The refractive index of a medium 'X' with respect to 'Y' is $\dfrac{2}{3}$ and the refractive index of the medium ‘Y’ with respect to medium ‘Z’ is $\dfrac{4}{3}$. Calculate the refractive index of the medium ‘Z’ with respect to medium ‘X’.
Answer
Verified
458.4k+ views
Hint: The refractive index is defined as the ratio of velocity of light in vacuum or free space to the velocity of light in the given medium. Here we have to find the refractive index of a medium ‘X’ with respect to ‘Y’ and refractive index of a medium ‘Y’ with respect to ‘Z’.
Complete step by step solution:
Given:
Refractive index of a medium ‘X’ with respect to ‘Y’ i.e., ${N_{xy}} = \dfrac{2}{3}$
Refractive index of a medium ‘Y’ with respect to ‘Z’ i.e, ${N_{yz}} = \dfrac{4}{3}$
The refractive index is defined as the ratio of velocity of light in vacuum or free space to the velocity of light in the given medium and it is given by the formula:
\[{\text{Refractive index = }}\dfrac{{\left( {{\text{Velocity of light in vacuum}}} \right)}}{{\left( {{\text{Velocity of light in that medium}}} \right)}}\]
Refractive index of a medium ‘X’ with respect to ‘Y’ is:
${N_{XY}} = \dfrac{{{N_x}}}{{{N_y}}}$
Refractive index of a medium ‘Y’ with respect to ‘Z’ is:
${N_{yz}} = \dfrac{{{N_y}}}{{{N_z}}}$
The refractive index of ‘Z’ with respect to ‘X’ is given as:
${N_{zx}} = \dfrac{{{N_z}}}{{{N_x}}} = \dfrac{{{N_y}}}{{{N_x}}} = \dfrac{{{N_z}}}{{{N_y}}}$
\[\Rightarrow {\text{ }}{N_{zx}}{\text{ = }}\dfrac{1}{{{N_{xy}}}} \times \dfrac{1}{{{N_{yz}}}}\] …… (i)
On substituting the given values of ${N_{xy}} = \dfrac{2}{3}$ and ${N_{yz}} = \dfrac{4}{3}$ in equation (i) we get,
\[\therefore {\text{ }}{N_{zx}}{\text{ = }}\dfrac{9}{8}\]
So, the refractive index of medium ‘Z’ with respect to medium ‘X’ is equal to $\dfrac{9}{8}$.
Note: The phase velocity is the speed at which the crests of the wave move and can be faster than the speed of light in vacuum, and thereby give a refractive index below one. This can occur close to resonance frequencies, for absorbing media, in plasmas, and for X-rays.
Complete step by step solution:
Given:
Refractive index of a medium ‘X’ with respect to ‘Y’ i.e., ${N_{xy}} = \dfrac{2}{3}$
Refractive index of a medium ‘Y’ with respect to ‘Z’ i.e, ${N_{yz}} = \dfrac{4}{3}$
The refractive index is defined as the ratio of velocity of light in vacuum or free space to the velocity of light in the given medium and it is given by the formula:
\[{\text{Refractive index = }}\dfrac{{\left( {{\text{Velocity of light in vacuum}}} \right)}}{{\left( {{\text{Velocity of light in that medium}}} \right)}}\]
Refractive index of a medium ‘X’ with respect to ‘Y’ is:
${N_{XY}} = \dfrac{{{N_x}}}{{{N_y}}}$
Refractive index of a medium ‘Y’ with respect to ‘Z’ is:
${N_{yz}} = \dfrac{{{N_y}}}{{{N_z}}}$
The refractive index of ‘Z’ with respect to ‘X’ is given as:
${N_{zx}} = \dfrac{{{N_z}}}{{{N_x}}} = \dfrac{{{N_y}}}{{{N_x}}} = \dfrac{{{N_z}}}{{{N_y}}}$
\[\Rightarrow {\text{ }}{N_{zx}}{\text{ = }}\dfrac{1}{{{N_{xy}}}} \times \dfrac{1}{{{N_{yz}}}}\] …… (i)
On substituting the given values of ${N_{xy}} = \dfrac{2}{3}$ and ${N_{yz}} = \dfrac{4}{3}$ in equation (i) we get,
\[\therefore {\text{ }}{N_{zx}}{\text{ = }}\dfrac{9}{8}\]
So, the refractive index of medium ‘Z’ with respect to medium ‘X’ is equal to $\dfrac{9}{8}$.
Note: The phase velocity is the speed at which the crests of the wave move and can be faster than the speed of light in vacuum, and thereby give a refractive index below one. This can occur close to resonance frequencies, for absorbing media, in plasmas, and for X-rays.
Recently Updated Pages
The correct geometry and hybridization for XeF4 are class 11 chemistry CBSE
Water softening by Clarks process uses ACalcium bicarbonate class 11 chemistry CBSE
With reference to graphite and diamond which of the class 11 chemistry CBSE
A certain household has consumed 250 units of energy class 11 physics CBSE
The lightest metal known is A beryllium B lithium C class 11 chemistry CBSE
What is the formula mass of the iodine molecule class 11 chemistry CBSE
Trending doubts
The reservoir of dam is called Govind Sagar A Jayakwadi class 11 social science CBSE
What problem did Carter face when he reached the mummy class 11 english CBSE
Proton was discovered by A Thomson B Rutherford C Chadwick class 11 chemistry CBSE
In China rose the flowers are A Zygomorphic epigynous class 11 biology CBSE
What is Environment class 11 chemistry CBSE
Nucleolus is present in which part of the cell class 11 biology CBSE