# Explain the concept of Young's double-slit experiment and formulas related to it.

## Question:

Explain the concept of Young's double-slit experiment and formulas related to it.

## The Young's Double Slit Experiment

The Young's double-slit arrangement is devised to ensure that light intensities add up coherently. In the arrangement, the two interfering waves having the same amplitude, the same frequency, and a fixed phase difference are allowed to superpose and a screen is inserted in this region of superposition. Then the resultant intensity on the screen is no longer the sum of the individual intensities. There is an additional interference term. This leads to a redistribution of the incident energy in the form of alternating high and low intensity regions.

The arrangement consists of a monochromatic source of radiation placed behind a slit plate with two tiny slits at a small separation. The slit separation is of the order of millimeters. Far from the slit plate at a distance of several meters a screen is placed.

Light falls on the screen from the two slits. Both the beams of light ave the same frequency, amplitude, and a fixed phase difference. Thus at a point on the screen where the waves arrive in phase, there will be constructive interference. If the path difference between the waves is an integral multiple of the wavelength then this condition is attained. Such points have a high intensity. Recall that the intensity is proportional to the square of the electric field strength of the wave. At points where the waves arrive out of phase, there is destructive interference and the waves cancel each other out. These points are dark. When the path difference between the waves is an odd multiple of half wavelengths then this condition is attained.

If the origin is chosen at the center of the screen and the position on the screen is denoted by {eq}\displaystyle {y} {/eq} then the path difference between the waves arriving at this point will be,

{eq}\displaystyle {s=\frac{yd}{D}} {/eq}

If this is an integral multiple of {eq}\displaystyle {\lambda} {/eq} then we have constructive interference.

That is,

{eq}\displaystyle { \frac{yd}{D}=n \lambda} {/eq} implies that {eq}\displaystyle {y} {/eq} is bright.

Here n=0,1,2,3,....

At the center n=0 as there is no path difference from the slits. So it is always bright.

Again substituting n=1 this gives,

{eq}\displaystyle {y=\frac{\lambda D}{d}} {/eq}.

So the next bright fringe is located at the y value corresponding to n=1. Hence this must be the fringe width {eq}\displaystyle {\beta} {/eq}.

Thus,

{eq}\displaystyle {\beta=\frac{\lambda D}{d}} {/eq}. 