Question

A, B and C are three optical media of respective critical angles ${C_1}$, ${C_2}$ and ${C_3}$. Total internal reflection of light occurs from A to B and also from B to C but not from C to A. Then, the correct relation between the critical angles is
A. ${C_1} < {C_2} < {C_3}$
B. ${C_3} < {C_1} < {C_2}$
C. ${C_3} < {C_2} < {C_1}$
D. ${C_2} < {C_3} < {C_1}$

Answer 1

Hint: The total internal reflection happens only when a ray goes from denser medium to rarer medium. Since it is given that the total internal reflection happens from A to B and from B to C, the refractive index of A is greater than that of B and refractive index of B is greater than that of C. The sine of critical angle is equal to the reciprocal of the refractive index of the medium. By using this we can find the correct relationship between the critical angles.

Complete step by step answer:
It is given that three optical media A, B and C have critical angles ${C_1}$, ${C_2}$ and ${C_3}$ respectively.
The total internal reflection occurs from A to B and from B to C but not from C to A.
Total internal reflection is the reflection of light rays when it travels from a denser medium to rarer medium at an angle greater than the critical angle.
 Since, total internal reflection will occur only when a ray of light goes from denser medium to rarer medium, if total internal reflection happens from A to B, it means A is denser than B.
Then, we can write
$ \Rightarrow {\mu _A} > {\mu _B}$ ……………...(1)
Where, ${\mu _A}$ is the refractive index of A and ${\mu _B}$ is the refractive index of B.
Total internal reflection from B to C is also possible which means B is denser compared to C.
$ \Rightarrow {\mu _B} > {\mu _C}$ ………….(2)
Where, ${\mu _C}$ is the refractive index of C.
Combining equations 1 and 2, we can write
$ \Rightarrow {\mu _A} > {\mu _B} > {\mu _C}$ ……….(3)
The critical angle is the angle at which the angle of refraction will be 90 degrees such that the refracted ray will grace the surface separating the two media. For any angle, greater than critical angle the ray will be totally internally reflected.
The relationship between the critical angle and refractive index of the medium is given as
$\sin C = \dfrac{1}{\mu }$
Where, C is the critical angle and $\mu $ is the refractive index.
$ \Rightarrow \mu = \dfrac{1}{{\sin C}}$
Hence, for the optical medium A we can write
$ \Rightarrow {\mu _A} = \dfrac{1}{{\sin {C_1}}}$ ………...(4)
For the optical medium B, the relationship between refractive index and critical angle can be written as
$ \Rightarrow {\mu _B} = \dfrac{1}{{\sin {C_2}}}$ ………...(5)
For optical medium C, the relationship between refractive index and critical angle can be written as
$ \Rightarrow {\mu _C} = \dfrac{1}{{\sin {C_3}}}$ …………..(6)
Substituting these values from equation 4, 5 and 6 in equation 3, we get
$ \Rightarrow \dfrac{1}{{\sin {C_1}}}\, < \dfrac{1}{{\sin {C_2}}} < \dfrac{1}{{\sin {C_3}}}$
$ \Rightarrow \sin {C_1} < \sin {C_2} < \sin {C_3}$
$ \Rightarrow {C_1} < {C_2} < {C_3}$

Therefore, the correct answer is option A.

Note:
Always remember that total internal reflection will not happen if a ray of light goes from a rarer medium to a denser medium. Total internal reflection takes place only if the ray travels from denser medium to rarer medium. At the critical angle the angle of refraction will be 90 degree such that the refracted ray will grace the surface separating the two media. For any angle, greater than critical angle the ray will be totally internally reflected.