
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
572.4k+ 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
Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

Discuss the various forms of bacteria class 11 biology CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

State the laws of reflection of light

Explain zero factorial class 11 maths CBSE

