Question

A coherent parallel beam of microwaves of wavelength $\lambda =0.05mm$ falls on a Youngs double slit apparatus. The separation between maxima is measured on a screen placed parallel to the plane of the slits at a distance of $1.0m$ from it as shown in figure. The separation between the slits is $2d=1mm$.
(a) If the incident beam falls normally on the double slit apparatus, find the $y-$coordinates of all the interface minima of the screen.
(b) If the incident beam makes an angle of ${30}^o$ with the $x-$axis (as shown in fig). find the $y-$coordinates of the first minima on either side of the central maximum.

Answer 1

(a) Given separation between two slits $=2d=1.0mm,D=1.0m$
Path difference $=2d\sin \theta$
For minima, $2d\sin \theta =(2n-1)\frac{\lambda }{2}$
$\therefore 1\times {10}^{-3}=\frac{(2n-1)\times (0.5\times {10}^{-3})}{2}$ as $\theta ={90}^o$ for the highest possible order of minimum
or $(2n-1)\frac{2\times {10}^{-3}}{0.5\times {10}^{-3}}=4\Rightarrow n=2.5$
Thus, only two minima are possible on either side of central maxima. Thus total number of minima $=2+2=4$
For minima $y=\pm \frac{(2n-1)\lambda D}{4d}=(2n-1)0.25$
$y_1=0.25m,y_2=0.75m,y_3=0.25m$ and $y_4=-0.75m$
(b) Initial path difference $=2d\sin {30}^o$
For the central maxima
$\frac{y}{D}\times 2d=2d\sin {30}^o$
$y=D\sin 30=0.5m$
For first minima.
$\frac{y}{D}\times 2d-2d\sin {30}^o=\pm \frac{\lambda }{2}$
or $y=\pm 0.25+0.5$
or $y_1=0.75m$ and $y_2=0.25m$.