Question

An optical fibre consists of a core having refractive index ${\mu }_1$ surrounded by a cladding of refractive index ${\mu }_2$ $({\mu }_2<{\mu }_1)$. A beam of light enters from the air at an angle of $\alpha$ with the axis of the fibre. The highest value of $\alpha$ for which ray can be travelled through fiber is

• A. ${\cos }^{-1}\sqrt{{\mu }_2^2-{\mu }_1^2}$
• B. ${\sin }^{-1}\sqrt{{\mu }_1^2-{\mu }_2^2}$
• C. ${\tan }^{-1}\sqrt{{\mu }_1^2-{\mu }_2^2}$
• D. ${\cos }^{-1}\sqrt{{\mu }_1^2-{\mu }_2^2}$

Answer 1

The correct option is B ${\sin }^{-1}\sqrt{{\mu }_1^2-{\mu }_2^2}$


This is only possible when TIR occurs inside the core.

So for TIR (total internal reflection),
$i>i_c$
where, $i_c$ is the critical angle

$\Rightarrow \sin i>(\sin i_c=\frac{{\mu }_2}{{\mu }_1}).......\left(1\right)$

Now apply Snell's law at the end surface of fibre

${\mu }_{air}\sin \alpha ={\mu }_1\sin r$

$\Rightarrow \sin \alpha ={\mu }_1\sin r.....\left(2\right)$

From above diagram,

$i+r={90}^{\circ }$

$\Rightarrow r={90}^{\circ }-i$

putting in eq.$\left(2\right)$, we get

$\Rightarrow \sin \alpha ={\mu }_1\sin ({90}^{\circ }-i)$

$\Rightarrow \sin \alpha ={\mu }_1\cos i={\mu }_1\sqrt{1-{\sin }^{2}i}$

For maximum value of $\alpha ,\sin i=\frac{{\mu }_2}{{\mu }_1}$

$\therefore \sin \alpha ={\mu }_1\sqrt{1-({\mu }_2/{\mu }_1{)}^2}$

$\Rightarrow \alpha ={\sin }^{-1}\sqrt{({\mu }_1^2-{\mu }_2^2)}$

Hence, option $\left(b\right)$ is correct.