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Here's a question I got in my final exam this morning. "If in a Young's double slit experiment setup, the ratio of intensity of the bright spot to the dark spot is 25:9, what is the ratio of the width of the slits?"

Here's what I did. Since the ratio of intensity at the bright and dark spots is 25:9, the ratio of amplitudes there must 5:3. Which means the amplitude of one wave is 4 times the other.

Now, knowing that the amplitude of light through the wider slit is 4 times the amplitude of light through the narrower slit, how can I determine the ratio of the slits' width?

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Here's a derivation - but you may not be able to pass the paywall.… – Carl Witthoft Mar 5 '14 at 16:50
up vote 4 down vote accepted

I hope you know that intensity $(I)$ of light at any point on the screen due to interference in the Young's Double Slit experiment can be given as


where $a_1, a_2$ are the amplitudes of the light waves with constant phase difference of $\phi$, $A$ is the amplitude of the resultant displaement at the point on the screen. For simplicity, we can assume that intensity of light to be equal to square of the amplitude as given above.

Thus, $$I_{max}=a_1^2+a_2^2+2a_1a_2(1)=(a_1+a_2)^2$$


Therefore, $\frac{I_max}{I_min}=\frac{(a_1+a_2)^2}{(a_1-a_2)^2}=\frac{25}{9}$

Thus, $a_1+a_2=5, a_1-a_2=3$

Thus, $a_1=8/2=4, a_2=1$

The intensity of light due to a slit (source of light) is directly proportional to width of the slit. Therefore, if $w_1$ and $w_2$ are widths of the tow slits $S_1$ and $S_2$; $I_1$ and $I_2$ are intensities of light due to the respective slits on the screen, then


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I worked it all out till the last part of your answer which related the slit width with intensity. Thanks a lot anyway. Really appreciate your help. – Bolt64 Mar 6 '14 at 3:28
Excellent Response. I am just trying to find a source that shows that the intensity of light passing through a slit is proportional to the width. Do you know of any? – Clement Decker Jan 2 at 6:50

The amplitude should be proportional to the 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 a constant phase difference is taken as existing between adjacent segments. The resultant amplitude is found by superposition at the point of consideration.

So amplitude should be proportional to slit width. The intensity is proportional to the square of slit width, as intensity is proportional to the square of the amplitude.

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Wait, this is confusing. I thought that intensity depends on power divided by area in which case changing the slit width would not affect the intensity, or at least they would not be directly proportional – Clement Decker Mar 9 at 1:32

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