Question

An optical signal (light ray) is incident from air at an angle $i$ with normal. If refractive index of core and cladding are, respectively, $n_{1}$ and $n_{2}$ , then what should be the maximum value of $sin\left(\right.i\left.\right)$ so that the ray gets total internally reflected at core-cladding interface? (The maximum value of $sin\left(\right.i\left.\right)$ is called numerical aperture of the optical fibre.)

• A. $\sqrt{n_{1}^{2} - n_{2}^{2}}$
• B. $\sqrt{\frac{n_{1}^{2} - n_{2}^{2}}{n_{2}}}$
• C. $\sqrt{n_{1} n_{2}}$
• D. $\sqrt{\frac{n_{1}^{2} - n_{2}^{2}}{n_{1}}}$

Answer 1

Answer: (A)

$1sini=n_{1}sinr\Rightarrow sinr=\frac{1}{n_{1}}sini$
For total internal reflection to take place,
$n_{1}sin\left(\right.90-r\left.\right)=n_{2}sin90^\circ $
$n_{1}cosr=n_{2}$
$n_{1}^{2}cos^{2}r=n_{2}^{2}$
$\Rightarrow n_{1}^{2}\left(1 - \left(sin\right)^{2} r\right)=n_{2}^{2}\Rightarrow n_{1}^{2}\left(1 - \frac{1}{n_{1}^{2}} \left(sin\right)^{2} i\right)=n_{2}^{2}$
$\Rightarrow 1-\frac{1}{n_{1}^{2}}\left(sin\right)^{2}=\left(\frac{n_{2}}{n_{1}}\right)^{2}\Rightarrow \left(1 - \frac{n_{2}^{2}}{n_{1}^{2}}\right)=\frac{1}{n_{1}^{2}}\left(sin\right)^{2}i$
$\frac{n_{1}^{2} \left(n_{1}^{2} - n_{2}^{2}\right)}{n_{1}^{2}}=\left(sin\right)^{2}i\Rightarrow sini=\sqrt{n_{1}^{2} - n_{2}^{2}}$