Question

Critical angle of water is ${\theta _1}$ and that of glass is ${\theta _2}$. Then the critical angle of water glass would be:
(${\mu _g} = 3/2$ and ${\mu _w} = 4/3$)
A) Less than ${\theta _2}$.
B) Between ${\theta _1}$ and ${\theta _2}$.
C) Less than ${\theta _1}$.
D) Greater than ${\theta _1}$.

Answer 1

Critical angle is defined as the angle of incidence for which angle of refraction is equal to ${90^0}$. Angle of refraction is ${90^0}$ when incident rays go from rarer medium to denser. Critical angle of water and glass surface depends on the optical densities of water and glass. Critical angle of a surface is given by $\theta = {\sin ^{ - 1}}(\dfrac{{{\mu _R}}}{{{\mu _D}}})$, where $\theta$ is critical angle and ${\mu _R}$ is optical density of rarer medium and ${\mu _D}$ is optical density of denser medium.

Complete step by step solution:
Given, optical density of water is ${\mu _w} = 4/3$ and optical density of glass is ${\mu _g} = 3/2$. Critical angle of water is ${\theta _1}$ and that of glass is ${\theta _2}$.
Then, ${\theta _1} = {\sin ^{ - 1}}\left( {\dfrac{1}{{{\mu _w}}}} \right) = {\sin ^{ - 1}}\left( {\dfrac{3}{4}} \right) = {48.59^0}$ and ${\theta _2} = {\sin ^{ - 1}}\left( {\dfrac{1}{{{\mu _g}}}} \right) = {\sin ^{ - 1}}\left( {\dfrac{2}{3}} \right) = {41.81^0}$ , here rarer medium is air and its optical density is 1.
Critical angle of water glass surface is given by
${\theta _{wg}} = {\sin ^{ - 1}}\left( {\dfrac{{{\mu _w}}}{{{\mu _g}}}} \right) = {\sin ^{ - 1}}\left( {\dfrac{8}{9}} \right) = {62.72^0}$.
Then, we get that the critical angle of water glass is greater than critical angle of water and critical angle of glass.

Hence, the correct answer is option D.

Note: Angle of refraction is ${90^0}$ if light rays are incident at critical angle from rarer medium but is not applicable if rays indent from denser medium. If rays are coming from denser medium and incident normal on the surface then the angle of refraction is a critical angle for that surface.