# Young's double slit equation problem

• sincosine
In summary, the conversation discusses the formula for destructive interference, which states that for destructive interference to occur, there needs to be an integer plus a half wavelengths worth of path difference. This stems from the phase difference of light needing to be an odd number of pi's for destructive interference. The formula for diffraction is also mentioned, with the correct formula for the first minimum being a/2 sinθ = λ/2, and the formula for the m-th order minimum being a sinθ = mλ. The conversation also mentions the use of a constant in these formulas.
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.

sincosine said:
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.

sincosine said:
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.

etotheipi said:
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.

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.

sincosine said:
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
etotheipi said:
rom each of the two slits
Thank you, i get it now.

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

etotheipi said:
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.

vela said:
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.

etotheipi said:
I thought the context here was that of double slit interference.
sincosine said:
And also for light diffraction d/2sinθ =λ/2, this one is correct but how do i insert constant here.

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!

## 1. What is the Young's double slit equation problem?

The Young's double slit equation problem is a physics problem that involves determining the interference pattern created by light passing through two parallel slits. It is named after Thomas Young, who first described the phenomenon in the early 1800s.

## 2. How is the Young's double slit equation problem solved?

The problem is solved using Young's double slit equation, which is:
I = I0cos2(πd sinθ/λ),
where I is the intensity of the light at a given point, I0 is the maximum intensity, d is the distance between the slits, θ is the angle of observation, and λ is the wavelength of the light.

## 3. What is the significance of the Young's double slit equation problem?

The Young's double slit equation problem is significant because it demonstrates the wave nature of light. The interference pattern created by the two slits is a result of the superposition of two waves, showing that light behaves as a wave rather than a particle in certain situations.

## 4. What factors affect the interference pattern in the Young's double slit equation problem?

The interference pattern is affected by several factors, including the distance between the slits, the wavelength of the light, and the angle of observation. Changing any of these parameters will result in a different interference pattern.

## 5. How is the Young's double slit equation problem used in real-world applications?

The Young's double slit equation problem has many applications in modern technology, such as in the design of diffraction gratings used in optical instruments like spectrometers and telescopes. It is also used in the study of quantum mechanics and the behavior of particles at the subatomic level.

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