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In a double-slit interference experiment carried out with monochromatic light of wavelength lambda, the intensity at a screen point where the path difference equals lambda is K units. At another point where the path difference is lambda/6, the intensity is nK/12. Find the integer n.

1. 1
2. 2
3. 3
4. 4

Correct answer: 3

Solution

In Young's experiment, intensity I = 4*I0*cos²(delta/2) where delta is the phase difference. Phase difference delta = (2*pi/lambda)*path_difference. When path difference = lambda, delta = 2*pi, so I = 4*I0*cos²(pi) = 4*I0 = K, meaning I0 = K/4. When path difference = lambda/6, delta = (2*pi/lambda)*(lambda/6) = pi/3. So I = 4*(K/4)*cos²(pi/6) = K*(sqrt(3)/2)² = K*(3/4) = 3K/12. Comparing with nK/12 gives n = 3.

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