Young's double slit equation problem

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  • #1
sincosine
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Homework Statement:
help confused guy
Relevant Equations:
Δx=(2k+1)*λ/2
Hello guys i have one problem and i can't find right solution. In some notebooks for destructive interference it says Δx=(2k+1)*λ/2, and it doesn't make sense for me because if i insert k=1 it will be 3λ/2 and it should be λ/2. And also for light diffraction d/2sinθ =λ/2, this one is correct but how do i insert constant here.
 

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  • #2
In some notebooks for destructive interference it says Δx=(2k+1)*λ/2, and it doesn't make sense for me because if i insert k=1 it will be 3λ/2 and it should be λ/2.

What happens if you try ##k=0##?

That formula just means that for destructive interference, you need an integer plus a half wavelengths worth of path difference. A path difference of ##\frac{\lambda}{2}## will give you a dark fringe, as will one of ##\frac{3\lambda}{2}##, as will ##\frac{5\lambda}{2}##.

It stems from the fact that the phase difference of the light, ##\frac{2\pi \Delta O}{\lambda}## (if ##\Delta O## is the optical path difference = ##\sum nd##), needs to be an odd number of ##\pi##'s for destructive interference.

And also for light diffraction d/2sinθ =λ/2, this one is correct but how do i insert constant here.

I'm not sure I understand, perhaps you could explain. Usually we say ##d\sin{\theta} = n\lambda## for constructive interference.
 
  • #3
sincosine
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What happens if you try ##k=0##?

That formula just means that for destructive interference, you need an integer plus a half wavelengths worth of path difference. A path difference of ##\frac{\lambda}{2}## will give you a dark fringe, as will one of ##\frac{3\lambda}{2}##, as will ##\frac{5\lambda}{2}##.

It stems from the fact that the phase difference of the light, ##\frac{2\pi \Delta O}{\lambda}## (if ##\Delta O## is the optical path difference = ##\sum nd##), needs to be an odd number of ##\pi##'s for destructive interference.



I'm not sure I understand, perhaps you could explain. Usually we say ##d\sin{\theta} = n\lambda## for constructive interference.
For example if i need to find distance between centre and third minimum(third dark spot)k=3, i know i have to use

Δx= 5λ/2 and then i can derive it from this formula, but if i try to write it using this formula

Δx=(2k+1)*λ/2, i get that Δx=7λ/2.
 
  • #4
You're getting confused, the condition for destructive interference is that the difference between the path lengths from each of the two slits to the screen is a half number of wavelengths, e.g. ##0.5\lambda, 1.5\lambda## etc.

To actually find fringe spacing, you can derive from that another relation which is usually stated as ##w = \frac{\lambda D}{s}##, where ##w## is the fringe spacing (between two adjacent maxima, or two adjacent minima), ##D## the distance to the screen and ##s## the slit spacing. If you draw a diagram of the fringes, that should help you to then find the distance between the centre and third minimum.

Δx= 5λ/2 and then i can derive it from this formula, but if i try to write it using this formula

Δx=(2k+1)*λ/2, i get that Δx=7λ/2.

##\Delta x = \frac{5\lambda}{2}## is correct for the third minimum, but you should be saying ##\Delta x = (2k+1) \frac{\lambda}{2}## and setting ##k=2##. Because ##k=0## corresponds to the first minimum.
 
  • #5
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  • #6
vela
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And also for light diffraction d/2sinθ =λ/2, this one is correct but how do i insert constant here.

I'm not sure I understand, perhaps you could explain. Usually we say ##d\sin{\theta} = n\lambda## for constructive interference.

For single-slit diffraction, the first minimum occurs where ##\frac a 2 \sin\theta = \frac{\lambda} 2##, where ##a## is the size of the slit. The formula for the ##m##-th-order minimum ends up being ##a \sin\theta = m\lambda##. It looks like the formula for maxima for two-slit interference, but in the case of diffraction, it's the condition for minima.
 
  • #7
For single-slit diffraction, the first minimum occurs where ##\frac a 2 \sin\theta = \frac{\lambda} 2##, where ##a## is the size of the slit. The formula for the ##m##-th-order minimum ends up being ##a \sin\theta = m\lambda##. It looks like the formula for maxima for two-slit interference, but in the case of diffraction, it's the condition for minima.

I thought the context here was that of double slit interference. And the equation ##d\sin{\theta} = n\lambda## follows via consideration of a triangle with hypotenuse ##d## and short side ##n\lambda##, with the approximation that ##D## is sufficiently large for ##\sin^{-1}(\frac{n\lambda}{d}) \approx \sin^{-1}(\frac{w}{D})##.

But if it is single slit interference, then it is indeed correct that we need an integer multiple of wavelengths path difference to a point on the screen for a minimum, which follows from reasoning with Huygen's principle.
 
  • #8
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I thought the context here was that of double slit interference.
And also for light diffraction d/2sinθ =λ/2, this one is correct but how do i insert constant here.
 
  • #9
Point taken 😁. That part must have gone right over my head, I've seen both double slit interference and double slit diffraction used but I think yes you're right in this case. Sorry!
 

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