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In a Young's double slit experiment with light of wavelength $\lambda$, fringe pattern on the screen has fringe width $\beta$. When two thin transparent glass (refractive index $\mu$ ) plates of thickness $t_1$ and $t_2$ $\left(t_1>t_2\right)$ are placed in the path of the two beams respectively, the fringe pattern will shift by a distance
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Verified Answer
The correct answer is:
$\frac{\beta(\mu-1}{\lambda}\left(t_1-t_2\right.$
$\frac{\beta(\mu-1}{\lambda}\left(t_1-t_2\right.$
$$
\text { } \begin{aligned}
\text { Shift } & =\frac{\beta(\mu-1)}{\lambda} t_1-\frac{\beta(\mu-1)}{\lambda} t_2 \\
& =\frac{\beta(\mu-1)}{\lambda}\left(t_1-t_2\right)
\end{aligned}
$$
\text { } \begin{aligned}
\text { Shift } & =\frac{\beta(\mu-1)}{\lambda} t_1-\frac{\beta(\mu-1)}{\lambda} t_2 \\
& =\frac{\beta(\mu-1)}{\lambda}\left(t_1-t_2\right)
\end{aligned}
$$
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