Question

The core of an optical fiber has an index of refraction of 1.35 , while the index of refraction of the cladding surrounding the core is 1.21 . What is the critical angle θc for total internal reflection at the core‑cladding interface?

Answer 1

Answer:

The  critical angle is  $\theta_c = \ 63.68^o$

Explanation:

From the question we are told that

   The refractive index of the core is  $n_c = 1.35$

   The refractive index of the cladding  is $n_s = 1.21$

Generally according to Snell's law

      $\frac{sin i }{sin r } = \frac{n_s}{n_c }$

Here for total internal reflection the refractive angle is  $r = 90^o$ and  the critical angle is equal to the critical angle so  $i = \theta_c$

      $\frac{sin \theta_c }{sin (90) } = \frac{n_s}{n_c }$

substituting values

       $\frac{sin \theta_c }{sin (90) } = \frac{1.21}{1.35 }$

       $\theta_c = sin^{-1} [\frac{1.21}{1.35} ]$

      $\theta_c = \ 63.68^o$