Question

(a) Derive an expression for path difference in Youngs double slit experiment and obtain the conditions for constructive and destructive interference at a point on the screen.
(b) The intensity at the central maxima in Youngs double slit experiment is $I_0$. Find out the intensity at a point where the path difference is $\frac{\lambda }{6},\frac{\lambda }{4}$ and $\frac{\lambda }{3}$.

Answer 1

(a)

Path difference in Young's Double slit experiment at point P is given by:

$\Delta x=S_{2}P-S_{1}P$

$S_2{P}^2-S_1{P}^2=[D^2+(x+\frac{d}{2}{)}^2]-[D^2+(x-\frac{d}{2}{)}^2]$

$=2xd$

$(S_{2}P-S_{1}P)(S_{2}P+S_{1}P)=2xd$

Assuming $S_{2}P+S_{1}P\approx 2D$ as $x<<D$ and $d<<D$

$\Delta x\approx \frac{xd}{D}$

For constructive interference, $\Delta x=n\lambda$

Position of nth bright fringe is: $x_n=\frac{n\lambda D}{d}$

and for destructive interference, $\Delta x=(2n+1)\frac{\lambda }{2}$

Position of nth dark fringe is: $x_n=\frac{(2n+1)\lambda D}{2d}$

(b)

Let intensity of light sources from slits be $I$.

Resultant intensity at a point is $I^{\prime }=I+I+2Icos\varphi$

where $\varphi$ is the phase difference at the point.

Path difference is given by:

$\Delta x=\frac{\lambda \varphi }{2\pi }$

Hence, $I^{\prime }=I+I+2Icos\left(2\pi \frac{\Delta x}{\lambda }\right)$

Given intensity at central maximum is $I_o$

Hence, $I_o=4I$

At $\Delta x=\frac{\lambda }{6},I^{\prime }=3I=\frac{3}{4}{I}_o$

At $\Delta x=\frac{\lambda }{4},I^{\prime }=2I=\frac{1}{2}{I}_o$

At $\Delta x=\frac{\lambda }{3},I^{\prime }=I=\frac{1}{4}{I}_o$