Does bringing the slits together affect the intensity of maxima for double slit experiment and also the diffraction grating? Does decreasing slit width affect the intensity of the maxima for a single slit experiment.
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1$\begingroup$ Do the math and see. $\endgroup$– Jon CusterMay 7, 2017 at 15:22
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2$\begingroup$ please give an explanation, even a crude one. I will accept it $\endgroup$– user86425May 7, 2017 at 17:15
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$\begingroup$ I can understand that the fringe spacing decreases cannot make sense of the intensity $\endgroup$– user86425May 7, 2017 at 17:23
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$\begingroup$ yeak ok, I have researched that intensity is $cos^2(\frac{d\pi sin\theta}{\lambda})$. So for small theta, (the central maxima), reducing $d$ will get you a greater intensity. Problem is 1 i don't understand it and 2 it's the wrong answer on a multiple choice in an exam question i have. So I struggle to verify the correct answer and neither can i explain it $\endgroup$– user86425May 7, 2017 at 17:32
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1 Answer
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First, let me tell you that amplitude is proportional to slit width because as slit width increases, more light gets through, so more energy is transmitted, and therefore, the amplitude must also get bigger (by the same factor).
Furthermore, intensity is proportional to amplitude squared.
By applying the 2 above statements, we get that intensity is proportional to slit width squared.
For the picture below, the x and y only represent that there is some kind of part in the equation, it is just not shown here, instead, it is denoted as x and y.
Answering below to: Does decreasing slit width affect the intensity of the maxima for a single slit experiment?
So, if we for example, reduce the width of the slit by a factor of two (we halve the width), what is going to happen, is that intensity will decrease by a factor of four.
$I$ = the central intensity before the width of the slit is reduced.
$I_1$ = the central intensity after the width of the slit is reduced by 2.
And since intensity is proportional to slit width squared, we will have $$I_1 = z\left(\frac d2\right)^2 = z(\frac{d^2}{4})$$ So, $d^2$ is reduced by 4, and because $d^2$ is directly proportional to intensity, $I$ (intensity) is reduced by 4. (The z represents some part of the equation).
So, yes, reducing the width of a slit decreases the intensity of the central maxima in a single slit experiment.
I hope that this helped! :)
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$\begingroup$ Hello! I have edited your answer using MathJax (LaTeX) math typesetting. For future posts, you can refer to MathJax basic tutorial and quick reference. Thanks! $\endgroup$– jng224May 8, 2021 at 12:45
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