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Find wavelength in double slit experiment

  • Thread starter k77i
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  • #1
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Homework Statement



Two narrow slits are 0.4 mm apart. The dark fringe of order 5 is 1.2 degrees from the central bright fringe. What is the wavelength of the light in nm?


Homework Equations



dsin(theta) = (m +1/2)lambda
where d = distance between silts
m=order
lambda = wavelength

The Attempt at a Solution



It seems simple enough right? I converted the 0.4mm into meters and plugged all the values into the equation which gives me:

(4x10^-4)(sin1.2) = (5 + 1/2)lambda

then lambda = (4x10^-4)(sin1.2)/5.5

Since the answer i get is in meters i have to multiply it by 10^9 right? but that answer is way too big. What am I doing wrong?
 

Answers and Replies

  • #2
ehild
Homework Helper
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1523 nm is in the infrared range of the optical spectrum. Why do you think that it is too big?

ehild
 
  • #3
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1523 nm is in the infrared range of the optical spectrum. Why do you think that it is too big?

ehild
Because according to my homework set that answer's not right. The wavelengths we're using in class is visible light between 400nm and 700nm and the answer I'm getting from my calculations is in the thousands.
 
  • #4
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I find that your solution is correct. I obtained the same result, 1523 nm for the wavelength. You never know, sometimes homework sets/ exercises in books are not 100% fine.
 
  • #5
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Well ok but i didnt get 1523 nm as an answer either. Did I use a wrong equation or wrong values somewhere?
 
  • #6
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No, I don't think so. the distance between slits (a) multiplied by the sin of the angle given (sinQm) equals wavelength (landa) x (order). In this case, you are refering to dark fringes and the order gets a 1/2 added to it.
So a.sinqm=landa.(5+0.5)
comes out to be 1523 x 10^-9 m.
The equation is okay, but whether or not the values are wrong I can't know, you have the exercise.
 
  • #7
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Yes all the values I posted are exactly the ones from the exercise. I set up the equation exactly like that too. There must be something really obvious that I'm not getting. Let me do it step by step to see exactly where I'm messsing up.

dsin(angle) = lambda(m+0.5)
lambda = [dsin(angle)]/(m+0.5)
lambda = [(4*10-4)(sin1.2)]/5.5
lambda = 3.73*10^-4/5.5

This gives me lambda = 6.78 * 10^-4 m and so when I convert it to nm its definitely not 1523 nm. So exactly what's my mistake?
 
  • #8
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(4*10-4)(sin1.2) actually comes up to ~ 8.377 x 10^-6 which divided by 5.5 gives ~ 1.523 x 10 ^ - 6m or 1523 nm. Remember, you are working with degrees, not radians.
 
  • #9
28
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Oh wow i get it. Ya it really was something obvious after all
 

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