
The speed of light in medium ${M_1}$ and ${M_2}$ are $1.5 \times {10^8}m/s$ and $2 \times {10^8}m/s$ respectively. A ray travels from medium ${M_1}$ to the medium ${M_2}$ with an angle of incidence $\theta $. The ray suffers total internal reflection. Then the value of angle of incidence $\theta $ is:
Answer
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Hint
Total internal reflection occurs only when the angle of incidence of a ray on a surface is greater than the critical angle for that surface. If angle of incidence ray is equal to critical angle and refracted ray travel normal to surface. The velocity of light in a medium depends on the refractive index of that medium.
Complete step by step answer
Let ${\mu _1}$ and ${\mu _2}$ are refractive index of medium ${M_1}$ and ${M_2}$.
Given the speed of light in medium ${M_1}$ and ${M_2}$ are $1.5 \times {10^8}m/s$ and $2 \times {10^8}m/s$ respectively.
We know that the product of speed of light in a medium and refractive index is equal to the speed of light in free space.
The speed of light in free space is $C$.
For medium ${M_1}$,
$C = {\mu _1} \times 1.5 \times {10^8}$ -(1)
For medium ${M_2}$,
$C = {\mu _2} \times 2 \times {10^8}$ -(2)
Here medium ${M_2}$ is rarer and medium ${M_1}$ is denser medium.
From equation (1) and (2),
$\dfrac{{{\mu _2}}}{{{\mu _1}}} = \dfrac{{1.5 \times {{10}^8}}}{{2 \times {{10}^8}}} = \dfrac{3}{4}$.
Critical for surface between ${M_1}$ and ${M_2}$ is given by ${\sin ^{ - 1}}\left( {\dfrac{{{\mu _2}}}{{{\mu _1}}}} \right) = {\sin ^{ - 1}}\left( {\dfrac{3}{4}} \right)$.
We know that for total internal reflection, the angle of incidence is greater than critical angle.
Then the incidence angle $\theta > {\sin ^{ - 1}}\left( {\dfrac{3}{4}} \right)$.
Hence the correct answer is option (A).
Note
When refraction occurs some part of light is reflected and the remaining part of light refract through a surface. There is always some reflection of light but when the angle of incidence is greater than critical angle then it blocks refraction of light through the surface and light reflects completely and this phenomenon is called total internal reflection.
Total internal reflection occurs only when the angle of incidence of a ray on a surface is greater than the critical angle for that surface. If angle of incidence ray is equal to critical angle and refracted ray travel normal to surface. The velocity of light in a medium depends on the refractive index of that medium.
Complete step by step answer
Let ${\mu _1}$ and ${\mu _2}$ are refractive index of medium ${M_1}$ and ${M_2}$.
Given the speed of light in medium ${M_1}$ and ${M_2}$ are $1.5 \times {10^8}m/s$ and $2 \times {10^8}m/s$ respectively.
We know that the product of speed of light in a medium and refractive index is equal to the speed of light in free space.
The speed of light in free space is $C$.
For medium ${M_1}$,
$C = {\mu _1} \times 1.5 \times {10^8}$ -(1)
For medium ${M_2}$,
$C = {\mu _2} \times 2 \times {10^8}$ -(2)
Here medium ${M_2}$ is rarer and medium ${M_1}$ is denser medium.
From equation (1) and (2),
$\dfrac{{{\mu _2}}}{{{\mu _1}}} = \dfrac{{1.5 \times {{10}^8}}}{{2 \times {{10}^8}}} = \dfrac{3}{4}$.
Critical for surface between ${M_1}$ and ${M_2}$ is given by ${\sin ^{ - 1}}\left( {\dfrac{{{\mu _2}}}{{{\mu _1}}}} \right) = {\sin ^{ - 1}}\left( {\dfrac{3}{4}} \right)$.
We know that for total internal reflection, the angle of incidence is greater than critical angle.
Then the incidence angle $\theta > {\sin ^{ - 1}}\left( {\dfrac{3}{4}} \right)$.
Hence the correct answer is option (A).
Note
When refraction occurs some part of light is reflected and the remaining part of light refract through a surface. There is always some reflection of light but when the angle of incidence is greater than critical angle then it blocks refraction of light through the surface and light reflects completely and this phenomenon is called total internal reflection.
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