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: $\Delta x = \frac{(\mu - 1) t D}{d}$.We know that

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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: $\Delta x = \frac{(\mu - 1) t D}{d}$.We know that the fringe width is $\beta = \frac{\lambda D}{d}$,so $\frac{D}{d} = \frac{\beta}{\lambda}$.Substituting this,the shift is $\Delta x = \frac{(\mu - 1) t \beta}{\lambda}$.Given that the shift $\Delta x = 5 \beta$,we have: $5 \beta = \frac{(\mu - 1) t \beta}{\lambda}$.Canceling $\beta$ from both sides: $5 = \frac{(\mu - 1) t}{\lambda}$.Substituting the given values $\mu = 1.5$ and $t = 6 	imes 10^{-6} \ m$:$5 = \frac{(1.5 - 1) 	imes 6 	imes 10^{-6}}{\lambda}$.$5 = \frac{0.5 	imes 6 	imes 10^{-6}}{\lambda}$.$5 = \frac{3 	imes 10^{-6}}{\lambda}$.$\lambda = \frac{3 	imes 10^{-6}}{5} = 0.6 	imes 10^{-6} \ m = 6 	imes 10^{-7} \ m$.Converting to $\mathring{A}$: $\lambda = 6000 \ \mathring{A}$.

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. Find the wavelength of light in $\mathring{A}$.

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