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In a Young's double-slit experiment with two identical slits, a thin perfectly transparent mica sheet is placed over the upper slit. The intensity at the centre of the screen then falls to 75% of its original (uncovered) value. Also, the second minimum lies just above the centre and the third maximum just below it. Which of the following could be the phase difference introduced by the mica sheet?

1. pi/3
2. 13*pi/3
3. 17*pi/3
4. 11*pi/3

Correct answer: 13*pi/3

Solution

Centre intensity I = I_max*cos²(phi/2) = 0.75 I_max => cos(phi/2) = sqrt(3)/2 => phi = 2*(pi/6 + n*pi) family, i.e. phi could be pi/3, 11pi/3, 13pi/3, etc. The condition that the second minimum is just above and the third maximum just below the centre fixes the magnitude/sign of fringe shift, selecting phi = 13*pi/3 (which corresponds to a path difference of about 13/6 wavelengths, placing the centre between the 2nd minimum and 3rd maximum). Hence 13*pi/3.

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