## Two source interference determining wavelength

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
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?
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 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?)

 Quote by 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?)
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.

## Two source interference determining wavelength

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.

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

 Tags interference, light, wavelengths