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The intensity at the maximum in a Young's double slit experiment is I₀. Distance between two slits is d = 5λ, where λ is the wavelength of light used in the experiment. What will be the intensity in front of one of the slits on the screen placed at a distance D = 10d is

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$\frac{{{I_0}}}{2}\$

$\frac{3}{4}{I_0}\$

$\frac{{{I_0}}}{4}\$

${I_0}\$

answer is D.

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Detailed Solution

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$y = \frac{d}{2} = \frac{{5\lambda }}{2}\,;\Delta x = d\sin \theta = d\frac{y}{D}\$

$\Delta x = \frac{{5\lambda }}{2}\frac{{5\lambda }}{{50\lambda }} = \frac{\lambda }{4};\phi = \frac{{2\pi }}{\lambda } \times \frac{\lambda }{4} = \frac{\pi }{2}\$

$\large I = {I_0}{\cos ^2}\,\frac{\phi }{2} = {I_0}{\cos ^2}\left( {\frac{\pi }{4}} \right) = \frac{{{I_o}}}{2}\$

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