Question

(a) In Young's double slit experiment, derive the condition for
(i) constructive interference and
(ii) destructive interference at a point on the screen.
(b) A bream of light consisting of two wavelengths, $800nm$ and $600nm$ is used to obtain the interference fringes in a Young's double slit experiment on a screen placed $1.4m$ away. If the two slits are separated by $0.28nm$, calculate the least distance from the central bright maximum where the bright fringes of the two wavelengths coincide.

Answer 1

Young double slit experiment: Let $S_1$ and $S_2$ are the two narrow rectangular slits placed perpendicular to the plane of paper. Screen is placed on the perpendicular bisector of$S_1{S}_2$ and is illuminated with monochromatic light.

The slits are separated by a small distance d. A screen is placed at a distance D from $S_1{S}_2.$

Consider a point P on the screen at a distance $x$ from O.

The path difference between the waves reaching P from $S_1$ and $S_2$ is

$P=S_{2}P-S_{1}P$ , draw a perpendicular $S_{1}N$ on $S_{2}P$

$P=S_{2}P-S_{1}P=S_{2}P-NP=S_{2}N$

From the triangle $\frac{S_{2}N}{S_2{S}_1}=sin\theta$

$\therefore P=S_{2}N=S_2{S}_{1}sin\theta =dsin\theta$

when $\theta$ is small,

$sin\theta \simeq \theta =tan\theta =\frac{x}{D}$

$\therefore P=\frac{xd}{D}$

For constructive interference

$\frac{xd}{D}=n\lambda ;n=0,1,\mathrm{2.....}$

Position of $n^{th}$ bright fringe

$X_n=\frac{nD\lambda }{d}$

when n=0 $X_n$= 0, central bright fringe is formed at O.

For destructive interference ,

$\frac{xd}{D}=(2n-1)\frac{\lambda }{2}$

$X_n=(2n+1)\frac{\lambda D}{2}$

Thus alternative bright fringe and dark fringe are formed on the screen.

(b)

Let the $n_1^{th}$ maximum corresponds to $n_1$ and $n_2^{th}$ maximum corresponds $n_2$, then

$\frac{n_1{\lambda }_{1}D}{d}=\frac{n_2{\lambda }_{2}D}{d}$

$⟹\frac{n_1}{n_2}=\frac{600}{800}=\frac{3}{4}$

$y_{min}=\frac{N_1{\lambda }_{1}D}{d}=\frac{3\times 800\times {10}^{-9}\times 1.4}{0.28\times {10}^{-3}}=12nm$