anonymous
  • anonymous
Consider a double-slit experiment with the spacing between the slits set at 3.50×10−3 cm, light of 630 nm, and a screen at a distance of 3.00 meters. Assume the dimensions are such that you can use the small angle approximation: sin θ ≈ tan θ ≈ θ (a) How high above the centerline from the slits to the screen is the 3rd bright fringe (counting the centerline as the 0th fringe)? (b) How high above the centerline to the screen is the 5th dark fringe? (c) How do the answers to (a) and (b) change if the measurements were made underwater (n=1.33)?
Physics
chestercat
  • chestercat
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anonymous
  • anonymous
a)\[n lambdaD/d=height on the screen\] n=no of fringe D=distance frome the screen d=spacing between slits
anonymous
  • anonymous
b)(2*n-1)*lambda*D/(2*d)=height on the screen 4 dark fringe c)answer remains the same if we use same wavelength

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