In a Young's double-slit experiment with wave | vedclass

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(A) The shift in the fringe pattern due to a glass plate of thickness $t$ and refractive index $\mu$ is given by $\Delta x = \frac{\beta}{\lambda} (\m

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$\frac{\beta (\mu - 1)}{\lambda} (t_1 - t_2)$

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(A) The shift in the fringe pattern due to a glass plate of thickness $t$ and refractive index $\mu$ is given by $\Delta x = \frac{\beta}{\lambda} (\mu - 1) t$.When two plates of thicknesses $t_1$ and $t_2$ are placed in the paths of the two beams,the shift produced by the first plate is $\Delta x_1 = \frac{\beta}{\lambda} (\mu - 1) t_1$ and the shift produced by the second plate is $\Delta x_2 = \frac{\beta}{\lambda} (\mu - 1) t_2$.Since the plates are placed in the paths of the two different beams,the net shift is the difference between the individual shifts:$\Delta x_{net} = |\Delta x_1 - \Delta x_2| = \left| \frac{\beta}{\lambda} (\mu - 1) t_1 - \frac{\beta}{\lambda} (\mu - 1) t_2 \right|$.Simplifying this,we get $\Delta x_{net} = \frac{\beta}{\lambda} (\mu - 1) (t_1 - t_2)$.

In a Young's double-slit experiment with wavelength $\lambda$,the fringe width is $\beta$. When two glass plates of thicknesses $t_1$ and $t_2$ $(t_1 > t_2)$ and refractive index $\mu$ are placed in the paths of the two light beams respectively,by what distance will the fringe pattern shift?

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