Single slit diffraction maxima

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Homework Help Overview

The discussion revolves around the calculation of the distance between adjacent maxima in single slit diffraction patterns, specifically in terms of the slit width (a), wavelength (λ), and distance to the screen (D). Participants are exploring the conditions for constructive and destructive interference in this context.

Discussion Character

  • Conceptual clarification, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants are attempting to relate the conditions for maxima to those for minima in single slit diffraction, questioning why the same reasoning does not apply. There are discussions about the geometric derivation of interference conditions and the lack of a straightforward formula for maxima.

Discussion Status

The discussion is active, with participants raising questions about the reasoning behind the absence of a simple argument for maxima compared to minima. Some guidance has been offered regarding the conditions for interference, but there is no consensus on a clear approach to finding the maxima.

Contextual Notes

Participants are navigating the complexities of single slit diffraction, including the distinction between constructive and destructive interference. There is an acknowledgment of varying educational levels among participants, which may influence the depth of discussion.

ruku320
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Hi, I was wondering if anyone could help me with this problem.

Calculate the distance y between adjacent maxima in single slit diffraction patterns. Your answer should be given in terms of a, λ and D. (a is the length of the slit, D is the distance between the slit and the screen and λ is the wavelength of the light).

Ok, so I know how to get the minima of single slit diffraction. You just break it down into many rays going through the slit to get a*sin O=nλ. So is it just the same for the maxima? Say for two rays you would get (a/2)*sin O=nλ (nλ since you want the two rays to constructively interfere and when its nλ they are perfectly in phase) so in general it would be be a*sin O = 2nλ. This doesn't really make much sense though...cause the 2nλ distances are just multiples of nλ which is the minima.
 
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For a single slit experiment, the condition for constructive interference is:

dsin\theta = (n + \frac{1}{2})\lambda

which is derived geometrically.
 
While it's easy to find the minima for the single slit diffraction pattern, there's no simple formula for the maxima. The positions of the minima are given by:
dsin\theta = n\lambda

The maxima are not found exactly in the middle between the minima. See here for details: http://scienceworld.wolfram.com/physics/FraunhoferDiffractionSingleSlit.html

(Depending upon the level of the course, they may want you to use mezarashi's formula as a rough approximation, but realize that this is not really correct.)
 
Hi everyone. Could anyone please explain why there is no simple argument to understand where the maxima of a single slit diffraction pattern are found? Why doesn't the same argument as for the minima work? I'm a freshman in university btw so you know what level of education I have.Thanks!
 


Hi,I have been facing some problem in single slit diffraction and need precise description.

we know from single slit diffraction,in term of destructive interfere a sinθ=nλ and constructive
interfere a sinθ=(2n+1)λ/2.Here (a is the length of the slit, D is the distance between the slit and the screen and λ is the wavelength of the light and θ is the diffraction angle).

But we know from the constructive interference in term of sound or light, path difference=nλ (even multiples of λ) and in case of destructive interference,path difference=(2n+1)λ/2(uneven multiples of λ) .

Could anybody describe thoroughly, in term of single slit diffraction,why the condition of constructive and destructive interference place opposite to each other. Please anyone solve my problem.
 

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