Young's double slit equation problem

[sincosine]

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.

 

[etotheipi]

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.

 

[sincosine]

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.

 

[etotheipi]

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.

 

[sincosine]

rom each of the two slits

Thank you, i get it now.

 

[vela]

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.

 

[etotheipi]

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.

 

[vela]

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.

 

[etotheipi]

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!