In a YDSE,light of wavelength lambda = 5000 ; | vedclass

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(B) When a transparent sheet of thickness $t$ and refractive index $\mu$ is placed in front of one of the slits,an additional path difference is intro

Vedclass · Accepted Answer

$4.9^{\circ}$ and $\frac{D(\mu-1) t}{d}$

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(B) When a transparent sheet of thickness $t$ and refractive index $\mu$ is placed in front of one of the slits,an additional path difference is introduced.The path difference introduced by the sheet is $\Delta x = (\mu - 1)t$.For the central maxima to shift to a new position at an angle $	heta$,the total path difference must be zero at that point. Thus,the path difference due to the geometry of the slits must compensate for the path difference introduced by the sheet:$d \sin 	heta = (\mu - 1)t$$\sin 	heta = \frac{(\mu - 1)t}{d}$Substituting the given values:$\sin 	heta = \frac{(1.17 - 1) 	imes 1.5 	imes 10^{-7}}{3 	imes 10^{-7}}$$\sin 	heta = \frac{0.17 	imes 1.5}{3} = \frac{0.255}{3} = 0.085$$	heta = \sin^{-1}(0.085) \approx 4.875^{\circ} \approx 4.9^{\circ}$.For small angles,$\sin 	heta \approx 	an 	heta = \frac{y}{D}$.Therefore,$\frac{y}{D} = \frac{(\mu - 1)t}{d}$,which gives the linear shift $y = \frac{D(\mu - 1)t}{d}$.

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