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 $n+1$.
 A: 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.
A: 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.
