Question

In young's double slit experiment, first slit has width four times the width of the second slit. The ratio of the maximum intensity to the minimum intensity in the interference fringe system is

Answer 1

Heya!!
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given,

w1/ w2 = 4:1

And as we know,

width is directly proportional to Intensity
and intensity is directly proportional to square of amplitude

let , A and B be the amplitude respectively

w1/w2 = I1/I2 = (A/B)^2

4:1 =( A/B)^2

A/B =2:1

i.e A =2×B

→I max/ Imin = (A+B)^2/(A-B)^2 = (2B+B)^2/(2B-B)^2 = 9:1

hope it helps!!

#Phoenix

Answer 2

Assertion: In Young's double slit experiment using white light, the central fringe is white, the violet coloured fringe will be seen nearest to the central fringe. Reason: Red coloured fringe will be seen near central fringe.