# Young double slit experiment

Can we say that the central maximum is the brightest fringe? If so, to find the width of the central maximum, why can't we say that it is twice the width of the brightest fringe (due to 2 times $\theta$)?

• You seem to be confusing the interference fringes with the diffraction envelope which modulates the intensity of the interference fringes? Jun 21 '18 at 5:33

## 1 Answer

For Young's Double Slit Experiment, the intensity is given by:
$$I = 4I_ocos^2(\delta/2)$$ where δ is the phase difference between the two light rays (emanating from the two individual slits) at a given point on a screen due to path difference, initial phase difference, or optical phase difference. Therefore intensity will be maximum at points where δ is an even integral multiple of π (0,2π,4π...). These are points of constructive interference. Thus the central maximum is one but not the only point of maximum intensity.

Your logic for finding the fringe width is therefore not applicable in YDSE setups. The correct way to find fringe width in YDSE can be found here:

What is the fringe separation in Young's double slit experiment?

Given below are the graphs of intensity for YDSE and diffraction, clearly you can notice here that for diffraction the central maximum has the highest intensity throughout the distribution, but the same does not hold true for YDSE.