Wave Optics Question 3
Question 3 - 2024 (01 Feb Shift 2)
In Young’s double slit experiment, monochromatic light of wavelength $5000 \mathring{A}$ is used. The slits are $1.0 mm$ apart and screen is placed at $1.0 m$ away from slits. The distance from the centre of the screen where intensity becomes half of the maximum intensity for the first time is $\times 10^{-6} m$.
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Answer: (125)
Solution:
Let intensity of light on screen due to each slit is $I _0$
So internity at centre of screen is $4 I _0$
Intensity at distance y from centre-
$I=I _0+I _0+2 \sqrt{I _0 I _0} \cos \phi$
$I _{\max }=4 I _0$
$\frac{I _{\text {max }}}{2}=2 I _0=2 I _0+2 I _0 \cos \phi$
$\cos \phi=0$
$\phi=\frac{\pi}{2}$
$K \Delta x=\frac{\pi}{2}$
$\frac{2 \pi}{\lambda} d \sin \theta=\frac{\pi}{2}$
$\frac{2}{\lambda} d \times \frac{y}{D}=\frac{1}{2}$
$y=\frac{\lambda D}{4 d}=\frac{5 \times 10^{-7} \times 1}{4 \times 10^{-3}}$
$=125 \times 10^{-6}$
$=125$
