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 $sin(i)$ so that the ray gets total internally reflected at core-cladding interface? (The maximum value of $sin(i)$ 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,
$n_{1}sin(90-r)=n_{2}sin90^{\circ }$
$n_{1}cosr=n_2$
$n_1^{2}co{s}^{2}r=n_2^2$
$\Rightarrow n_1^2\left(1-{\left(sin\right)}^{2}r\right)=n_2^2\Rightarrow n_1^2\left(1-\frac{1}{n_1^2}{\left(sin\right)}^{2}i\right)=n_2^2$
$\Rightarrow 1-\frac{1}{n_1^2}{\left(sin\right)}^2={\left(\frac{n_2}{n_1}\right)}^2\Rightarrow \left(1-\frac{n_2^2}{n_1^2}\right)=\frac{1}{n_1^2}{\left(sin\right)}^{2}i$
$\frac{n_1^2\left(n_1^2-n_2^2\right)}{n_1^2}={\left(sin\right)}^{2}i\Rightarrow sini=\sqrt{n_1^2-n_2^2}$