Light: Fringes and Interference

  • #1
Coherent light with wavelength 597 nm passes through two very narrow slits, and the interference pattern is observed on a screen a distance of 3.00 m from the slits. The first-order bright fringe is a distance of 4.84 mm from the center of the central bright fringe.

For what wavelength of light in micrometers will the first-order dark fringe be observed at this same point on the screen?

To solve this, I must use the formula for destructive interference:
d*sin(theta) = (m + 1/2)*lambda, where m= 0 or -1 since it is the first dark fringe,

but I am stuck on using small-angle approximations where sin(theta) = tan(theta) to find d.

Could someone please explain this method to me clearly?

After finding d, what formula must I use for the 2nd beam's interference pattern to find the 2nd wavelength?

Thanks.
 

Answers and Replies

  • #2
Doc Al
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Bright fringes are found where [itex]d \sin\theta = m\lambda[/itex]; dark fringes where [itex]d \sin\theta = (m + 1/2)\lambda[/itex]. m = 0, 1, 2, etc.

The position (y) of the fringes is given by [itex]\tan\theta = y/L[/itex]. For small angles [itex]\tan\theta = \sin\theta = \theta[/itex]. So the bright fringes are found at [itex]y = m\lambda L /d[/itex]; the dark fringes at [itex]y = (m + 1/2)\lambda L /d[/itex].

Note that d, L, and y stay the same for the two wavelengths. Hint: Solve for the new wavelength symbolically before plugging in any numbers.
 
  • #3
What value is y in the problem? 4.84 mm?
 
  • #4
Doc Al
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Soaring Crane said:
What value is y in the problem? 4.84 mm?
That's correct. Note that the same value of y is used for both wavelengths:
For the first, y = 4.84mm is the postion of the first bright fringe
For the second, y = 4.84mm is the postion of the first dark fringe​

Hint: You don't really need to plug in that y value! What you need to do is find the second wavelength in terms of the first.
 
  • #5
How do I do this? Do I use the formula for bright fringes for the second beam as a starting point?
 
  • #6
Doc Al
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Here's what I suggest: Call the first wavelength [itex]\lambda_1[/itex] and the second [itex]\lambda_2[/itex]. Now write the appropriate equation for each and compare. I'll write the equation for the first wavelength. Since it gives the location of the first bright fringe, I'll set m = 1:
[itex]y = \lambda_1 L /d[/itex]
Rearranging that gives you: [itex]\lambda_1 = yd/L[/itex]

Now you write the equation that applies to the second wavelength and compare.
 
  • #7
Well, for the second, is it this?

y = [(m + 1/2)*lambda_2*L]/[d], where m = 0

lambda_2 = (2*y*d)/L
 
  • #8
Doc Al
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Exactly. So... how does [itex]\lambda_2[/itex] compare to [itex]\lambda_1[/itex]?
 
  • #9
It's twice the value of lambda_1.
 
  • #10
Doc Al
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Correct. Now express the answer in micrometers and you're done.
 

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