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Question

(C) When a light wave enters a medium of refractive index $\mu$,its wavelength becomes $\lambda^{\prime} = \frac{\lambda}{\mu}$.Fringe width in $	ext

Vedclass · Accepted Answer

$1.35$

Vedclass · Answer

(C) When a light wave enters a medium of refractive index $\mu$,its wavelength becomes $\lambda^{\prime} = \frac{\lambda}{\mu}$.Fringe width in $	ext{YDSE}$ is given by $\beta = \frac{\lambda^{\prime} D}{d}$.Given: $\lambda = 550 	imes 10^{-9} \ m$,$D = 0.6 \ m$,$d = 0.2 	imes 10^{-3} \ m$,and $\mu = \frac{11}{9}$.Substituting the values:$\beta = \frac{(\lambda / \mu) D}{d} = \frac{\lambda D}{\mu d} = \frac{550 	imes 10^{-9} 	imes 0.6}{(11/9) 	imes 0.2 	imes 10^{-3}}$.$\beta = \frac{550 	imes 10^{-9} 	imes 0.6 	imes 9}{11 	imes 0.2 	imes 10^{-3}} = \frac{330 	imes 10^{-9} 	imes 9}{2.2 	imes 10^{-3}} = \frac{2970 	imes 10^{-9}}{2.2 	imes 10^{-3}} = 1350 	imes 10^{-6} \ m = 1.35 \ mm$.

Similar Questions

In Young's double-slit experiment,the position of the $5^{\text{th}}$ bright fringe from the central maximum is $5\,cm$. The distance between the slits and the screen is $1\,m$ and the wavelength of the monochromatic light used is $600\,nm$. The separation between the slits is $........\mu m$.

In Young's double slit experiment,the amplitudes of the two waves incident on the two slits are $A$ and $2A$. If $I_{0}$ is the maximum intensity,then the intensity at a spot on the screen,where the phase difference between the two interfering waves is $\phi$,is:

In $YDSE$ experiment,when two light rays form the third minima,they have:

The fringe width in Young's double-slit experiment increases when:

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