Wave Optics

JEE Main level, numerical answer, medium

51 of 191

Question 51 of 191

Monochromatic light of wavelength 500 nm500 \mathrm{~nm} is used in Young's double slit experiment. An interference pattern is obtained on a screen. When one of the slits is covered with a very thin glass plate (refractive index =1.5=1.5 ), the central maximum is shifted to a position previously occupied by the 4th 4^{\text {th }} bright fringe. The thickness of the glass-plate is ______ μm\mu \mathrm{m}.

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Answer: 4

(μ−1)t=nλ(1.5−1)t=4×500×10−9 mt=4000×10−9 mt=4μm\begin{aligned} & (\mu-1) \mathrm{t}=\mathrm{n} \lambda \\ & (1.5-1) \mathrm{t}=4 \times 500 \times 10^{-9} \mathrm{~m} \\ & \mathrm{t}=4000 \times 10^{-9} \mathrm{~m} \\ & \mathrm{t}=4 \mu \mathrm{m}\end{aligned}

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