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In a Young's double slit experiment, the fringe width is found to be $2 \mathrm{~mm}$, when light of wavelength $6000 Å$ is used. Find the change in fringe width if the whole apparatus is immersed in water of refractive index $1.33 .$
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The correct answer is:
$1.5 \mathrm{~mm}$
Fringe width $\beta=2 \mathrm{~mm}=2 \times 10^{-3} \mathrm{~m}$
$$
\begin{aligned}
\lambda &=6000 Å \\
\mu &=1.33 \\
\beta^{\prime} &=? \\
\beta^{\prime} &=\frac{\beta}{\mu} \\
&=\frac{2}{1.33}=1.5 \mathrm{~mm}
\end{aligned}
$$
$$
\begin{aligned}
\lambda &=6000 Å \\
\mu &=1.33 \\
\beta^{\prime} &=? \\
\beta^{\prime} &=\frac{\beta}{\mu} \\
&=\frac{2}{1.33}=1.5 \mathrm{~mm}
\end{aligned}
$$
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