Here's a question I got in my final exam this morning. "If in Young's double-slit experiment setup, the ratio of the 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?

  • $\begingroup$ Here's a derivation - but you may not be able to pass the paywall. opticsinfobase.org/… $\endgroup$ Commented Mar 5, 2014 at 16:50
  • $\begingroup$ This is an old post, but unfortunately I think R C Mishra is correct here and the accepted answer is not correct. I explain my reasoning here. $\endgroup$ Commented Jan 27, 2020 at 22:07
  • $\begingroup$ @MaximalIdeal And the right answer doesn't even contain a single mathematical formula. Who says that math is important to physics? Most physicists... $\endgroup$ Commented Aug 21, 2020 at 18:30

2 Answers 2


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 a constant phase difference of $\phi$, $A$ is the amplitude of the resultant displacement at the point on the screen. For simplicity, we can assume that intensity of the light to be equal to the 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 the 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


  • $\begingroup$ 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. $\endgroup$ Commented Mar 6, 2014 at 3:28
  • $\begingroup$ 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? $\endgroup$ Commented Jan 2, 2016 at 6:50
  • $\begingroup$ Can you provide reasoning why you said $I \propto w$ as opposed to $a \propto w$? From my analysis, I see that the amplitude is what is directly proportional to slit width. $\endgroup$ Commented Jan 27, 2020 at 8:34
  • $\begingroup$ I have a dispute of the claim "the intensity of light due to a slit (source of light) is directly proportional to width of the slit" here. $\endgroup$ Commented Jan 27, 2020 at 10:51

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.

  • $\begingroup$ 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 $\endgroup$ Commented Mar 9, 2016 at 1:32

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