Question

Light travels in two media A and B with speeds $1.8 \times {10^8}m{s^{ - 1}}$ and $2.4 \times {10^8}m{s^{ - 1}}$ respectively. Then the critical angle between them is:
(A) ${\sin ^{ - 1}}(\dfrac{2}{3})$
(B) ${\tan ^{ - 1}}(\dfrac{3}{4})$
(C) ${\tan ^{ - 1}}(\dfrac{2}{3})$
(D) ${\sin ^{ - 1}}(\dfrac{3}{4})$

Answer 1

The speed of light in two different mediums are given as $1.8 \times {10^8}m{s^{ - 1}}$ and $2.4 \times {10^8}m{s^{ - 1}}$. We know that critical angle is the ratio of refractive indices. Calculate the refractive indices using the speed of the light in air and speed of light in medium and conclude your answer.
Formula used :
$n = \dfrac{c}{v}$
$\theta = {\sin ^{ - 1}}(\dfrac{{{n_2}}}{{{n_1}}})$

Complete step by step answer:
Critical angle is defined as the angle of incidence after which the light rays passing through a denser medium from the rarer medium are not refracted but totally reflected. This angle is called the critical angle. Critical angle is the concept behind total internal reflection and is vital in fiber optics. Mathematically, it is given as
$\theta = {\sin ^{ - 1}}(\dfrac{{{n_2}}}{{{n_1}}})$, where ${n_1}$ is refractive index of the first medium and ${n_2}$is refractive index of the second medium. Now, in order to find the critical angle, we need to first identify the refractive indices using the speed. Refractive index is given as the ratio of speed of light to the speed of light in medium. Mathematically given as ,
$n = \dfrac{c}{v}$
Now, substituting values in the above equation , we get
$\Rightarrow {n_1} = \dfrac{c}{{{v_1}}}$
$\Rightarrow {n_1} = \dfrac{{3 \times {{10}^8}}}{{1.8 \times {{10}^8}}}$
Cancelling out the common terms and simplifying , we get
$\Rightarrow {n_1} = 1.67$
Applying the same formula and technique for the second medium , we get
$\Rightarrow {n_2} = \dfrac{c}{{{v_2}}}$
Substituting the known values , we get,
$\Rightarrow {n_2} = \dfrac{{3 \times {{10}^8}}}{{2.4 \times {{10}^8}}}$
Cancelling out the common term and simplifying we get,
$\Rightarrow {n_2} = 1.25$
Now, we can obtain critical angle value using, the above mentioned formula,
$\Rightarrow \theta = {\sin ^{ - 1}}(\dfrac{{1.25}}{{1.67}})$
On simplifying we get,
$\Rightarrow \theta = {\sin ^{ - 1}}(\dfrac{3}{4})$

Hence, option (d) is the right answer for the given question.

Note: Snell’s law is a basic formula that can be used to relate the angle of incident with angle of refraction. The law states that the ratio of refractive index of two different mediums is equal to the sine of incident angles in the medium.