Question

If ${n_2} > {n_1}$ then,
$(a)$ $\sin {\theta _1} > \sin {\theta _2}$
(b)$$$${\theta _1} > {\theta _2}
$(c)$ $\cos {\theta _1} > \cos {\theta _2}$
$(d)$ Refractive angle > Incident angle

Answer 1

We will use Snell’s law to derive the correct answer as the question is related to incident angle and refractive angle.

Formula Used:
In order to follow the quickest path from one medium to another medium, a ray changes its direction of travel. The refractive angle and the incident angle depend on the refractive index and the incident index of this system.
Snell’s law states that the ratio of the sine of the angles of incidence and refraction is equivalent to the reciprocal of the ratio of the indices of refraction.
So, according to Snell’s law we can derive the following equation:
$\dfrac{{\sin {\theta _1}}}{{\sin {\theta _2}}} = \dfrac{{{n_1}}}{{{n_2}}}$.
Where, ${\theta _1}$ and ${\theta _2}$ are angle of incidence and angle of refraction respectively, and, ${n_1}$ and ${n_2}$ are the index of the incident medium and refractive medium respectively.

Complete step by step answer:
So, angle of incidence and angle of refraction depends upon the density of the medium by which the light passes through.
It means that if the medium is more densely, the quality of its indexes changes.
So, if ${n_1} < {n_2}$, the angle of refraction is always smaller than the angle of incidence and if ${n_1} > {n_2}$, the angle of refraction is always larger than the angle of incidence.
Let's imagine the index of the refractive medium is larger than the index of the incident medium, in such cases the light makes the larger angle in refracted medium than the angle made by the ray in incident medium as the density of the medium changed, the critical angle made by the ray changes accordingly.
So, if the incident angle is too large after reflecting in the system then the refracted angle will be very small as the index of the refractive medium is larger, thus we can say that ${\theta _2} = {90^ \circ }$ almost.
Now, again using the Snell’s law:
$\dfrac{{\sin {\theta _1}}}{{\sin {{90}^ \circ }}} = \dfrac{{{n_1}}}{{{n_2}}}$
Or, $\sin {\theta _i} = \dfrac{{{n_1}}}{{{n_2}}}$
Or, ${\theta _i} = {\sin ^{ - 1}}\left( {\dfrac{{{n_1}}}{{{n_2}}}} \right)$.
So, critical angle will get smaller when incident angle get large.
So, if the index of refraction is greater than the index of the incidence of a ray, then the angle of the incidence will be greater than the angle of the refraction.

$\therefore$ Option $(b)$, ${\theta _1} > {\theta _2}$ is the correct choice.

Note: Refractive index of light is the ratio of speed of light and phase velocity of the light.
$n = \dfrac{c}{v}$ .
The refractive index of water is $1.33$ means that light travels $1.33$ times slower in water than in a vacuum space.
Phase velocity is the velocity at which the phase of any $1$ frequency of particle of the wave travels.