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PHYSICS 1251 HW 12 #9

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I get part a, how do you do part b? I'm very confused. There is no D.
asked Nov 15, 2014 in Intro College Physics by anonymous

1 Answer

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For my version of the question:

Light of wavelength = 585 nm passes through a pair of slits that are 15 μm wide and 240 μm apart.
How many bright interference fringes are there in the central diffraction maximum?
How many bright interference fringes are there in the whole pattern?​

Bright fringes of interference patterns happen at ||dsin\theta=n\lambda||,   ||n=0, \pm 1,\pm 2 ...||.

The first order dark fringes of a diffraction pattern are at ||asin\theta=\lambda  ||.

The largest n for bright fringe within the first maximum of the diffraction pattern has to agree with $$sin\theta=\frac {n\lambda}{d}=\frac {\lambda}{a}$$.

Therefore, ||n=d/a=16||. Plus the center maximum where n=0, we have a total of 33.

For the second part, we need to consider for the entire ||\pm \pi/2||. $$dsin(\theta=\pi/2)=n\lambda$$
Then ||n=d/\lambda=410||. Here the largest possible integer is taken. Plus the central maximum, we have a total of 821 bright fringes.

-- the above seem to be the key of the problem but these have missed the consideration of overlapping a bright interference fringe with a dark diffraction fringe. If the overlaps are considered the number will be smaller. The first part will be 31 and the second part will need to calculate the additional overlapping locations -- should be about 40 fewer.
answered Nov 16, 2014 by PhysicsDude (21,750 points)
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