Wave Optics Question 3
Question 3 - 2024 (01 Feb Shift 2)
In Young’s double slit experiment, monochromatic light of wavelength $5000 \AA$ 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 matho $\times 10^{-6} \mathrm{~m}$.
Show Answer
Answer: (125)
Solution:
Let intensity of light on screen due to each slit is $\mathrm{I}_{0}$
So internity at centre of screen is $4 I_{0}$
Intensity at distance y from centre-
$\mathrm{I}=\mathrm{I}{0}+\mathrm{I}{0}+2 \sqrt{\mathrm{I}{0} \mathrm{I}{0}} \cos \phi$
$\mathrm{I}{\max }=4 \mathrm{I}{0}$
$\frac{\mathrm{I}{\text {max }}}{2}=2 \mathrm{I}{0}=2 \mathrm{I}{0}+2 \mathrm{I}{0} \cos \phi$
$\cos \phi=0$
$\phi=\frac{\pi}{2}$
$\mathrm{K} \Delta \mathrm{x}=\frac{\pi}{2}$
$\frac{2 \pi}{\lambda} \mathrm{d} \sin \theta=\frac{\pi}{2}$
$\frac{2}{\lambda} \mathrm{d} \times \frac{\mathrm{y}}{\mathrm{D}}=\frac{1}{2}$
$\mathrm{y}=\frac{\lambda \mathrm{D}}{4 \mathrm{~d}}=\frac{5 \times 10^{-7} \times 1}{4 \times 10^{-3}}$
$=125 \times 10^{-6}$
$=125$
