Wave Optics

JEE Main level, single correct, medium

101 of 191

Question 101 of 191

In a Young's double slit experiment, the path difference, at a certain point on the screen, betwen two interfering waves is 18\frac{1}{8} th of wavelength. The ratio of the intensity at this point to that at the centre of a bright fringe is close to:

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Answer: (B) 0.85

Given, path difference, Δx=λ8\Delta x=\frac{\lambda}{8}

Phase difference (Δϕ)(\Delta \phi) is given by

Δϕ=2πλ(Δx)\Delta \phi=\frac{2 \pi}{\lambda}(\Delta \mathrm{x})

Δϕ=(2π)λλ8=π4\Delta \phi=\frac{(2 \pi)}{\lambda} \frac{\lambda}{8}=\frac{\pi}{4}

For two sources in different phases,I=I0cos⁡2(π8)1I0=cos⁡2(π8)=1+cos⁡π42=1+122=0.85\begin{array}{l} I=I_{0} \cos ^{2}\left(\frac{\pi}{8}\right) \\ \frac{1}{I_{0}}=\cos ^{2}\left(\frac{\pi}{8}\right) \\ =\frac{1+\cos \frac{\pi}{4}}{2}=\frac{1+\frac{1}{\sqrt{2}}}{2}=0.85 \end{array}

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