In Young's double slit experiment using two identical slits, the intensity of the maximum at the centre of the screen is I. What will be the intensity at the centre of the screen if one of the slits is closed?

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$\frac{I}{2}$

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$\frac{I}{4}$

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None of these

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Solution

The correct option is C
$\frac{I}{4}$

Let $I_{0}$ be the intensity at the centre of the screen due to each slit. Then, for the central maximum, the intensity is
$I = I_{0} + I_{0} + 2 \sqrt{I_{0} I_{0}} = 4 I_{0} o r I_{0} = \frac{I}{4} .$

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In Young's double slit experiment using two identical slits, the intensity of the maximum at the centre of the screen is I. What will be the intensity at the centre of the screen if one of the slits is closed?

The correct option is C I4 Let I0 be the intensity at the centre of the screen due to each slit. Then, for the central maximum, the intensity is I=I0+I0+2√I0I0=4I0or I0=I4.

The correct option is C
$\frac{I}{4}$

Let $I_{0}$ be the intensity at the centre of the screen due to each slit. Then, for the central maximum, the intensity is
$I = I_{0} + I_{0} + 2 \sqrt{I_{0} I_{0}} = 4 I_{0} o r I_{0} = \frac{I}{4} .$