Answer
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Hint: The refractive index of a medium is a measure of the speed of light in the medium. In a medium with a higher refractive index (optically denser), light travels slower than that in a medium of lower refractive index (optically rarer). It can also be written as the ratio of the speed of light in vacuum (or air) to the speed of light in that medium.
Formula used: The refractive index $n$ of a medium in which the speed of light is $v$ is given by
$n=\dfrac{c}{v}$
where $c=3\times {{10}^{8}}m/s$ is the speed of light in vacuum (or air).
Complete step by step answer:
We will solve this problem by using the formula for the refractive index of a medium and plugging in the information given in the question. Hence, let us proceed to do that.
The refractive index $n$ of a medium in which the speed of light is $v$ is given by
$n=\dfrac{c}{v}$ -(1)
where $c=3\times {{10}^{8}}m/s$ is the speed of light in vacuum (or air).
Now, let the speed of light in diamond be $v=121000km.{{s}^{-1}}=121000\times {{10}^{3}}m.{{s}^{-1}}=1.21\times {{10}^{8}}m.{{s}^{-1}}$ $\left( \because 1km.{{s}^{-1}}={{10}^{3}}m.{{s}^{-1}} \right)$
Let the refractive index of diamond be $n$.
Putting this information in (1), we get,
$n=\dfrac{3\times {{10}^{8}}}{1.21\times {{10}^{8}}}=\dfrac{3}{1.21}\approx 2.47$
Hence, the refractive index of diamond is $2.47$.
Therefore, the correct option is B) $2.47$.
Note: Students sometimes get confused in the ratio for the formula for the refractive index of a medium. A good way to remember is that theoretically nothing and travel faster than the speed of light and the refractive index of a medium cannot be greater than one in general. Hence, the speed of light in the ratio must be in the numerator so that the ratio (or the refractive index of the medium) comes out to be greater than one.
Formula used: The refractive index $n$ of a medium in which the speed of light is $v$ is given by
$n=\dfrac{c}{v}$
where $c=3\times {{10}^{8}}m/s$ is the speed of light in vacuum (or air).
Complete step by step answer:
We will solve this problem by using the formula for the refractive index of a medium and plugging in the information given in the question. Hence, let us proceed to do that.
The refractive index $n$ of a medium in which the speed of light is $v$ is given by
$n=\dfrac{c}{v}$ -(1)
where $c=3\times {{10}^{8}}m/s$ is the speed of light in vacuum (or air).
Now, let the speed of light in diamond be $v=121000km.{{s}^{-1}}=121000\times {{10}^{3}}m.{{s}^{-1}}=1.21\times {{10}^{8}}m.{{s}^{-1}}$ $\left( \because 1km.{{s}^{-1}}={{10}^{3}}m.{{s}^{-1}} \right)$
Let the refractive index of diamond be $n$.
Putting this information in (1), we get,
$n=\dfrac{3\times {{10}^{8}}}{1.21\times {{10}^{8}}}=\dfrac{3}{1.21}\approx 2.47$
Hence, the refractive index of diamond is $2.47$.
Therefore, the correct option is B) $2.47$.
Note: Students sometimes get confused in the ratio for the formula for the refractive index of a medium. A good way to remember is that theoretically nothing and travel faster than the speed of light and the refractive index of a medium cannot be greater than one in general. Hence, the speed of light in the ratio must be in the numerator so that the ratio (or the refractive index of the medium) comes out to be greater than one.
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