Please help me with..Destructive Interference

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Homework Statement



Light of wavelength 460 nm falls on two slits spaced .3 mm apart. What is the required distance from the slits to a screen if the spacing between the first and second dark fringes is to be 4mm?

Homework Equations



Ydark= (([tex]\lambda[/tex]L)/d)(m+1/2)
Using
ydark= .004 m
d= .0003 m
[tex]\lambda[/tex]=4.6e-7 m
L= ? (must find)
m= ? (don't understand)

The Attempt at a Solution


It's basically plug n' chug, but I don't understand what m is supposed to be at all!
The answer is 2.61.

My friend says (m + 1/2) must turn into (3/2 - 1/2) but I don't understand why. . .

Thank you in advance for any help! I really appreciate it. My textbook does a bad job of explaining this.
 
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Thank you for your response, but I'm sorry...I still don't understand how that information translates into what I actually need to put into the equation.

If I'm just using the m value for the second fringe, I have to put m=1 into the equation, which does not yield the correct answer.
 
It is.
However, I still don't have a value for m nor understand what it should be.

Ydark= (wavelength*L)/d (m+1/2)
 
Last edited:
imatreyu said:
It is.
However, I still don't have a value for m nor understand what it should be.

Ydark= (wavelength*L)/d (m+1/2)

m is an integer: 0, 1, 2, 3...

You get the first dark band when m=0, and the second one when m=1. Plug in m=0 , to get an equation for Y1 and m=1 for Y2. The difference between Y2 and Y1 equals 4 mm. Find L. Show your work.

ehild
 
From my understanding, I already know Y from the problem?
L= (Ydark * d)/ lambda(m+1/2)
using .004 m for Ydark, .0003 m for d, 4.6e-7 for lambda

When m= 0, L= 5.217...
When m= 1, L=1.7391...

L1-L2= 3.532 m

Which is, unfortunately, not correct. (Correct answer is 2.61 m) Which is weird because it seems very intuitive, and I am disappointed that it does not correct according to the textbook.


I did find a way to do it correctly: .004*.003/ (4.6 e-7*(3/2 - 1/2)) =2.608=2.61 m
However, I don't understand why the (3/2-1/2) part is correct. . .