(a) Let us first derive the condition for total internal reflection. Critical angle for the interface of medium 1 and medium 2. sin C = $\frac{1}{\,\!{}^{1}\mu_2}$ = $\frac{\mu_2}{\mu_1}$ = $\frac{1.44}{1.68}$ = 0.8571 So, critical angle C = 59° Condition for total internal reflection from core to cladding $i_2$ > 59° or r ≤ $\frac{\pi}{2}$ - 59° or r ≤ 31° Now, for refraction at first surface air to core. Snell’s law, $\frac{\sin i_1}{\sin r}$ = $\,\!{}^{a}\mu_1$ sin $i_1$ = $\,\!{}^{a}\mu_1$ sin r = 1.68 sin 31° or $i_1$ = 60° Thus all incident rays which makes angle of incidence between 0° and 60° will suffer total internal reflection in the optical fibre. (b) When there is no outer covering critical angle from core to surface. sin C = $\frac{1}{\,\!{}^{1}\mu_a}$ = $\frac{\mu_a}{\mu_1}$ sinC = $\frac{1}{1.68}$ ⇒ C = $\sin^{-1}\left(\frac{1}{1.68}\right)$ = 36.5° So, condition for total internal reflection from core to surface $i_2$ > 36.5° or r < $\frac{\pi}{2}$ - 36.5° or r < 53.5° Let us find range of incident angle at first surface air to core. $\,\!{}^{a}\mu_1$ = $\frac{\sin i_1}{\sin r}$ sin $i_1$ = 1.68 × sin 53.5° = 1.68 × 0.8039 sin $i_1$ = 1.35 or i ≈ 90°. Thus all incident rays at first surface 0° to 90° will suffer total internal reflection inside core.