Answer
(a) $13.3^{\circ}$
(b) No
(c) $2.24$
Work Step by Step
(a) We know that
$sin\theta_c=\frac{1}{n}$
$\implies \theta=sin^{-1}(\frac{1}{1.66})=37.04^{\circ}$
Now $r_1=45^{\circ}-\theta_c$
$r_1=45-37.04=7.957^{\circ}$
As $sin \space i=nsir_1$
$\implies i=sin^{-1}(1.66sin(7.957^{\circ}))$
$i=13.3^{\circ}$
(b) We know that if the angle of incidence (i) is increased then the angle of refraction (r) increases as well. Thus, $\theta_c\lt 37.04^{\circ}$ and total internal reflection did not take place.
(c) We know that
$n=\frac{sini}{sin(45^{\circ}-\theta_c)}$
$\implies sin(45-\theta_c)=sini(sin\theta_c)$ ($because \frac{1}{n}=sin\theta_c$)
$\implies sin45cos\theta_c-cos45sin\theta_c=sini(sin\theta_c)$
$\implies \frac{1}{\sqrt2}cos\theta-\frac{1}{\sqrt2}sin\theta_c=\frac{1}{\sqrt2}sin\theta_c$
$\implies tan\theta_c=\frac{1}{2}$
$\theta_c=26.565$
$n=\frac{1}{sin\theta_c}=\frac{1}{26.565}=2.24$
