How is the amplitude of a wave related to the slit width?

AI Thread Summary
The amplitude of a wave is directly proportional to the width of the slit through which it passes, meaning that increasing the slit width results in a greater amplitude. This relationship is rooted in the principle that intensity, which is proportional to the square of amplitude, also increases with wider slits. In single slit diffraction, the amplitude is derived from dividing the slit into segments, with the total amplitude being the superposition of these segments. The original question's reference to 'slit separation' may have caused confusion, as it likely pertains to the intensity of the central maximum in a double slit experiment. Understanding these principles clarifies how slit width impacts wave amplitude and intensity.
Faiq
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Homework Statement


How is the amplitude of a wave affected if the slit separation remains same but slit width is
1.)Increased?
2.)Decreased?
and why?

Homework Equations


Intensity is proportional to the square of amplitude

The Attempt at a Solution


I consider the amplitude as allowed "fatness" of a wave. The more slit width there is, the more wave can be "fat" and so it will have a greater amplitude.
 
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As you wrote intensity is proportional to the square of amplitude.
Do you know the relation between intensity of wave and the width of slits?
The intensity of light due to a slit (source of light) is directly proportional to width of the slit.
Therefore if the two slits ##S_1## & ##S_2## have widths ##W_1## & ##W_2## respectively and ##I_1## & ##I_2## are intensities of light waves having amplitude ##a_1## and ##a_2## from respective slits then,

##\frac{I_1}{1_2}## ∝ ##\frac{a_1^2}{a_2^2}## ∝ ##\frac{W_1}{W_2}##
 
Yes, I know that. I am asking why does the amplitude increases with increasing slit width
 
Faiq said:
Yes, I know that. I am asking why does the amplitude increases with increasing slit width
In single slit diffraction calculations, the resultant amplitude is obtained by dividing the slit width into a large number of equal segments. For each segment, the amplitude is taken proportionally equal and the resultant amplitude is found by superposition of amplitudes of all the segments at the point of consideration.
So amplitude should be proportional to slit width.
 
diffraction.png


Can you see division of the slit width into a large number of equal segments?
 
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Then (amplitude)^2 is proportional to slit width or it is amplitude proportional to slit width??
 
The amplitude of the wave passing through a slit is directly proportional to its width. If the slit width is x and amplitude of the wave is A, then A ∝ x.
 
HI @ube. For information, you are replying to a very old (April 2016) thread.

Note, the original question is unclear/incomplete. The Homework Statement refers to 'slit separation' - I would guess the question is actually about the intensity of the central maximum in a Young's double slit experiment.

If you have your own question, your best bet is to post a new thread and include your own attempt/reasoning at the answer.

Edit - typo' corrected.
 
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