Question

An optical fibre is made of quartz filaments of refractive index 1. 70 and it has a coating of material whose refractive index is 1.45. The range of angle of incidence for one laser beam to suffer total internal reflection is

• A. $0^{\circ }$ to ${56.8}^{\circ }$
• B. $0^{\circ }$ to ${62.6}^{\circ }$
• C. $0^{\circ }$ to ${90}^{\circ }$
• D. $0^{\circ }$ to ${180}^{\circ }$

Answer 1

Answer: (B)

$0^{\circ }$ to ${62.6}^{\circ }$

$i$-angle of incidence of the laser beam

$r$-angle of refraction

$i^{\prime }$-angle of incidence of the laser beam inside the fibre

$i_c$-critical angle

By definition of critical angle

$\sin i_c=\frac{1}{{}_l{\mu }_g}=\frac{1}{\frac{1.70}{1.45}}=0.856$

$⟹i_c={\sin }^{-1}0.856={58.5}^{\circ }$

Thus if $i^{\prime }>{58.5}^{\circ }\to r=90-r^{\prime }$

or $r<{90}^{\circ }-{58.5}^{\circ }$

$⟹r<{31.5}^{\circ }$

By snell's law, $\frac{\sin i}{\sin r}{=}_a{\mu }_g$

$⟹\sin i=1.70\times \sin {31.5}^{\circ }=1.70\times 0.524=0.89$

$⟹i={\sin }^{-1}0.89={62.6}^{\circ }$

$\therefore$ range is $0^{\circ }$ to ${62.6}^{\circ }$