Question

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

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

Answer 1

Answer: (D)

Here the requirement is that $i>c$
$\Rightarrow \sin i>\sin c\Rightarrow \sin i>\frac{{\mu }_2}{{\mu }_1}$ ....(i)
From Snell's law ${\mu }_1=\frac{\sin \alpha }{\sin r}$ ....(ii)
Also in $△OBA$
$r+i={90}^o\Rightarrow r=(90-i)$
Hence from equation (ii)
$\sin \alpha ={\mu }_1\sin (90-i)$
$\Rightarrow \cos i=\frac{\sin \alpha }{{\mu }_1}$
$\sin i=\sqrt{1-{\cos }^{2}i}=\sqrt{1-{\left(\frac{\sin \alpha }{{\mu }_1}\right)}^2}$ ....(iii)
From equation (i) and (iii) $\sqrt{1-{\left(\frac{\sin \alpha }{{\mu }_1}\right)}^2}>\frac{{\mu }_2}{{\mu }_1}$
$\Rightarrow {\sin }^2\alpha <({\mu }_1^2-{\mu }_2^2)\Rightarrow \sin \alpha <\sqrt{{\mu }_1^2-{\mu }_2^2}$
${\alpha }_{max}={\sin }^{-1}\sqrt{{\mu }_1^2-{\mu }_2^2}$