Wave Optics

JEE Main level, single correct, medium

107 of 191

Question 107 of 191

In Young's double-slit experiment, the distance between slits and the screen is 1m1m and monochromatic light of wavelength 600nm600nm is being used. A person standing near the slits is looking at the fringe pattern. When the separation between the slits is varied, the interference pattern disappears for a particular distance d0{d}_{0} between the slits. If the angular resolution of the eye is 160∘\dfrac{1}{60}^{\circ}, then the value of d0{d}_{0} is close to

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Answer: (C) 2mm2mm

When the angular fringe width in the interference pattern becomes less than the angular resolving power of the person's eye, the fringes will disappear.
The angular fringe width θ0=βD=λd0{\theta }_{0}=\dfrac{\beta }{D}=\dfrac{\lambda }{{d}_{0}}

d0=λθ0=600×10−9(160×π180){d}_{0}=\dfrac{\lambda }{{\theta }_{0}}=\dfrac{600\times {10}^{-9}}{(\dfrac{1}{60}\times \dfrac{\pi }{180})}
d0=600×60×180×10−9πm=2mm{d}_{0}=\dfrac{600\times 60\times 180\times {10}^{-9}}{\pi }m=2\text{mm}

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