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A light beam illuminating a double slit consists of two wavelengths, 620 nm and an unknown wavelength λ. The 8th bright fringe of the unknown wavelength overlaps with the 7th dark fringe of the 620 nm light. What is λ?

I understand you just set the constructive and destructive formulas equal to each other and solve for the unknown λ, but I don't understand why we use (m-1/2)λ and not (m+1/2)λ.

When I did mλ (bright spot) = (m+1/2)λ (dark spot) and solved for the λ for the dark spot, I got 584 nm, but when I checked the steps to solve it, my professor did (m-1/2)λ and got 504 nm.

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You can use either as long as you know what you mean. The first dark fringe occurs when the path length difference is equal to $\lambda/2$, so as long as your equation gets that then you're good to go.

If you use $m+1/2$, then $m$ needs to be offset by $1$, e.g. the first dark fringe is $m=0$, the second is $m=1$, etc.

If you use $m-1/2$, then $m$ does not need to be offset by $1$, e.g. the first dark fringe is $m=1$, the second is $m=2$, etc.

So the latter seems more intuitive, but either is fine as long as you know what you are doing. Sometimes the condition is even given as $m/2$ for $m=1,3,5,\dots$, which is also correct but more work to think about the "nth" dark fringe.

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  • $\begingroup$ Ah, I see. So if I were to do what I was doing earlier by setting mλ = (m+1/2)λ, I would have to set m as 6 for the dark fringe instead of 7 since it needs to be offset by one. Thank you very much! Is there anything like that for bright fringes, or would I always just set m as whatever nth bright fringe it is in the problem for something like this? $\endgroup$
    – Cimaster
    Commented Aug 4 at 1:26
  • $\begingroup$ @Cimaster It's the same thing; you want integer multiples of the wavelength. There are multiple functions that can give you that. Also, your work is imprecise in your whole problem since you're using the variable $m$ to label two different things $\endgroup$
    – BioPhysicist
    Commented Aug 4 at 3:09
  • $\begingroup$ I see, thanks. And yes, I realize that it's imprecise; I just didn't bother with proper subscripts to label correctly. Moving forward, I'll make sure to do that so I don't mess up my questions on here. $\endgroup$
    – Cimaster
    Commented Aug 4 at 18:37

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