# When does the bright fringes of two rays of different wavelengths coincide in a double slit?

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 $$n+1$$.

• Photons act on their own directed by the EM field. The red pattern and green patterns would just overlap, i.e. take the separate color patterns and just add them. While its a good math question as posed ... it's not physics! Commented Apr 17 at 1:40

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.

• So they do not coincide when one has nth fringe and the other n+1. Rather they do at other point? Commented May 18, 2022 at 17:45
• @SamyakMarathe One needs to be a multiple of the other. Commented May 18, 2022 at 23:31

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.

• 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 Commented May 19, 2022 at 9:13
• 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. Commented May 19, 2022 at 9:21
• 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. Commented May 19, 2022 at 9:25
• I think they use direct result and don't make any reference for that which causes confusion. Commented May 19, 2022 at 9:30