# Number of fringes obtained on interference of light of different wavelengths [closed]

The Question goes as follows:

I am stuck at the part where we obtain general equation for path difference of two interfering lights. Since wavelength of those interfering lights are different, so I cannot apply the equation $$\Delta \psi = 2\pi\Delta x/\lambda$$, and thus cannot find the general condition for the position of dark fringes.

Would be really grateful if someone suggests me how to tackle these kind of problem.

• I think the question has no definite answer. Since each wavelength generates a (formally) infinite number of dark fringes, the only thing that reduces that number is either 1) screen size or 2) diffraction by non-zero width slits. Commented May 6, 2022 at 11:27
• I think the idea is you need to count how many places are there where both banks produce a dark fine at the same place. Commented May 6, 2022 at 14:30
• @ThePhoton Even then, there's an infinite number of those (they're evenly spaced). Commented May 6, 2022 at 19:00
• @Miyase You do actually obtain some sort of Limiting Condition by ∆x = d sinθ, where d is the Distance btw the slits (since sinθ cannot be >1), and then you can relate ∆x=(2n+1)λ/2 for dark fringes, and thus you get Limiting Condition for n (Number of Dark Fringes). But since this equation can be used for Interference of Lights of same Wavelength only, I am having some difficulty using that, which is basically my doubt. Commented May 7, 2022 at 15:15
• Since these are different wavelength beams, they are incoherent (assuming no extreme measures were taken to produce coherent beams). Therefore for practical purposes the intensities of the beams add, rather than the electric fields (which is usually what we mean when we talk about the superposition principle in optics). Commented May 7, 2022 at 15:21

The condition for a maximum is $$n \lambda = d\sin \theta$$ where $$d$$ Is the slit separation and $$n$$ is the order of the fringe. Using this formula with $$\theta = \pm \pi/2$$ will give you the number of bright fringes either side of the central maximum and this eliminates two of the answers.
The wavelengths have been chosen so that they are in a simple ratio $$(9/7)$$ which makes finding the position of the minima relatively easy.
For coincidence of minima at a certain position you must have $$(n+1/2)420 = (m+1/2)540 = d\sin \theta$$ where $$m$$ and $$n$$ are integers which you need to find.