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(C) In a Young's double slit experiment, the intensity at the central maximum is given by $I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos(\phi)$.Since the slits

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$I/4$

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(C) In a Young's double slit experiment, the intensity at the central maximum is given by $I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos(\phi)$.Since the slits are identical, let the intensity due to each slit be $I_0$. Thus, $I_1 = I_2 = I_0$.At the central maximum, the phase difference $\phi = 0$, so $I = I_0 + I_0 + 2\sqrt{I_0 I_0} \cos(0) = 4I_0$.Given that the intensity at the centre is $I$, we have $4I_0 = I$, which implies $I_0 = I/4$.If one of the two slits is covered, only one slit remains, and the intensity measured by the detector will be the intensity due to a single slit, which is $I_0 = I/4$.

In a Young's double slit experiment, a small detector measures an intensity of illumination of $I$ units at the centre of the fringe pattern. If one of the two (identical) slits is now covered, the measured intensity will be

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