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QJEEM24Phb330: JEE Main 2024 Physics - Class 12 JEEM24Phb330 Solution: In Young's double slit experiment, monochromatic light of…

In Young's double slit experiment, monochromatic light of wavelength 5000 Å is used. The slits are 1.0 mm1.0 \mathrm{~mm} apart and screen is placed at 1.0 m1.0 \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 _________ ×10−6\times 10^{-6} m\mathrm{m} .

Step-by-Step Verified Solution

Let intensity of light on screen due to each slit is I0\mathrm{I}_0

So internity at centre of screen is 4I04 \mathrm{I}_0

Intensity at distance y from centre-

amp;I=I0+I0+2I0I0cos⁡ϕamp;Imax⁡=4I0amp;Imax⁡2=2I0=2I0+2I0cos⁡ϕ\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}

amp;cos⁡ϕ=0amp;ϕ=π2amp;KΔx=π2amp;2πλdsin⁡θ=π2amp;2λd×yD=12\begin{aligned} & \cos \phi=0 \\\\ & \phi=\frac{\pi}{2} \\\\ & 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}\end{aligned}

amp;KΔx=π2amp;2πλdsin⁡θ=π2amp;2λd×yD=12amp;y=λD4 d=5×10−7×14×10−3amp;=125×10−6amp;=125\begin{aligned} & \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\end{aligned}

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