In Young's double slit experiment,the two sli | vedclass

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(A) Given: Slit separation $d = 0.6 \; mm = 0.6 	imes 10^{-3} \; m$. Screen distance $D = 80 \; cm = 0.8 \; m$.The position of the first dark fringe

Vedclass · Accepted Answer

$450$

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(A) Given: Slit separation $d = 0.6 \; mm = 0.6 	imes 10^{-3} \; m$. Screen distance $D = 80 \; cm = 0.8 \; m$.The position of the first dark fringe from the central axis is given by $y = \frac{D \lambda}{2d}$.According to the problem,the first dark fringe is observed directly opposite to one of the slits,which means its distance from the central axis is $y = \frac{d}{2}$.Equating the two expressions: $\frac{D \lambda}{2d} = \frac{d}{2}$.Solving for wavelength $\lambda$: $\lambda = \frac{d^2}{D}$.Substituting the values: $\lambda = \frac{(0.6 	imes 10^{-3})^2}{0.8} = \frac{0.36 	imes 10^{-6}}{0.8} = 0.45 	imes 10^{-6} \; m$.Converting to nanometers: $\lambda = 450 	imes 10^{-9} \; m = 450 \; nm$.

In Young's double slit experiment,the two slits are $0.6 \; mm$ apart. The interference pattern is observed on a screen at a distance of $80 \; cm$ from the slits. The first dark fringe is observed on the screen directly opposite to one of the slits. The wavelength of light is $\dots \; nm$.

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