Question

Light from a source located in a medium (refractive index $={\mu }_0$ ) enters an optical fibre with core refractive index ${\mu }_1$ and clad refractive index ${\mu }_2$ , as shown in the figure. The maximum value of incident angle $\theta$ which undergo total internal reflection in the fibre is

• A. $\theta ={\cos }^{-1}\left(\frac{\sqrt{{\mu }_1^2-{\mu }_2^2}}{{\mu }_0}\right)$
• B. $\theta ={\tan }^{-1}\left(\frac{{\mu }_1-{\mu }_2}{{\mu }_0}\right)$
• C. $\theta ={\sin }^{-1}\left(\frac{\sqrt{{\mu }_1^2-{\mu }_2^2}}{{\mu }_0}\right)$
• D. None of the above

Answer 1

Answer: (C)

${\mu }_0\sin \theta ={\mu }_1\sin {\theta }_1$
$\sin \theta =(\frac{{\mu }_1}{{\mu }_0})\sin {\theta }_1$
If ${\theta }_c$ is critical angle, then $\sin {\theta }_c=\frac{{\mu }_2}{{\mu }_1}$
$\sin \theta =\frac{{\mu }_1}{{\mu }_2},\sin {\theta }_1=(\frac{{\mu }_1}{{\mu }_0})\sin (90^{\circ }-{\theta }_c)$
$=\frac{{\mu }_1}{{\mu }_2}\cos {\theta }_c=(\frac{{\mu }_1}{{\mu }_2})\sqrt{1-(\frac{{\mu }_2}{{\mu }_1}{)}^2}$
$\sin \theta =\frac{{\mu }_1^2-{\mu }_2^2}{{\mu }_0}$