Data Structure
Networking
RDBMS
Operating System
Java
MS Excel
iOS
HTML
CSS
Android
Python
C Programming
C++
C#
MongoDB
MySQL
Javascript
PHP
Physics
Chemistry
Biology
Mathematics
English
Economics
Psychology
Social Studies
Fashion Studies
Legal Studies
- Selected Reading
- UPSC IAS Exams Notes
- Developer's Best Practices
- Questions and Answers
- Effective Resume Writing
- HR Interview Questions
- Computer Glossary
- Who is Who
State Snell's law of refraction of light. Write an expression to relate refractive index of a medium with speed to light in vacuum.The refractive index of a medium 'a' with respect to medium 'b' is 2/3 and the refractive index of medium 'b' with respect to medium 'c' is 4/3. Find the refractive index of medium 'c' with respect to medium 'a'.
Snell's laws of refraction of light.
1. The incident ray, refracted ray and normal to the refracted surface lie in the same plane.
2. The ratio of the sine of the angle of incidence to the sine of the angle of refraction is constant.
Mathematically, it can be given as-
$\frac {sin\ i}{sin\ r}=n_{ab}$
Here, $n_{ab}$ is the relative refractive index of medium $a$ (incident medium) with respect to medium $b$ (refracted medium).
An expression to relate refractive index of a medium with speed to light in vacuum is given below:
If the light ray going from vacuum to medium $a$ then the refractive index of medium a can be written as,
${n_a}=\frac {\text {Speed of light in vacuum}}{\text {Speed of light in medium a}}=\frac {c}{v}$
Where, $c=\text {speed of light in vacuum}$, and $v=\text {speed of light in medium a}$
Given:
The refractive index of a medium 'a' with respect to medium 'b'= $n_{ab}=\frac {n_a}{n_b}=\frac {2}{3}$
The refractive index of a medium 'b' with respect to medium 'c'=$n_{bc}=\frac {n_b}{n_c}=\frac {4}{3}$
To find: Refractive index of medium $'c'$ with respect to medium $'a'$, $n_{ca}=\frac {n_c}{n_a}$.
Solution:
To find refractive index of medium $'c'$ with respect to medium $'a'$, $(n_{ca})$ we divide the reciprocal of $n_{bc}$ by $n_{ab}$.
$\frac {n_c}{n_b}\div \frac {n_a}{n_b}=\frac {3}{4}\div \frac {2}{3}$
$\frac {n_c}{n_b}\times {\frac {n_b}{n_a}}=\frac {3}{4}\times {\frac {3}{2}}$
Thus,
$\frac {n_c}{n_a}=\frac {9}{8}$
Hence, refractive index of medium $'c'$ with respect to medium $'a'$, $n_{ca}=\frac {n_c}{n_a}=\frac {9}{8}$.
55 Views
