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On replacing a thin film of mica of thickness $12 \times 10^{-5} \mathrm{~cm}$ in the path of one of the interfering beams in Young's double slit experiment using monochromatic light, the fringe pattern shifts through a distance equal to the width of bright fringe. If $\lambda=6 \times 10^{-5} \mathrm{~cm}$, the refractive index of mica is
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The correct answer is:
1.5
The equation for the shift in the fringe pattern is given as,
$\mathrm{n} \lambda=(\mu-1) \mathrm{t}$
$\therefore \quad$ Refractive index of mica is:
$\begin{array}{ll}
& \mu=\frac{\mathrm{n} \lambda}{\mathrm{t}}+1 \\
\therefore \quad & \mu=\frac{1 \times 6 \times 10^{-5}}{12 \times 10^{-5}}+1=0.5+1 \\
\therefore \quad & \mu=1.5
\end{array}$
$\mathrm{n} \lambda=(\mu-1) \mathrm{t}$
$\therefore \quad$ Refractive index of mica is:
$\begin{array}{ll}
& \mu=\frac{\mathrm{n} \lambda}{\mathrm{t}}+1 \\
\therefore \quad & \mu=\frac{1 \times 6 \times 10^{-5}}{12 \times 10^{-5}}+1=0.5+1 \\
\therefore \quad & \mu=1.5
\end{array}$
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