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Question

In a Young's double slit interference experiment the fringe pattern is observed on a screen placed at a distance D from the slits. The slits are separated by a distance d and are illuminated by monochromatic light of wavelength λ. Find the distance from the central point where the intensity falls to (a) half the maximum, (b) one fourth of the maximum.

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Solution

a)As shown in the figure BP0AP0=λ/3

(D2+d2)D=λ/3D2+d2=D2+(λ2/9)+(2λD)/3

d=(2λD)/3 (neglecting the term λ2/9 as it is very small)


b)To find the intensity at P0, we habe to consider the interference of light waves coming from all the three slits.
Here. CP0AP0=D2+4d2D
=D2+8λD3D=D{1+8λ3}1/2D
=D{1+8λ3D×2+.....}D=4λ3 [using binomial expansion]

So, the corresponding phase difference between waves from C and A is
ϕC=2πxλ=2π×4λ3λ=8π3=(2π+2π3)=2π3 ...


Again, ϕB=2πx3λ=2π3So, it can be said that light from B and C are in same phase as thry have some phase differencewith respect to A

So, R=(2r)2+r2+2×2r×rcos(2π/6)=4r2+r22r2=3r

IP0K(3r)2=3Kr2=3I

As the resulting amplitude is 3 times, the intensity will be three times the intensity due to individual slits.

1664624_1195457_ans_a4e6e16bc17d47ba95d93218350db17a.jpg

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