Two source interference determining wavelength

[Alchemist90]

Homework Statement

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 ?

Homework 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

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?

 

[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?)

 

[Alchemist90]

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 that's not right either.

 

[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.

 

[Alchemist90]

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

 

[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.

 

[Alucinor]

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