$(a)$ Figure shows a cross-section of a 'light pipe' made of a glass fibre of refractive index $1.68$. The outer covering of the pipe is made of a material of refractive index $1.44$. What is the range of the angles of the incident rays with the axis of the pipe for which total internal reflections inside the pipe take place,as shown in the figure.
$(b)$ What is the answer if there is no outer covering of the pipe?

(N/A) Refractive index of the glass fibre,$\mu_{1} = 1.68$.
Refractive index of the outer covering of the pipe,$\mu_{2} = 1.44$.
Let $i$ be the angle of incidence at the air-core interface and $r$ be the angle of refraction.
Let $i'$ be the angle of incidence at the core-cladding interface.
For total internal reflection $(TIR)$ to occur at the core-cladding interface,the angle of incidence $i'$ must be greater than the critical angle $i_c$.
$\sin i_c = \frac{\mu_2}{\mu_1} = \frac{1.44}{1.68} \approx 0.8571$.
Thus,$i_c = \sin^{-1}(0.8571) \approx 59^{\circ}$.
For $TIR$,$i' > 59^{\circ}$.
In the right-angled triangle formed at the entry,$r = 90^{\circ} - i'$.
Since $i' > 59^{\circ}$,$r < 90^{\circ} - 59^{\circ} = 31^{\circ}$.
Applying Snell's law at the air-core interface: $\frac{\sin i}{\sin r} = \mu_1 = 1.68$.
$\sin i = 1.68 \sin r$.
For maximum $i$,$r = 31^{\circ}$.
$\sin i_{\max} = 1.68 \sin 31^{\circ} = 1.68 \times 0.515 = 0.8652$.
$i_{\max} = \sin^{-1}(0.8652) \approx 60^{\circ}$.
So,the range of incident angles is $0 < i < 60^{\circ}$.
$(b)$ If there is no outer covering,the cladding is air,so $\mu_2 = 1.0$.
$\sin i_c = \frac{1.0}{1.68} \approx 0.5952$.
$i_c = \sin^{-1}(0.5952) \approx 36.5^{\circ}$.
For $TIR$,$i' > 36.5^{\circ}$.
$r < 90^{\circ} - 36.5^{\circ} = 53.5^{\circ}$.
$\sin i_{\max} = 1.68 \sin 53.5^{\circ} = 1.68 \times 0.8038 \approx 1.35$.
Since $\sin i$ cannot exceed $1$,$i_{\max} = 90^{\circ}$.
Thus,all rays entering the pipe will undergo $TIR$.