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Question

Consider a plane inclined at an angle 45 with the horizontal has two slits S1andS2 separated by a distance d=2 mm. The screen is placed at a distance of D=10 m from the midpoint of the slits, as shown in the figure. A parallel monochromatic light beam of wavelength 5000 ˙A is incident on the slits as shown. If the fringe width of interference pattern on the screen is k×103 m, then the value of k is .


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Solution


Considering S2M as reference,

Path difference between S1 and S2 at P is
Δx=(S2N+NP)(MS1+S1P)
Δx=S2NMS1 (NP=S1P)

Since, the distance between the slits is very small compared to the distance between the slits and screen, θ is small and S1S2N=45θ

Δx=dcos(45θ)dsin45

Δx=d[cosθ2+sinθ2]d2

As θ is small, sinθ=θ,cosθ=1

Δx=dθ2

For bright fringe, Δx=nλ

θ=2nλd

As θ is small, θ=yD

y=2nλDd

Fringe width is given by
β=yn+1yn=2Dλd

β=2×10×5000×10102×103=5×103 m

Hence, k=5 is the correct answer.

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