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In Young's double slit experiment when a glass plate of refractive index 1.44 is introduced in the path of one of the interfering beams, the fringes are displaced by a distance ' $y$ '. If this plate is replaced by another plate of same thickness but of refractive index 1.66, the fringes will be displaced by a distance
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The correct answer is:
$\frac{3 y}{2}$
As a glass plate is used in one of the paths,
$$
\begin{aligned}
& \mathrm{y}_1=\frac{\beta}{\lambda}(1.44-1) \mathrm{t} \\
& \mathrm{y}_1=0.44 \mathrm{t} \times \frac{\beta}{\lambda}
\end{aligned}
$$
New displacement is:
$$
\begin{aligned}
\mathrm{y}_2 & =\frac{\beta}{\lambda}(1.66-1) \mathrm{t} \\
\mathrm{y}_2 & =0.66 \mathrm{t} \times \frac{\beta}{\lambda} \\
\frac{\mathrm{y}_2}{\mathrm{y}_1} & =\frac{0.66}{0.44} \\
\therefore \quad \mathrm{y}_2 & =\frac{3 \mathrm{y}}{2}
\end{aligned}
$$
$$
\begin{aligned}
& \mathrm{y}_1=\frac{\beta}{\lambda}(1.44-1) \mathrm{t} \\
& \mathrm{y}_1=0.44 \mathrm{t} \times \frac{\beta}{\lambda}
\end{aligned}
$$
New displacement is:
$$
\begin{aligned}
\mathrm{y}_2 & =\frac{\beta}{\lambda}(1.66-1) \mathrm{t} \\
\mathrm{y}_2 & =0.66 \mathrm{t} \times \frac{\beta}{\lambda} \\
\frac{\mathrm{y}_2}{\mathrm{y}_1} & =\frac{0.66}{0.44} \\
\therefore \quad \mathrm{y}_2 & =\frac{3 \mathrm{y}}{2}
\end{aligned}
$$
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