Maximum Intensity Between Dark Fringes, Diffraction

  • Thread starter JackFlash
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Homework Statement


Laser light of wavelength 632.8 nm falls normally on a slit that is 0.0210 mm wide. The transmitted light is viewed on a distant screen where the intensity at the center of the central bright fringe is 8.50 W/m2.

a) Find the maximum number of totally dark fringes on the screen, assuming the screen is large enough to show them all.

b) At what angle does the dark fringe that is most distant from the center occur?

c) What is the maximum intensity of the bright fringe that occurs immediately before the dark fringe in part (b)? Approximate the angle at which this fringe occurs by assuming it is midway between the angles to the dark fringes on either side of it.

Homework Equations


sinθ = mλ/a
Intensity = I0([itex]\frac{sin(πa*sin(θ)/λ)}{πa*sin(θ)/λ}[/itex])2
Intensity = [itex]\frac{Io}{((m+.5)^2 + π^2)}[/itex]

where
m = a number based on the fringe
λ = wavelength
a = distance between the slits
π = a delicious dessert of varying flavors, and 3.14159

The Attempt at a Solution


I found the first two solutions (66 and 83.9° respectively) with some ease. The last one is what is grinding my gears.

I've calculated the angle between the two dark fringes at the end, m=33 and m=32, and got 83.9335° and 74.63699° respectively. I got the angle between them, 79.29°, by taking the average. I plug it into the first equation for intensity and I get a wrong answer. I try using m=32.5, which would be the bright fringe the question refers to, in the second equation for intensity and still no luck. It might not seem like I've tried at all on this question, but I have run through several different combinations (using different values for m to get an angle, then plugging into the second equation or just using m in the third). The solution is 7.21*10-4, which is about .6*10-4 less than all the solutions I've gotten.

I've a suspicion I may be using the wrong m (32.5) or I'm doing something else odd. Any help is much appreciated.
 
Last edited:

Answers and Replies

  • #2
ehild
Homework Helper
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When you get π*(a/λ) *sin(θ) it is in radians, not degrees. Check if you use "RAD" when calculating sin(π*(a/λ) *sin(θ)) in the numerator of intensity. .

ehild
 
  • #3
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Ah yes. Radians. That was the issue.
That always seems the case. Radians is a reoccuring enemy of mine. Thanks.
 
  • #4
ehild
Homework Helper
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You are welcome and take care. Do not let you beat by the radians. :smile:

ehild
 

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