Question

In a Young's double slit experiment, the fringe pattern is observed on a screen placed at a distance $D$. The slits are separated by $d$ and are illuminated by light of wavelength $\lambda$. The distance from the central point, where the intensity falls to half the maximum is_______

• A. $\frac{\lambda D}{3d}$
• B. $\frac{\lambda D}{2d}$
• C. $\frac{\lambda D}{d}$
• D. $\frac{\lambda D}{4d}$

Answer 1

The correct option is D $\frac{\lambda D}{4d}$

Given: $I=\frac{I_{max}}{2}$
we know that
$I=I_{max}{\cos }^2\left(\frac{\varphi }{2}\right)$

$\frac{I_{max}}{2}=I_{max}co{s}^2\left(\frac{\varphi }{2}\right)$

${\cos }^2\left(\frac{\varphi }{2}\right)=\frac{1}{2}$

$\cos \left(\frac{\varphi }{2}\right)=\frac{1}{\sqrt{2}}$

$\frac{\varphi }{2}=\pi /4(\because \cos (\frac{\pi }{4})=1/\sqrt{2})$

$\varphi =\pi /2$

So, path difference $\Delta x=\frac{\lambda }{4}$
So, distance from central point $=\frac{\Delta \lambda D}{d}$

$=\frac{\lambda }{4}\frac{D}{d}$