Q.
Two slits s1 and s2 are on a plane inclined at an angle of 450 with horizontal. The distance between the slits is 2 mm. A monochromatic point source S of wavelength λ=5000A0 is placed at a distance 1/2 mm from the midpoint of slits as shown in figure. The screen is placed at a distance of 2 m. The fringe width of interference pattern on the screen is n10mm. Find the integer closest to n
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answer is 7.
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Detailed Solution
For small angles,sinθ=θ and cosθ=1∠S1S2M=45-θΔx1=dcos(45-θ)Δx=Δxi+dcos(45−θ)Δx=Δxi+d(cos45cosθ+sin45sinθ)Δx=Δxi+2×10−3121+12θ For central & first maxima Δx is 0,λ∴0=Δxi+2×10−312+12θ1λ=Δxi+2×10−312+12θ2 Subtracting λ=2×10−312Δθ∵Δθ=βD5×10−7=10−3β2β=52×10−4β=5210mm=7.0710mm
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