Question

When the orange light passes from air (\[n=1\]) into glass ($n=1.5$), what is its new wavelength?
$A.4\times {{10}^{-7}}m$
$B.$$4\times {{10}^{-6}}m$
$C.2.5\times {{10}^{-7}}m$
$D.6\times {{10}^{-7}}m$
$E.9\times {{10}^{-7}}m$

Answer 1

Hint: We will use the concept of absolute refractive index and formula of wavelength in terms of frequency and speed of light. We also know that when a light ray travels from one medium to another then the wavelength and speed of light change but frequency remains the same.
Formula Used: We will use the following formula to solve the given problem:-
$\dfrac{c}{n}=\nu \lambda $.

Complete step by step answer:
We know that the absolute refractive index of a medium is defined as the ratio of speed of light in vacuum to the speed of light in the medium. If $c$ is speed of light in vacuum, $v$ is speed of light in the medium and $n$ is the absolute refractive index of the medium, then mathematically
$n=\dfrac{c}{v}$………………. $(i)$
From $(i)$ we can write
$v=\dfrac{c}{n}$……………… $(ii)$
But for velocity of light in a medium can also be represented with the help of wavelength $(\lambda )$ and frequency $(\nu )$ by following equation:-
$v=\nu \times \lambda $…………………. $(iii)$
For air we have velocity of light, $v$equal to $c$ and wavelength of orange light, ${{\lambda }_{a}}=600nm=600\times {{10}^{-9}}m$.
From equation $(ii)$ and $(iii)$ for air we get
$\dfrac{c}{{{n}_{a}}}=\nu \times {{\lambda }_{a}}$……………. $(iv)$ where the refractive index of air is denoted as ${{n}_{a}}=1$.
Putting respective values in $\left( iv \right)$, we get
$\dfrac{c}{1}=\nu \times 600\times {{10}^{-9}}$
$\Rightarrow \nu =\dfrac{c}{600\times {{10}^{-9}}}$………………… $(v)$
For glass we have wavelength,${{\lambda }_{g}}$ which is to be calculated. Refractive index, ${{n}_{g}}=1.5$ and frequency of the orange light will remain the same as it does not change in different media. We have the following equation on the basis of equation $(ii)$ and $\left( iii \right)$:-
$\dfrac{c}{{{n}_{g}}}=\nu \times {{\lambda }_{g}}$…………………….. $(vi)$
Putting respective values in $(vi)$, we get
$\dfrac{c}{1.5}=\nu \times {{\lambda }_{g}}$
$\Rightarrow \nu =\dfrac{c}{1.5\times {{\lambda }_{g}}}$……………….. $(vii)$
We know that the frequency of light is same in both the cases and therefore from equations $(v)$ and $(vii)$, we get
$\dfrac{c}{600\times {{10}^{-9}}}=\dfrac{c}{1.5\times {{\lambda }_{g}}}$
Solving further we get
$600\times {{10}^{-9}}=1.5\times {{\lambda }_{g}}$
$\Rightarrow {{\lambda }_{g}}=\dfrac{600\times {{10}^{-9}}}{1.5}$
$\Rightarrow {{\lambda }_{g}}=400\times {{10}^{-9}}m$
$\Rightarrow {{\lambda }_{g}}=4\times {{10}^{-7}}m$

Hence, the correct option is $(A)$.

Note:
We should keep in mind that the frequency of light does not change with change in medium. We should be clear in the concept of absolute refractive index and should not consider it as a relative refractive index. Correct formula should be used and colour of the light should also be taken care of.