# Homework Help: Two source interference determining wavelength

1. Apr 30, 2012

### Alchemist90

1. The problem statement, all variables and given/known data
A laser with wavelength d/8 is shining light on a double slit with slit separation 0.300 . This results in an interference pattern on a screen a distance L away from the slits. We wish to shine a second laser, with a different wavelength, through the same slits.

What is the wavelength λ2 of the second laser that would place its second maximum at the same location as the fourth minimum of the first laser, if d= 0.300 ?

2. Relevant equations

dsinθ=mλ
dsinθ=(m+1/2)λ
λ1=d/8 <---- not sure about this one but problem statement suggests it.
d=3*10-4m

3. The attempt at a solution
sinθ=((m+1/2)λ1)/d where m=-4
sinθ=((m+1/2)λ2)/d where m= 2

set equal and i got -0.0525mm.

im not sure where I'm misunderstanding but it could be my understanding of m values, reading the problem wrong or simply a misunderstanding of the whole question posed. Any takers?

Last edited: Apr 30, 2012
2. Apr 30, 2012

### Alucinor

Check your equations for maxima location versus minima location, there is a small difference between the two that might help you solve your problem. You're trying to put a minimum where a maximum should be, however in your attempt you're working only with maxima. You also want to be sure you don't end up with that sign error, there is a symmetry that you should be looking for that makes all of those m's positive.

Also, be careful with your units, you're saying d=0.300 (whats?)

3. Apr 30, 2012

### Alchemist90

yea I've gotten closer.

d=0.3mm (millimeter)

oh and i was being stupid about m values

since m starts at m=0 its max at m=1 and min at m=3

now i tried dsinθ=(3+1/2)λ1 where λ1=d/8
then sub in for dsinθ=λ2
thus I get (3.5/8)d=λ2

which gives 0.13125mm but thats not right either.

Last edited: Apr 30, 2012
4. Apr 30, 2012

### Alucinor

You're still saying that the "4th minimum" is a maximum in your equation.
$$dsin\theta=\left(m+\frac{1}{2}\right)\lambda$$ is a maximum location.

5. Apr 30, 2012

### Alchemist90

That is the condition for single slit light diffraction where dsinθ=mλ is a local minimum. This is double slit diffraction. where maximum locations are dsinθ=mλ where m=0,1,2,3......etc
minimum locations are dsinθ=(m+1/2) where m=0,1,2,3.....etc

6. Apr 30, 2012

### Alchemist90

i got it.............after requesting the answer. your analysis was incorrect.
dsinθ=mλ is a maximum where m=1,2,3....etc <----- watch for this
dsinθ=(1+1/2)λ is a minimum where m=0,1,2,3.....etc

Don't make my mistakes.

Replier: don't listen to him he's wrong.

7. Apr 30, 2012

### Alucinor

Oh goodness, my mistake. I was reading it as a single slit.