Single Slit Diffraction and Interference Maxima

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Discussion Overview

The discussion revolves around the conditions for interference maxima and diffraction minima in a single slit diffraction experiment. Participants explore the relationships between slit width, wavelength, and the resulting intensity patterns on a screen, focusing on the mathematical expressions governing these phenomena.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant proposes that the condition for interference maxima is given by the equation asinΘ=nλ, where n is a positive integer.
  • Another participant questions the clarity of the initial query and suggests that the intensity of interference maxima may decrease with distance from the center, indicating a need for specificity in the question.
  • A participant seeks clarification on the condition for interference minima, presenting two potential equations: asinΘ=(2n-1)λ/2 and asinΘ=(2n-1)λ, and asks which is correct based on different path difference considerations.
  • Another participant reiterates the conditions for minima and maxima, stating that for minima, the equation asinΘ=nλ applies, while for maxima, the equation is asinΘ=n(λ+1/2).
  • A later reply confirms the expression for diffraction minima but distinguishes it from the double-slit interference maxima equation, which involves a different variable for slit separation.

Areas of Agreement / Disagreement

Participants express differing views on the conditions for interference minima and maxima, with no consensus reached on the correct equations or their implications for intensity. The discussion remains unresolved regarding the specific conditions and their effects.

Contextual Notes

Participants reference different equations for minima and maxima without resolving the assumptions or conditions under which each applies. There is also a lack of clarity regarding the path difference considerations in the context of single slit diffraction.

Das apashanka
In a single slit experiment if the condition of interference maxima be asinΘ=nλ where n=1,2,3,4...
and the condition of diffraction minima is also the same
Will it not cause any effect on the intensity of the interference maxima as both are at a same distance from the central maximum on the screen?
a is the length of the slit
 
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Das apashanka said:
Will it not cause any effect on the intensity of the interference maxima as both are at a same distance from the central maximum on the screen?
I don't understand your question.

Is your question referring to the decreasing intensity of the interference maxima as the distance increases from the center? Could you please be more specific or clear?

By the way, the interference and destruction maxima and minima have different equations for single slit interference. You can find the minima from m, but the maxima from [m+1/2].
 
Actually I want to clarify that for a single slit experiment if ,a be the slit width
Then what will be the condition of interference minima
Is it asinΘ=(2n-1)λ/2 ,n=1,2,3 4...taking path difference between rays from the two edges
Or
asinΘ=(2n-1)λ,n=1,2,3,4...taking path difference between two rays from the centre and from one edge
 
Das apashanka said:
Actually I want to clarify that for a single slit experiment if ,a be the slit width
Then what will be the condition of interference minima
Is it asinΘ=(2n-1)λ/2 ,n=1,2,3 4...taking path difference between rays from the two edges
Or
asinΘ=(2n-1)λ,n=1,2,3,4...taking path difference between two rays from the centre and from one edge

If a is the slit width, θ is the angle of the minima, λ is the wavelength of the light, and n is a value, then this equation will apply for the destruction minima:

a sinθ = nλ

For interference maxima, the following equation applies:

a sinθ = n(λ+1/2)
 
Das apashanka said:
In a single slit experiment if the condition of interference maxima be asinΘ=nλ where n=1,2,3,4...
and the condition of diffraction minima is also the samet

That's the correct expression for diffraction minima where ##a## is the slit width.

The expression for double-slit interference maxima is ##d \sin \theta = n \lambda## where ##d## is the slit separation distance (and ##n## can also equal ##0## for the central maximum).
 

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