Question
In Young's double slit experiment, the intensity on the screen at a point where path difference is $\lambda$ is $K.$ What will be the intensity at the point where path difference is $\lambda /4$
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Answer
(b)By using phase difference $\phi = \frac{{2\pi }}{\lambda }(\Delta )$ For path difference $\lambda$, phase difference ${\phi _1} = 2\pi $and for path difference $\lambda$/4, phase difference $\phi$$_2 $ = $\pi$$/2. $
Also by using $I = 4{I_0}{\cos ^2}\frac{\phi }{2}$ ==> $\frac{{{I_1}}}{{{I_2}}} = \frac{{{{\cos }^2}({\phi _1}/2)}}{{{{\cos }^2}({\phi _2}/2)}}$ ==> $\frac{K}{{{I_2}}} = \frac{{{{\cos }^2}(2\pi /2)}}{{{{\cos }^2}\left( {\frac{{\pi /2}}{2}} \right)}} = \frac{1}{{1/2}}$ ==> ${I_2} = \frac{K}{2}.$
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