In a Fresnel's biprism experiment,a mica shee | vedclass

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(B) The shift in the central fringe due to the introduction of a thin sheet is given by the formula: $\Delta x = \frac{D}{d} t(\mu - 1)$.Given that th

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$6000$

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(B) The shift in the central fringe due to the introduction of a thin sheet is given by the formula: $\Delta x = \frac{D}{d} t(\mu - 1)$.Given that the shift is equal to $5$ fringe widths,we have $\Delta x = 5\beta$,where $\beta = \frac{\lambda D}{d}$ is the fringe width.Equating the two expressions: $\frac{D}{d} t(\mu - 1) = 5 \left( \frac{\lambda D}{d} ight)$.Simplifying the equation: $t(\mu - 1) = 5\lambda$.Substituting the given values: $6 	imes 10^{-6} 	imes (1.5 - 1) = 5\lambda$.$6 	imes 10^{-6} 	imes 0.5 = 5\lambda$.$3 	imes 10^{-6} = 5\lambda$.$\lambda = \frac{3 	imes 10^{-6}}{5} = 0.6 	imes 10^{-6} \ m$.Converting to $\mathring{A}$: $\lambda = 0.6 	imes 10^{-6} 	imes 10^{10} \ \mathring{A} = 6000 \ \mathring{A}$.

In a Fresnel's biprism experiment,a mica sheet of refractive index $1.5$ and thickness $6 \times 10^{-6} \ m$ is placed in the path of one of the interfering beams. As a result,the central fringe shifts by a distance equal to $5$ fringe widths. The wavelength of light used is .... $\mathring{A}$

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