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Q.

The figure shows the interference pattern obtained in double slit experiment using light of wavelength 600 nm. P, Q, R, S and T are marked on five fringes-

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a

The fringe resulting from a phase difference of 4π is S

b

If ΔXA and ΔXC represent path differences between waves interfering at P and R respectively then ΔXCΔXA is equal to 300 nm.

c

The third order bright fringe is R.

d

The third order bright fringe is T

answer is B, C, D.

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Detailed Solution

Phase difference for the central maxima is zero path difference is also zero.

.Phase differnce for first order maxima is 2π and path difference is λ

Phase difference for first order minima is π and path difference is λ/2

Keeping this concept in mind we could say that T belongs to third order maxima

S belongs to 2nd order maxima so phase difference is 4π

P belongs to 1st mininma path difference is -λ2 and R belongs to 2nd minima path difference 3λ2 difference in their magnitude is λ2

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