Question

The refractive index of the core of an optical fiber is ${\mu }_2$​ and that of the cladding is ${\mu }_1$​. The angle of incidence on the face of the core so that the light ray just grazes at the cladding as shown in the figure is :

• A. ${\sin }^{-1}\left(\frac{{\mu }_1}{{\mu }_2}\right)$
• B. ${\sin }^{-1}\left(\frac{{\mu }_2}{{\mu }_1}\right)$
• C. ${\sin }^{-1}\sqrt{{\mu }_2^2-{\mu }_1^2}$
• D. ${\sin }^{-1}\sqrt{{\mu }_2^2+{\mu }_1^2}$

Answer 1

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


Applying Snell's law at second refraction,

${\mu }_2\sin ({90}^{\circ }-r)={\mu }_1\times \sin {90}^{\circ }$
$\therefore \cos r=\frac{{\mu }_1}{{\mu }_2}$ - - - - - $\left(1\right)$

Snell's law at first refraction gives,

$1\times \sin i={\mu }_2\sin r$
$\Rightarrow \sin i={\mu }_2\times \sqrt{1-{\cos }^{2}r}$
$(\because \sin r=\sqrt{1-{\cos }^{2}r})$
$\Rightarrow \sin i={\mu }_2\sqrt{1-{\left(\frac{{\mu }_1}{{\mu }_2}\right)}^2}$
$\left(\text{From (1)}\right)$
$\Rightarrow \sin i=\sqrt{{\mu }_2^2-{\mu }_1^2}$
Or, $i={\sin }^{-1}\sqrt{{\mu }_2^2-{\mu }_1^2}$

Hence, option $\left(c\right)$ is the correct answer.

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