Wave Optics

JEE Main level, single correct, medium

20 of 191

Question 20 of 191

Interference fringes are observed on a screen by illuminating two thin slits 1mm1mm apart with a light source (λ=632.8nm)\left(\lambda =632.8nm\right). The distance between the screen and the slits is 100cm100cm. If a bright fringe is observed on a screen at distance of 1.27mm1.27mm from the central bright fringe, then the path difference between the waves, which are reaching this point from the slits is close to :

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Answer: (A) 1.27μm1.27\mu m

ΔP=dsin⁡θ\Delta P=d\sin \theta
=dθ=d\theta
=dyD=10−3×1.270mm1m=1.27μm=\dfrac{dy}{D}=\dfrac{{10}^{-3}\times 1.270mm}{1m}=1.27\mu m
1.27μm1.27\mu m

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