Question

Green light of wavelength $5460A{}^\circ $ is incident on an air-glass interface. If the refractive index of glass is $1.5$, the wavelength of light in glass would be (given-velocity of light in air $c=3\times {{10}^{8}}m{{s}^{-1}}$)
$\begin{align}
  & A)3460A{}^\circ \\
 & B)5460A{}^\circ \\
 & C)4861A{}^\circ \\
\end{align}$
$D)$None of the above

Answer 1

Hint: Frequency of light is independent of the medium of propagation. Firstly, the frequency of green light is determined using the given wavelength of green light in air. Using the refractive index of glass, velocity of green light in glass is calculated. Finally, using the obtained frequency and velocity of green light in glass, the wavelength of green light inside the glass is calculated.
Formula used:
$1)f=\dfrac{c}{\lambda }$
$2)v=\dfrac{c}{n}$
$3)v=f\lambda $

Complete answer:
Frequency of light does not depend on the medium through which the light is travelling. It is dependent only on the source of light. Mathematically, frequency of light is given by
$f=\dfrac{c}{\lambda }$
where
$f$ is the frequency of light
$c$ is the speed of light in air
$\lambda $ is the wavelength of light in air
Let this be equation 1.
Coming to our question, we are given that green light of wavelength $5460A{}^\circ $ is incident on an air-glass interface. If the refractive index of glass is $1.5$ and velocity of light in air is $3\times {{10}^{8}}m{{s}^{-1}}$, we are required to determine the wavelength of green light in glass.
Let us call the wavelength of green light in air ${{\lambda }_{1}}$. Clearly,
${{\lambda }_{1}}=5460A{}^\circ =5460\times {{10}^{-10}}m$
If we call the frequency of green light $f$, using equation 1, we have
$f=\dfrac{c}{{{\lambda }_{1}}}=\dfrac{3\times {{10}^{8}}m{{s}^{-1}}}{5460\times {{10}^{-10}}m}=5.49\times {{10}^{14}}Hz=5.49\times {{10}^{14}}{{s}^{-1}}$
where
$f$ is the frequency of green light
$c$ is the speed of light in air ($=3\times {{10}^{8}}m{{s}^{-1}}$, as given in the question)
${{\lambda }_{1}}$ is the wavelength of green light in air
Let this be equation 2.
Now, we know that velocity of light in a particular material or medium is given by
$v=\dfrac{c}{n}$
where
$n$ is the refractive index of a material or a medium
$c$ is the speed of light in air
$v$ is the velocity of light in that material or that medium
Let this be equation 3.
Using equation 3, let us determine the velocity of green light in glass. Clearly,
$v=\dfrac{c}{n}=\dfrac{3\times {{10}^{8}}m{{s}^{-1}}}{1.5}=2\times {{10}^{8}}m{{s}^{-1}}$
where
$n=1.5$, is the refractive index of glass (as given in the question)
$c=3\times {{10}^{8}}m{{s}^{-1}}$, is the speed of light in air (as given in the question)
$v$ is the velocity of green light in glass
Let this be equation 4.
Velocity of light in a material is also equal to the product of frequency of light propagating through the material and the wavelength of light propagating through the material. Clearly,
$v=f\lambda $
where
$v$ is the velocity of light propagating through a particular material
$f$ is the frequency of light
$\lambda $ is the wavelength of light in that particular material
Let this be equation 5.
Using equation 5, let us calculate the wavelength of green light in glass. If wavelength of green light in glass is denoted as ${{\lambda }_{2}}$, then, using equation 5, ${{\lambda }_{2}}$ is given by
$v=f{{\lambda }_{2}}\Rightarrow 2\times {{10}^{8}}m{{s}^{-1}}=5.49\times {{10}^{14}}{{s}^{-1}}\times {{\lambda }_{2}}\Rightarrow {{\lambda }_{2}}=\dfrac{2\times {{10}^{8}}m{{s}^{-1}}}{5.49\times {{10}^{14}}{{s}^{-1}}}=3642A{}^\circ $
Therefore, the wavelength of green light in glass is equal to $3642A{}^\circ $.

Since this answer is not there in the first three options provided, option $D$ is the correct answer.

Note:
Students need to be thorough with conversion formulas. Conversion formulas used in the solution given above are:
$\begin{align}
  & 1A{}^\circ ={{10}^{-10}}m \\
 & 1Hz=1{{s}^{-1}} \\
\end{align}$
Students should also try to remember that frequency of light is independent of the medium of propagation whereas wavelength of light is dependent on the medium of propagation.