Question

With the help of a diagram, describe Young's double slit experiment to demonstrate
interference of light. Derive an expression for intensity of fringes.​

Answer 1

When light waves from two illuminated slits is incident on the screen, the path traveled by each light wave is different. This path difference leads to a phase difference in the two light waves. The path difference is different for each point on the screen and hence, intensity is different for all the points. This leads to the formation and bright and dark fringes on the screen.

Consider point P on the screen as shown in the figure.

S2P2=S2F2+PF2

S2P=D2+(x+2d)2

Similarly,

S1P=D2+(x−2d)2

Path difference is given by:

S2P−S1P=D2+(x+2d)2−D2+(x−2d)2

Using binomial expansion,

S2P−S1P=D(1+21(Dx+2D

Position of Fringes In Young’s Double Slit Experiment

Position of Bright Fringes

For maximum intensity at P

Δz = nλ (n = 0, ±1, ±2, . . . .)

Or d sin θ = nλ (n = 0, ±1, ±2, . . . .)

The bright fringe for n = 0 is known as the central fringe. Higher order fringes are situated symmetrically about the central fringe. The position of nth bright fringe is given by

y (bright) = (nλ\d)D (n = 0, ±1, ±2, . . . .)

Position of Dark Fringes

For minimum intensity at P,

\Delta z=\left( 2n-1 \right)\frac{n\lambda }{2}\left( n=\pm 1,\pm 2,….. \right)Δz=(2n−1)2nλ(n=±1,±2,…..) d\sin \theta =\left( 2n-1 \right)\frac{n\lambda }{2}dsinθ=(2n−1)2nλ

The first minima are adjacent to the central maximum on either side. We will obtain the position of dark fringe as

{{y}_{dark}}=\frac{\left( 2n-1 \right)\lambda D}{2d}\left( n=\pm 1,\pm 2,….. \right)ydark=2d(2n−1)λD(n=±1,±2,…..)