In a Young's double slit experiment with slit separation 0.1mm, one observes a bright fringe at angle 140rad by using light of wavelength λ1. When the light of wavelength λ2 is used a bright fringe is seen at the same angle in the same set up. Given that λ1 and λ2 are in visible range (380nm to 740nm), their values are
380nm,500nm
625nm,500nm
380nm,525nm
400nm,500nm
A
380nm,500nm
B
625nm,500nm
C
400nm,500nm
D
380nm,525nm
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Solution
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The correct option is B625nm,500nm Path difference =dsinθ≈dθ =0.1×140mm=2500nm or bright fringe, path difference must be integral multiple of λ. ∴2500=nλ1=mλ2 ∴λ1=625,λ2=500 (from m=5) (for n=4).
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