Question

If a Young's double slit experiment with light of wavelength $\lambda$ the separation of slits is d and distance of screen is $D$ such that $D>>d>>\lambda$. If the Fringe width is $\beta$, the distance from point of maximum intensity to the point where intensity falls to half of maximum intensity on either side is

• A. $\frac{\beta }{4}$
• B. $\frac{\beta }{3}$
• C. $\frac{\beta }{6}$
• D. $\frac{\beta }{2}$

Answer 1

The correct option is A $\frac{\beta }{4}$

The intensity is given by $I_{resultant}=I_1+I_2+2\sqrt{I_1{I}_2}cos\theta$...(i)
and $\theta$ is phase difference which is given as $\theta =\frac{2\pi }{\lambda }\Delta x$ and $\Delta x$ is path difference.
at center the path difference is zero i.e.maximum intensity:
$I_{\circ }+I_{\circ }+2I_{\circ }cos0=4I_{\circ }$...(ii)
The phase difference for intensity to be half of maximum value
$2I_{\circ }=I_{\circ }+I_{\circ }+2I_{\circ }cos\theta \Rightarrow \theta =\frac{\pi }{2}$
$\Delta x=\frac{\lambda }{2\pi }\frac{\pi }{2}=\frac{\lambda }{4}$
in YDSE $\Delta x$ is given as $\Delta x=\frac{yd}{D}$
$y=\frac{\lambda D}{4d}$...(iii)
The fringe width is given as $\beta =\frac{\lambda D}{d}$ substituting value of $\frac{\lambda D}{d}$ in equation (iii)
$y=\frac{\beta }{4}$ correct answer is option A.