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The refractive index of a medium ‘x’ with respect to a medium ‘y’ is $\frac {2}{3}$ and the refractive index of medium ‘y’ with respect to medium ‘z’ is $\frac {4}{3}$. Calculate the refractive index of medium ‘z’ with respect to medium ‘x’. Find the speed of light in medium ‘y’ if the speed of light in medium ‘x’ is $3\times {10^8}ms^{-1}$.
Given:
The refractive index of medium 'x' with respect to medium 'y', $n_{xy}, or\ {\frac {n_x}{n_y}}$ = $\frac {2}{3}$
The refractive index of medium 'y' with respect to medium 'z', $n_{yz}, or\ {\frac {n_y}{n_z}}$ = $\frac {4}{3}$
To find: The refractive index of medium 'z' with respect to medium 'x', $n_{zx}$.
Solution:
We know that the refractive index of medium 1 with respect to medium 2 is reciprocal to the refractive index of medium 2 with respect to medium 1.
Therefore,
The refractive index of medium 'z' with respect to medium 'x', $n_{zx}$ is given as-
$n_{zx}=\frac {n_z}{n_x}=n_{zy}\times {n_{yz}}=\frac {n_z}{n_y}\times {\frac {n_y}{n_x}}$
Therefore,
$n_{zx}=\frac {1}{n_{yz}}\times {\frac {1}{n_{xy}}}$ $(\because {n_{zy}=\frac {1}{n_{yz}}})$
$n_{zx}=\frac {1}{\frac {n_y}{n_z}}$ $\times$ $\frac {1}{\frac {n_x}{n_y}}$
$n_{zx}=\frac {1}{\frac {4}{3}}$ $\times$ $\frac {1}{\frac {2}{3}}$
$n_{zx}=\frac {3}{4}$ $\times$ $\frac {3}{2}$
$n_{zx}=\frac {9}{8}$
Thus, the refractive index of medium 'z' with respect to medium 'x', $n_{zx}$ is $\frac {9}{8}$.
Let the speed of light in medium 'y' be 'V', and the speed of light in medium 'x' be 'C' i.e. $3\times {10^8}ms^{-1}$.
Therefore,
$n_{yx},\ or\ {\frac {n_y}{n_x}}\Rightarrow \frac {1}{n_{xy}}\Rightarrow \frac {1}{{\frac {n_x}{n_y}}}\Rightarrow \frac {1}{{\frac {2}{3}}}\Rightarrow \frac {3}{2}$
$n_{yx},\ or\ {\frac {n_y}{n_x}}=\frac {3}{2}$ $(\because {n_{yx}=\frac {1}{n_{xy}}})$
$\frac {y}{x}=\frac {3}{2}$ ................... (1)
$\frac {C}{V}=\frac {3\times {10^8}}{V}$ ................... (2)
Equating (1) and (2) we get-
$\frac {3}{2}=\frac {3\times {10^8}}{V}$
$3\times V=2\times {3\times {10^8}}$
$3\times V=6\times {10^8}$
$V=\frac {6\times {10^8}}{3}$
$V=2\times {10^8}$
Thus, the speed of light in the medium 'y' is $2\times {10^8}$.
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