Please help me with..Destructive Interference

In summary, the required distance from the slits to a screen if the spacing between the first and second dark fringes is to be 4mm is 2.61 meters.
  • #1
imatreyu
82
0

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.
 
Last edited:
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  • #2
For the first dark fringe, m=0, and m=1 for the second one.

ehild
 
  • #3
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.
 
  • #4
Write the equation for Y-s of the first and second dark fringe in terms of d and L. The distance between these fringes should be 4 mm.

ehild
 
  • #5
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:
  • #6
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
 
  • #7
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. . .
 
  • #8
You need to know the meaning of the symbols in the formula you use. Tell me please what are λ, d, L and Y? Check your lecture notes of your textbook.

ehild
 

Related to Please help me with..Destructive Interference

1. What is destructive interference?

Destructive interference is a phenomenon that occurs when two waves of equal frequency and amplitude meet and cancel each other out. This results in a decrease in the overall amplitude of the wave.

2. How does destructive interference differ from constructive interference?

Unlike constructive interference, where waves of equal frequency and amplitude combine to increase the overall amplitude, destructive interference causes the waves to cancel each other out and decrease the amplitude.

3. What causes destructive interference to occur?

Destructive interference occurs when two or more waves meet and their individual amplitudes are out of phase, meaning they do not align with each other. This causes the waves to cancel each other out, resulting in destructive interference.

4. Can destructive interference be beneficial in any way?

While destructive interference is typically seen as a hindrance, it can be beneficial in some cases. For example, it is used in noise-cancelling technology, where two sound waves with opposite phases cancel each other out, reducing unwanted noise.

5. How is destructive interference measured?

Destructive interference is measured by the resulting amplitude of the wave after the interference occurs. If the amplitude decreases, it is a sign of destructive interference. The phase difference between the waves can also be measured to determine the degree of destructive interference.

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