Wave Optics

JEE Main level, single correct, medium

103 of 191

Question 103 of 191

In a Young's double slit experiment slit separation 0.1mm,0.1mm, one observes a bright fringe at angle 140rad\dfrac{1}{40}rad by using light of wavelength λ1.{\lambda }_{1}. When the light of wavelength λ2{\lambda }_{2} is used a bright fringe is seen at the same angle in the same set up. Given that λ1{\lambda }_{1} and λ2{\lambda }_{2} are in visible range (380nmto740nm),\left(380nmto740nm\right), their values are:

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Answer: (C) 625nm,500nm625nm,500nm

dsinθ=nλd sin\theta =n\lambda

∴λ=dsinθn=10−4×140×1n∴\lambda =\dfrac{dsin\theta }{n}={10}^{-4}\times \dfrac{1}{40}\times \dfrac{1}{n}

∴λ=2.5×10−6n=2500×10−9n=2500n(nm)∴\lambda =\dfrac{2.5\times {10}^{-6}}{n}=\dfrac{2500\times {10}^{-9}}{n}=\dfrac{2500}{n}\left(nm\right)

Putting n=1,λ1=2500nmn=1, {\lambda }_{1}=2500 nm

n=2,λ2=1250nmn=2, {\lambda }_{2}=1250 nm

n=3,λ3=833nmn=3, {\lambda }_{3}=833 nm

n=4,λ4=625nmn=4, {\lambda }_{4}=625 nm

n=5,λ5=500nmn=5, {\lambda }_{5}=500 nm

n=6,λ6=357nmn=6, {\lambda }_{6}=357 nm

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