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×10−3 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=S2N−MS1 (∵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=45∘−θ

⇒Δx=dcos(45∘−θ)−dsin45∘

⇒Δx=d[cosθ√2+sinθ√2]−d√2

As θ is small, sinθ=θ,cosθ=1

∴Δx=dθ√2

For bright fringe, Δx=nλ

⇒θ=√2nλd

As θ is small, θ=yD

⇒y=√2nλDd

Fringe width is given by

β=yn+1−yn=√2Dλd

∴β=√2×10×5000×10−10√2×10−3=5×10−3 m

Hence, k=5 is the correct answer.

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