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Young’s double slit experiment

  • Thread starter aChordate
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  • #1
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Homework Statement



In a version of Young’s double slit experiment, the slits are separated by a distance of
1.0 mm. The light from the slits is observed on a screen that is 12 m from the slits. If the
wavelength of the light is 550 nm, what is the linear distance on the screen between the
first order (m = 1) and second order (m = 2) dark fringes?


Homework Equations



sinθ=(m+1/2)λ/d

The Attempt at a Solution



sinθ=(1+1/2)550nm/1mm=8.25*10^-4
θ=0.04727°

sinθ=(2+1/2)550nm/1mm=0.001375
θ=0.07878°

tanθ=y/L

y = Ltanθ = 12tan0.04727
y=0.00990 m

y = Ltanθ = 12tan0.07878°
y=0.0165 m

0.165-0.00990=0.00660 m
 

Answers and Replies

  • #2
Simon Bridge
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Cool!
Was there a question in all that?

I see your attempted solution is incomplete - no reasoning provided, and no conclusion.
Why did you pick that particular equation?
 
  • #3
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What do you mean, "was there a question in all that?"
 
  • #4
Simon Bridge
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What do you mean, "was there a question in all that?"
Exactly what I said - you wrote quite a lot, but you have not asked a question.
The most important part of science is asking questions - you should practise.

In the English language, a question is a sentence that ends in a question mark and frames some sort of query.
In your entire post, the only question mark appears in the problem statement - that is a question being asked of you, as homework, so I am not allowed to answer it.

You have done the physics problem, and shown your working.
That's fine - but so what?

Perhaps you wanted to know something?
 
  • #5
ehild
Homework Helper
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aChordate,

Your solution is correct, I think, you just wanted us to check it. Simon misses the question "Is my solution correct?"

But a solution is really good if you explain what you do and why.
During the solution, do not use the same notation for different things. If y is the position of a dark spot, write y1 for the first one and y2 for the second one. At the end, give the result in a sentence: " the distance between the first and second dark fringes on the screen is d=y2-y1=..."

By the way, the first dark fringe that appears in the Young experiment corresponds to 1/2 λ/D, when m=0 (between the central maximum and the first maximum)

ehild
 
  • #6
Simon Bridge
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Simon misses the question "Is my solution correct?"
... which does not appear in post #1 at the time of writing this.
I could "infer" this question - but it could be that OP has a specific concern that should be addressed.
I could infer the question, "How can I improve on this?", which changes the style of the reply.

Like I said, getting an explicit question helps show how the OP is thinking about the problem.
Answers in related thread has apparently shown that OP does not understand why the equation chosen is correct.

Technically, the distance between any two adjacent dark fringes will be, given the geometry, the same - and that will be the same as the distance between two adjacent bright fringes too.

Note: 0.07878deg = 0.001375rad,
tan(0.001375rad)=0.001375 ...
sin(0.001375rad)=0.001375 ...
... should suggest an approximation that would help you in future.
 
Last edited:

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