A double-slit (Young experiment) problem

  • #1

Homework Statement



In Young's double-slit interference experiment with light of wavelength [tex]\lambda[/tex], two light waves come from the slits and are superposed at the point P on the screen. These waves have the following forms:

E1 = E0 sin(wt + π/4)
E2 = E0 sin(wt + Φ)

Find the possible values of Φ if we have destructive interference at the point P.

Homework Equations



27baa5fb481ec2dcaf6cbb38b56f2570.png


The Attempt at a Solution



I attempted the problem two different ways and I had two different results.

Φ = 2kπ - π/4
Φ = 2kπ + 5π/4

Help me out please.
 

Answers and Replies

  • #2
799
0
Can you show us how you got that?
 
  • #3
Phase difference is Φ - π/4.

So d/λ = (Φ - π/4)/2π

d = (2k+1)/2 λ for destructive interference

So, (2k+1)/2 λ = (Φ - π/4)/2π λ

Φ = 2kπ + 5π/4

-------------

The second way, I added 2 wave functions together and I equated the expression to zero.

I found Φ = 2kπ - π/4
 
  • #4
799
0
How did you arrive at Φ = 2kπ - π/4?
 
  • #5
Let's see. The sum of the two waves is:

E = 2E0 sin(wt + π/8 + Φ/2) cos(π/8 - Φ/2) = 0

That is satisfied when

wt + π/8 + Φ/2 = kπ

wt=2πft=2π, so we can drop it.

Φ = 2kπ - π/4
 
  • #6
799
0
Let's see. The sum of the two waves is:

E = 2E0 sin(wt + π/8 + Φ/2) cos(π/8 - Φ/2) = 0

That is satisfied when

wt + π/8 + Φ/2 = kπ

wt=2πft=2π, so we can drop it.

Φ = 2kπ - π/4

The condition E=0 must be satisfied at ALL TIME for destructive pattern, and the bold line is not true at all time.
Find another condition :wink:
 
  • #7
Hmm, thanks hikaru. How can I proceed? I have no idea.
 
  • #8
799
0
What about cos(π/8 - Φ/2)?
 
  • #9
Oh yes, I found it now. I must equate the cosine term (cos(π/8 - Φ/2)) to zero. But why exactly?
 
  • #10
799
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Because of this: "The condition E=0 must be satisfied at ALL TIME for destructive pattern" :wink:
 

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