Wave Optics

JEE Main level, numerical answer, medium

52 of 191

Question 52 of 191

In a Young's double slit experiment, the intensity at a point is (14)th \left(\frac{1}{4}\right)^{\text {th }} of the maximum intensity, the minimum distance of the point from the central maximum is ________ μm\mu \mathrm{m}.
(Given : λ=600 nm, d=1.0 mm,D=1.0 m\lambda=600 \mathrm{~nm}, \mathrm{~d}=1.0 \mathrm{~mm}, \mathrm{D}=1.0 \mathrm{~m} )

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Answer: 200

I=I0cos⁡2(Δϕ2)I04=cos⁡2(Δϕ2)Δϕ=2π32πλ(ydD)=2π3y=λD3 d=600×10−9×13×10−3=2×10−4 m\begin{aligned} & \mathrm{I}=\mathrm{I}_0 \cos ^2\left(\frac{\Delta \phi}{2}\right) \\ & \frac{\mathrm{I}_0}{4}=\cos ^2\left(\frac{\Delta \phi}{2}\right) \\ & \Delta \phi=\frac{2 \pi}{3} \\ & \frac{2 \pi}{\lambda}\left(\frac{\mathrm{yd}}{\mathrm{D}}\right)=\frac{2 \pi}{3} \\ & \mathrm{y}=\frac{\lambda \mathrm{D}}{3 \mathrm{~d}}=\frac{600 \times 10^{-9} \times 1}{3 \times 10^{-3}}=2 \times 10^{-4} \mathrm{~m}\end{aligned}

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