Question 197: Wave Optics

Question

In Young's double slit experiment, monochromatic light of wavelength 5000 Å is used. The slits are $1.0 \mathrm{~mm}$ apart and screen is placed at $1.0 \mathrm{~m}$ away from slits. The distance from the centre of the screen where intensity becomes half of the maximum intensity for the first time is _________ $	imes 10^{-6}$ $\mathrm{m}$.

Accepted Answer

Let intensity of light on screen due to each slit is $\mathrm{I}_0$
So internity at centre of screen is $4 \mathrm{I}_0$
Intensity at distance y from centre-
$$
\begin{aligned}
& I=I_0+I_0+2 \sqrt{I_0 I_0} \cos \phi \\
& I_{\max }=4 I_0 \\
& \frac{I_{\max }}{2}=2 I_0=2 I_0+2 I_0 \cos \phi
\end{aligned}
$$
$\begin{aligned} & \cos \phi=0 \\ & \phi=\frac{\pi}{2} \\ & K \Delta \mathrm{x}=\frac{\pi}{2} \\ & \frac{2 \pi}{\lambda} \mathrm{d} \sin 	heta=\frac{\pi}{2} \\ & \frac{2}{\lambda} \mathrm{d} 	imes \frac{\mathrm{y}}{\mathrm{D}}=\frac{1}{2}\end{aligned}$
$\begin{aligned} & \mathrm{K} \Delta \mathrm{x}=\frac{\pi}{2} \\ & \frac{2 \pi}{\lambda} \mathrm{d} \sin 	heta=\frac{\pi}{2} \\ & \frac{2}{\lambda} \mathrm{d} 	imes \frac{\mathrm{y}}{\mathrm{D}}=\frac{1}{2} \\ & \mathrm{y}=\frac{\lambda \mathrm{D}}{4 \mathrm{~d}}=\frac{5 	imes 10^{-7} 	imes 1}{4 	imes 10^{-3}} \\ & =125 	imes 10^{-6} \\ & =125\end{aligned}$

Wave Optics

Integer / Numeric Answer Type