Question

The speed of light in media ${M_1}$and ${M_2}$are $1.5 \times {10^8}m{s^{ - 1}}$and $2.0 \times {10^8}m{s^{ - 1}}$respectively. A ray of light enters from medium ${M_1}$to ${M_2}$ at an incidence angle$\theta $. If the ray suffers total internal reflection, then the value of the angle of incidence $\theta $ is
A. Equal to ${\sin ^{ - 1}}\left( {\dfrac{2}{3}} \right)$
B. Equal to or less than ${\sin ^{ - 1}}\left( {\dfrac{3}{5}} \right)$
C. Equal to or greater than${\sin ^{ - 1}}\left( {\dfrac{3}{4}} \right)$
D. Less than ${\sin ^{ - 1}}\left( {\dfrac{2}{3}} \right)$

Answer 1

Hint: According to the above question we have two different mediums such that we need to use the formula of refractive index for both the mediums. After that we need to apply the relation between the critical angle and the refractive index then we find out the value of angle of incidence.

Complete step by step answer:
From the given data
${v_1}$=$1.5 \times {10^8}m{s^{ - 1}}$
${v_2}$=$2.0 \times {10^8}m{s^{ - 1}}$
By using the Refractive index formula for medium${M_1}$ is
$\
  {\mu _1} = \dfrac{c}{{{v_1}}} \\
\implies {\mu _1} = \dfrac{{3 \times {{10}^8}}}{{1.5 \times {{10}^8}}} \\
\implies {\mu _1} = 2 \\
\ $
By using the Refractive index formula for medium ${M_2}$ is
$\
  {\mu _2} = \dfrac{c}{{{v_2}}} \\
\implies {\mu _2} = \dfrac{{3 \times {{10}^8}}}{{2.0 \times {{10}^8}}} \\
\implies {\mu _2} = \dfrac{3}{2} \\
\ $
From the total internal reflection,
$\sin i \geqslant \sin c$
Where i is the angle of incidence and C is the critical angle
By using the relation between the critical angle and the refractive index then
$\sin c = \dfrac{{{\mu _2}}}{{{\mu _1}}}$
$\
  \sin i \geqslant \dfrac{{{\mu _2}}}{{{\mu _1}}} \\
\implies \sin i \geqslant \dfrac{{\dfrac{3}{2}}}{2} \\
\implies \sin i \geqslant \dfrac{3}{4} \\
\implies i \geqslant {\sin ^{ - 1}}\dfrac{3}{4} \\
\ $

So, the correct answer is “Option C”.

Note:
To solve this kind of numerical, students should be aware of phenomena like total internal reflection and critical angles. The concept of critical angle can be understood by using Snell's law. At the critical angle, as we increase the incident angle then reflected ray goes upon the interface.