A parallel beam of light ( λ =5000 Å ) is incident at an angle α = 30° with the normal to the slit plane in a young’s double slit experiment. Assume that the intensity due to each slit at any point on the screen is Ι 0 . Point O is equidistant from S 1 & S 2 .The distance between slits is 1mm.
Text Solution
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(a,c)As d << D, ⇒ path difference = d sin θ (at 0) = 1mm × sin 30º = 0.5 mm
if it is a maxima ⇒ 10 –3 × 0.5 = (5000 × 10 –10 )m × (n)
n must be integer. get n = 1000.
Hence O is a maxima of intensity 4 I 0
Now
Now path difference at Q = d sin θ only QS 1 ≈ QS 2 .
d sin θ = 1 × 1/2 = 0.5 mm = integer multiple of λ . Hence maxima.
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