many interference maxima lie within the central diffraction peak?

Homework Equations



sin( degree) = m(lambda)/d

The Attempt at a Solution


I have no idea what the question is asking, but I started with:

arc sin ( 1 * 530e-9)/ .001
I got .03 degrees for the half width of the central diffraction.
Double this degree, .06, for the entire width of the central diffraction.

I then did:
[sin(.06) * .001]/530e-9 = m
m = 1.97

So wrong! The answer is 21.
Please help me understand this.
Thanks.
 
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The halfwidth of the diffraction pattern is about lambda/width of the slit, so it is 0.304°.
The maxima of the interference pattern of the two slits are apart by lambda/(distance between the slits).

ehild
 
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