Question

The critical angle for a medium is ${45^ \circ }$. What is the refractive index of the medium? The critical angle for a ray is ${38^ \circ }$, ${42^ \circ }$ for glass and ${42^ \circ }$ for glycerine. What is the refractive index of glass with respect to glycerine?

Answer 1

Hint The refractive index of the medium, when the critical angle is given which is determined by the relation between the critical angle and the refractive index. The sine of the critical angle and the refractive index both are inversely proportional to each other.

Useful formula
The relation between the refractive index and the critical angle is given by,
$\mu = \dfrac{1}{{\sin \,c}}$
Where, $\mu $ is the refractive index of the material and $c$ is the critical angle.

Complete step by step solution
Given that,
The critical angle of the medium is ${45^ \circ }$ ,
The critical angle of the glass is ${42^ \circ }$,
The critical angle of the glycerine is ${42^ \circ }$.
Now,
The refractive index of the medium is given by,
$\mu = \dfrac{1}{{\sin \,c}}\,......................\left( 1 \right)$
By substituting the critical angle of the medium in the above equation, then the above equation is written as,
$\mu = \dfrac{1}{{\sin \,{{45}^ \circ }}}$
The value of the $\sin {45^ \circ }$ is $0.707$, then the above equation is written as,
$\mu = \dfrac{1}{{0.707}}$
On dividing the above equation, then
$\mu = 1.414\,..................\left( 2 \right)$
The refractive index of the glass is given by,
By substituting the critical angle of the glass in the equation (1), then the equation (1) is written as,
$\mu = \dfrac{1}{{\sin \,{{42}^ \circ }}}$
The value of the $\sin {42^ \circ }$ is $0.669$, then the above equation is written as,
$\mu = \dfrac{1}{{0.669}}$
On dividing the above equation, then
$\mu = 1.494\,.................\left( 3 \right)$
The refractive index of the glycerine is given by,
By substituting the critical angle of the glycerine in the equation (1), then the equation (1) is written as,
$\mu = \dfrac{1}{{\sin \,{{42}^ \circ }}}$
The value of the $\sin {42^ \circ }$ is $0.669$, then the above equation is written as,
$\mu = \dfrac{1}{{0.669}}$
On dividing the above equation, then
$\mu = 1.494\,...............\left( 4 \right)$
Now, the refractive index of the glass with respect to the refractive index of the glycerine is given by,
$\dfrac{{{\mu _{glass}}}}{{{\mu _{glycerine}}}} = \dfrac{{1.494}}{{1.494}}$
On dividing the above equation, then
$\dfrac{{{\mu _{glass}}}}{{{\mu _{glycerine}}}} = 1$

Thus, the above equation shows the refractive index of the glass with respect to the refractive index of the glycerine is one.

Note: If the refractive index of the glycerine with respect to the refractive index of the glass is asked, then it is determined by dividing the refractive index of the glycerine to the refractive index of the glass, then the solution is determined.