Rankers Physics

Young's Double Slit Experiment: Practice Problem & Solution

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
2I
I
I/4
I/2

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Similar Practice Questions

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1. Assertion (A): In Young's double slit experiment if intensity of each source is \(I_0\) then minimum and maximum intensity is zero and \(4I_0\) respectively. Reason (R): In Young's double slit experiment energy conservation is not followed.  
2. Two identical light waves, propagating in the same direction, have a phase difference δ. After they superpose, the intensity of the resulting wave will be proportional to :
3. Interference was observed in interference chamber where air was present, now the chamber is evacuated, and if the same light is used, a careful observer will see (1993)
4. In Young's double slit experiment, one of the slit is wider than other, so that the amplitude of the light from one slit is double of that from other slit. If Im be the maximum intensity, the resultant intensity I when they interfere at phase difference Φ is given by :
5. Assertion (A): In Young's double slit experiment, if one of the slits is closed, intensity at the position of central fringe becomes half. Reason (R): Resultant intensity, being sum of intensities from individual slits, becomes half as one slit is closed.