A ray of light enters a glass optical fibre (refractive index 1.45) surrounded by air. Calculate the angle of incidence greater than which causes total internal reflection.

If the angle of incidence is greater than the critical angle then it leads to Total internal reflection of light.
Critical angle $\left({\theta }_c\right)={\sin }^{-1}\left(\frac{1}{\mu }\right)$
$={\sin }^{-1}\left(\frac{1}{1.45}\right)={43.6}^{\circ }$

1. 43.6

Q. 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

Q.

For total internal reflection to occur, incidence angle must be greater than the critical angle.

Q. An optical fibre $(\mu =1.72)$ is surrounded by a glass coating $(\mu =1.50)$. Find the critical angle for total internal reflection at the fibre-glass interface.

Q. A ray of light is incident from a denser to a rarer medium. The critical angle for total internal reflection is ${\theta }_{iC}$ and the Brewster's angle of incidence is ${\theta }_{iB}$, such that $sin{\theta }_{iC}/sin{\theta }_{iB}=n=1.28$. The relative refractive index of the two media is :

Q. Which of the following are the essential conditions for Total Internal Reflection to take place?
A. Ray of light should travel from rarer medium to denser medium.
B. Ray of light should travel from denser medium to rarer medium.
C. Angle of incidence should be greater than the critical angle.
D. Critical angle must be greater than the angle of incidence.