Question

In a YDSE, $D=1m,d=1mm$ and $\lambda =1/2mm$
Find the distance between the first and central maxima on the screen.

Answer 1

from given, we have,

$D>>d$

$\frac{d}{\lambda }=2\to n<<\frac{d}{\lambda }=2$

Path difference,

$\Delta P=d\sin \theta$

for 1st maxima,

$\Delta P=d\sin \theta =\lambda$

$\Rightarrow \sin \theta =\frac{\lambda }{D}=\frac{1}{2}$

$\Rightarrow \theta ={30}^{\circ }$

$y=D\tan \theta =\frac{1}{\sqrt{3}}m$