Question

In young's double slit experiment using a monochromatic light of wavelength . The path difference corresponding to any point having half the peak intensity is

Answer 1

(2n+1)lambda /4

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Answer 2

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We have:-



$I=4{I_0} \cos{2} \frac{\phi}{2}$



${2I_0}=4{I_0} \cos{2} \frac{\phi}{2}$



$\cos{\frac{\phi}{2}}=\frac{1}{\sqrt2}$



$\phi = \frac{\pi}{2}$



$Path\: difference \triangle x= \frac{\lambda}{2 \pi} \times \triangle \phi$



$\triangle x= \frac{\lambda}{4}$



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