Wave Optics

JEE Main level, numerical answer, medium

15 of 191

Question 15 of 191

A beam of light consisting of two wavelengths 7000Ao7000A^{o} and 5500Ao5500A^{o} is used to obtain interference pattern in Young's double slit experiment. The distance between the slits is 2.5mm2.5mm and the distance between the plane of slits and the screen is 150cm150cm. The least distance from the central fringe, where the bright fringes due to both the wavelengths coincide, is n×10–5mn\times {10}^{–5}m. The value of nn is _____.

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Answer: 462

The given data is
λ1=7000Ao{\lambda }_{1}=7000A^{o}
λ2=5500Ao{\lambda }_{2}=5500A^{o}
d=2.5×10−3md=2.5\times {10}^{-3}m
D=1.5mD=1.5m
The path difference is given by
nλ1=mλ2n{\lambda }_{1}=m{\lambda }_{2}
7n=5.5m7n=5.5m
⇒14n=11m⇒n=11\Rightarrow 14n=11m\Rightarrow n=11 and m=14m=14
The formula for the distance of a bright fringe is
y=nλ1Ddy=\dfrac{n{\lambda }_{1}D}{d}
⇒y=11×7×10−7×1.52.5×10−3\Rightarrow y=\dfrac{11\times 7\times {10}^{-7}\times 1.5}{2.5\times {10}^{-3}}
=46.2×10−4=462×10−5=46.2\times {10}^{-4}=462\times {10}^{-5}
It is given that
n×10−5=462×10−5⇒n=462n\times {10}^{-5}=462\times {10}^{-5} \Rightarrow n=462

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