
The refractive index of a medium $X$ with respect to medium $Y$ is $\dfrac{2}{3}$. and the refractive index of medium $Y$ with respect to medium $Z$ is $\dfrac{4}{3}$. Find the refractive index of the medium $'z'$ with respect to medium $'x'$. Calculate the speed of light in medium $'y'$, when the speed of light in medium $'x'$ is $3 \times 10^8 m/s$.
Answer
216.6k+ views
Hint: Refractive index is the ratio of the speed of light in vacuum to the speed of light in denser medium.
It is denoted as;
speed of light in vacuum/ speed of light in denser medium.
Using the above relations we will solve the given problem
Complete step by step solution:
Let us first describe the Refractive index in more detail and then we will do the calculation part of the problem.
Refractive index: It is a measure of the bending of ray of light when passing from one medium to another medium. Refractive index depends upon the frequency of light passing through, higher the frequency higher is the refractive index.
Now we will do the calculation part of the problem;
We have refractive index of X with Y and Y with Z respectively;
${N_{XY}} = \dfrac{2}{3},{N_{YZ}} = \dfrac{4}{3}$
We have to find the value of refractive index of medium Z with X
$ \Rightarrow {N_{ZX}} = \dfrac{{{N_{ZY}}}}{{{N_{YX}}}}$
$ \Rightarrow {N_{ZX}} = \dfrac{{\dfrac{1}{{\dfrac{4}{3}}}}}{{\dfrac{1}{{\dfrac{2}{3}}}}}$ (We have substituted the value of reciprocal of both the refractive indexes)
$ \Rightarrow {N_{ZX}} = \dfrac{1}{2}$ (refractive index of medium Z with respect to X)
Second part of the question is;
As per the definition of refractive index we can write;
${N_{YX}} = \dfrac{3}{2}$ or $\dfrac{3}{2} = \dfrac{{{V_y}}}{{3 \times {{10}^8}}}$ (${V_y}$ is the speed of light in medium Y, speed of light in medium X is given)
$ \Rightarrow {V_y} = \dfrac{{3 \times {{10}^8} \times 2}}{3}$
$ \Rightarrow {V_y} = 2 \times {10^8}$ m/s
Note: Refractive index is an important property of any optical instrument, refractive index determines the focusing power of the lens, it determines the dispersive power of the prism, the reflectivity of the lens coatings, and the light guiding nature of optical fiber, refractive index is used to find the purity of a particular substance.
It is denoted as;
speed of light in vacuum/ speed of light in denser medium.
Using the above relations we will solve the given problem
Complete step by step solution:
Let us first describe the Refractive index in more detail and then we will do the calculation part of the problem.
Refractive index: It is a measure of the bending of ray of light when passing from one medium to another medium. Refractive index depends upon the frequency of light passing through, higher the frequency higher is the refractive index.
Now we will do the calculation part of the problem;
We have refractive index of X with Y and Y with Z respectively;
${N_{XY}} = \dfrac{2}{3},{N_{YZ}} = \dfrac{4}{3}$
We have to find the value of refractive index of medium Z with X
$ \Rightarrow {N_{ZX}} = \dfrac{{{N_{ZY}}}}{{{N_{YX}}}}$
$ \Rightarrow {N_{ZX}} = \dfrac{{\dfrac{1}{{\dfrac{4}{3}}}}}{{\dfrac{1}{{\dfrac{2}{3}}}}}$ (We have substituted the value of reciprocal of both the refractive indexes)
$ \Rightarrow {N_{ZX}} = \dfrac{1}{2}$ (refractive index of medium Z with respect to X)
Second part of the question is;
As per the definition of refractive index we can write;
${N_{YX}} = \dfrac{3}{2}$ or $\dfrac{3}{2} = \dfrac{{{V_y}}}{{3 \times {{10}^8}}}$ (${V_y}$ is the speed of light in medium Y, speed of light in medium X is given)
$ \Rightarrow {V_y} = \dfrac{{3 \times {{10}^8} \times 2}}{3}$
$ \Rightarrow {V_y} = 2 \times {10^8}$ m/s
Note: Refractive index is an important property of any optical instrument, refractive index determines the focusing power of the lens, it determines the dispersive power of the prism, the reflectivity of the lens coatings, and the light guiding nature of optical fiber, refractive index is used to find the purity of a particular substance.
Recently Updated Pages
Wheatstone Bridge Explained: Working, Formula & Uses

Young’s Double Slit Experiment Derivation Explained

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

Electricity and Magnetism Explained: Key Concepts & Applications

Chemical Properties of Hydrogen - Important Concepts for JEE Exam Preparation

Trending doubts
JEE Main 2026: Application Form Open, Exam Dates, Syllabus, Eligibility & Question Papers

Derivation of Equation of Trajectory Explained for Students

Hybridisation in Chemistry – Concept, Types & Applications

Understanding the Angle of Deviation in a Prism

Understanding Collisions: Types and Examples for Students

How to Convert a Galvanometer into an Ammeter or Voltmeter

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Ideal and Non-Ideal Solutions Explained for Class 12 Chemistry

Degree of Dissociation: Meaning, Formula, Calculation & Uses

Understanding Electromagnetic Waves and Their Importance

Understanding the Electric Field of a Uniformly Charged Ring

Understanding Average and RMS Value in Electrical Circuits

