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A monochromatic beam of light is travelling from medium $A$ of refractive index $n_{1}$ to a medium $\mathrm{B}$ of refractive index $\mathrm{n}_{2}$. In the medium $A$, there are $x$ number of waves in certain distance. In the medium B, there are $y$ number of waves in the same distance. Then, refractive index of medium $A$ with respect to medium $B$ is
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Verified Answer
The correct answer is:
$\frac{x}{y}$
If $x$ number of waves of wavelength $\lambda_{1}$ and $y$ number of waves of wavelength $\lambda_{2}$ are present in same distance s, then
$$
\begin{array}{ll}
& \mathrm{s}=\mathrm{x} \lambda_{1}=\mathrm{y} \lambda_{2} \\
\text { or } \quad \frac{\mathrm{x}}{\mathrm{y}}=\frac{\lambda_{2}}{\lambda_{1}}=\frac{\mathrm{n}_{1}}{\mathrm{n}_{2}} \\
\therefore & \frac{\mathrm{n}_{1}}{\mathrm{n}_{2}}=\frac{\mathrm{x}}{\mathrm{y}}
\end{array}
$$
$$
\begin{array}{ll}
& \mathrm{s}=\mathrm{x} \lambda_{1}=\mathrm{y} \lambda_{2} \\
\text { or } \quad \frac{\mathrm{x}}{\mathrm{y}}=\frac{\lambda_{2}}{\lambda_{1}}=\frac{\mathrm{n}_{1}}{\mathrm{n}_{2}} \\
\therefore & \frac{\mathrm{n}_{1}}{\mathrm{n}_{2}}=\frac{\mathrm{x}}{\mathrm{y}}
\end{array}
$$
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