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Standard X

Physics

Lens Formula

If the refrac...

Question

If the refractive index of medium 2 with respect to medium 1 is $n_{21}$ then the refractive index of medium 1 with respect to medium 2 is

$n_{12} = n_{21}$
$n_{12} = \frac{1}{n_{21}}$
$n_{12} = - n_{21}$
$n_{12} = \frac{- 1}{n_{21}}$

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Solution

The correct option is B
$n_{12} = \frac{1}{n_{21}}$

The correct option is (B)
Consider the speed of light in medium 1 is $v_{1}$ and the speed of light in medium 2 is $v_{2}$
The relative refractive index $n_{12}$ is the ratio of $v_{1}$ and $v_{2}$ such that
$n_{12} = \frac{v_{1}}{v_{2}}$
Similarly, the relative refractive index $n_{21}$ is the ratio of $v_{2}$ and $v_{1}$ such that
$n_{21} = \frac{v_{2}}{v_{1}}$
Therefore, from both the relations mentioned above, we get
$n_{12} = \frac{1}{n_{21}}$

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If the refractive index of medium 2 with respect to medium 1 is n21 then the refractive index of medium 1 with respect to medium 2 isn12=n21n12=1n21n12=-n21n12=-1n21

If the refractive index of medium 2 with respect to medium 1 is $n_{21}$ then the refractive index of medium 1 with respect to medium 2 is

$n_{12} = n_{21}$
$n_{12} = \frac{1}{n_{21}}$
$n_{12} = - n_{21}$
$n_{12} = \frac{- 1}{n_{21}}$