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Two light rays of different wavelengths are allowed to pass through double slit. What is the least distance for which the two bright fringes coincide. I have seen a example where a ray makes a bright fringe for some value $n$. And the other ray coincides at $n+1$. I don't think it is correct. It is not compulsory that the two rays coincide at $n$ and $n+1$. It could be anything. May be $n$ and $n+3$.

For reference consider the following example:

A beam of light consisting of two wavelength 650 nm and 520 nm, is used to obtain fringes in a Young’s double slit experiment on a screen 1.2 m away. The separation between the slits is 2 mm. What is the least distance from the central maximum when the bright fringes due to both the wavelength coincide?

I don't think it is always that the ray coincide at $n$ and $n+1$.

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They cannot coincide at m=1 but after that they can coincide and form a beat pattern. In your setup you have two slits separated by 2,000,000nm center to center with a detection screen 1,200,000,000nm, beyond. (1) With 520nm wavelength light (Green) you get a bright spot every 312,000nm. (2) With 650nm wavelength light (Orange) you get a bright spot every 390,000nm. Because the math works out, they will coincide right away forming a beat pattern every 1,500,000 nanometers. With 520 wavelength the fifth bright fringe 5x312,000nm=1,500,000nm With 650 wavelength the forth bright fringe 4x390,000nm=1,500,000nm After that they will coincide every 1,500,000 nanometers. Keep in mind that single slit destructive interference coming from both wavelengths could and probably would disrupt the pattern.

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  • $\begingroup$ So they do not coincide when one has nth fringe and the other n+1. Rather they do at other point? $\endgroup$ May 18, 2022 at 17:45
  • $\begingroup$ @SamyakMarathe One needs to be a multiple of the other. $\endgroup$ May 18, 2022 at 23:31
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We can write the position of maxima for each wavelength light as

$$y_n=n\lambda_1 \frac{D}{d}$$ $$y_m=m\lambda_2 \frac{D}{d}$$

Now when the two maxima conincide we have, $$y_n = y_m$$ This gives $$n\lambda_1=m\lambda_2$$

In your example $\lambda_1$= 650 nm and $\lambda_2$= 520 nm So $$\frac{n}{m}=\frac{520}{650}=\frac{4}{5}$$ The smallest integral values of n and m are 4 and 5 respectively. In this particular example m = n + 1 but it is not necessarily true always. Let's say if $\lambda_2$ would have been 390 nm then n and m would be 3 and 5 respectively.

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  • $\begingroup$ Yeah. So we should aim in finding the LCM of both the distances rather than the absurd assumption that the coincide at n and n+1 $\endgroup$ May 19, 2022 at 9:13
  • $\begingroup$ Yes they don't necessarily coincide at n and n+1. And you don't need to find LCM of distances as the distances are itself unknown. $\endgroup$ May 19, 2022 at 9:21
  • $\begingroup$ yes but these websites like topper doubnutvedantu byjus are such stupid that they had considered n and n+1. May be they already know the answer. Stupid Indian Education System. And shame kn such websites who are not authorized for their accuracy. $\endgroup$ May 19, 2022 at 9:25
  • $\begingroup$ I think they use direct result and don't make any reference for that which causes confusion. $\endgroup$ May 19, 2022 at 9:30

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