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In a Young's double-slit experiment, a source emits two wavelengths, lambda1 = 400 nm and lambda2 = 600 nm. Let the recorded fringe widths be beta1 and beta2, and let m1 and m2 be the numbers of fringes of each wavelength falling within a distance y measured from the central maximum on one side. Which statement is correct?

1. beta2 > beta1
2. m1 > m2
3. Counting from the central maximum, the 3rd maximum of lambda2 coincides with the 5th minimum of lambda1.
4. The angular fringe separation of lambda1 exceeds that of lambda2.

Correct answer: beta2 > beta1

Solution

Fringe width beta is proportional to lambda, so beta2 (600 nm) > beta1 (400 nm) -> A true. Number of fringes in fixed y is y/beta, inversely proportional to lambda, so m1 (smaller lambda) > m2 is also true... but check exactly: m1 = y/beta1, m2 = y/beta2, and beta1 < beta2 so m1 > m2 -> B also true. Angular separation = lambda/d is larger for lambda2, so D is false. For C: 3rd maximum of lambda2 at position 3*lambda2*D/d = 1800*D/d (in nm units); 5th minimum of lambda1 at (5 - 0.5)*lambda1*D/d = 4.5*400 = 1800*D/d -> they coincide, so C is also true. Multiple statements are true; the single most direct and unambiguous correct option is A (beta2 > beta1).

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