Question

The refractive index of water is $1.33$ . What is the angle of refraction if light (in air) strikes water at an angle of incidence of a ${24^ \circ }$ ?

Answer 1

In this question, we are given light traveling from air to water and it strikes the surface of the water at an angle of incidence of a ${24^ \circ }$. To solve this question, we can use Snell’s law which states that if a ray of light passes through one medium to the other medium, then $\mu \sin i = {\text{constant}}$.

Formula used: ${\mu _1}\sin i = {\mu _2}\sin r$
Where ${\mu _1}$ and ${\mu _2}$ are the refractive indices of the two mediums
$i$ is the angle of incidence
$r$ is the angle of refraction

Complete step by step answer:
We have given,
The refractive index of water is $1.33$
The light is traveling from air to water i.e., from a rarer medium to a denser medium.
According to Snell’s law, when light travels from a medium of refractive index ${\mu _1}$ to a medium of refractive index ${\mu _2}$, then
${\mu _1}\sin i = {\mu _2}\sin r$
Where $i$ is the angle of incidence
$r$ is the angle of refraction
For us the angle of incidence is ${24^ \circ }$
The refractive index of air is ${\mu _1} = 1$
The refractive index of water is ${\mu _2} = 1.33$
Substituting the values in the equation,
$\sin {24^ \circ } = 1.33\sin r$
$\Rightarrow 0.41 = 1.33\sin r$
$\Rightarrow \sin r = 0.306$
Taking inverse to find angle of refraction,
$\Rightarrow r = {\sin ^{ - 1}}\left( {0.306} \right)$
$\Rightarrow r = {17.81^ \circ }$
Thus, the angle of refraction of the light in the water is ${17.81^ \circ }$

Note: When light travels from a rarer medium to denser medium i.e., ${\mu _2} > {\mu _1}$ then the light bends towards the normal, and the speed of light in the rarer medium is more than speed the speed of light in the denser medium i.e., the speed of light decreases as it travels from a rarer medium to a denser medium and vice-versa. Similarly, the wavelength of light is more in the denser medium compared to that of the denser medium.