A double slit experiment is immersed in water | vedclass

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(B) Given:Slit separation $d = 1 ~mm = 10^{-3} ~m$Distance between slit and screen $D = 1.33 ~m$Refractive index of water $\mu = 1.33$Wavelength of li

Vedclass · Accepted Answer

$6.3 	imes 10^{-4} ~m$

Vedclass · Answer

(B) Given:Slit separation $d = 1 ~mm = 10^{-3} ~m$Distance between slit and screen $D = 1.33 ~m$Refractive index of water $\mu = 1.33$Wavelength of light in air $\lambda = 6300 ~Å = 6.3 	imes 10^{-7} ~m$The wavelength of light in water is given by $\lambda_w = \frac{\lambda}{\mu} = \frac{6.3 	imes 10^{-7}}{1.33} ~m$.The fringe width $X$ is given by the formula:$X = \frac{\lambda_w D}{d}$Substituting the values:$X = \frac{(6.3 	imes 10^{-7} / 1.33) 	imes 1.33}{10^{-3}}$$X = \frac{6.3 	imes 10^{-7}}{10^{-3}} = 6.3 	imes 10^{-4} ~m$.Thus,the fringe width is $6.3 	imes 10^{-4} ~m$.

$A$ double slit experiment is immersed in water of refractive index $1.33$. The slit separation is $1 ~mm$ and the distance between the slit and the screen is $1.33 ~m$. The slits are illuminated by light of wavelength $6300 ~Å$. The fringe width is:

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