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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? I'd appreciate any help even though I skipped the question in my exam, it's been bugging me all day.

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Here's a derivation - but you may not be able to pass the paywall. opticsinfobase.org/… –  Carl Witthoft Mar 5 at 16:50

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up vote 2 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 at 3:28

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