Light is incident from glass ( μ = 1.50 ) to water ( μ = 1.33 )....

The critical angle $\theta_c$ for light traveling from glass to water is given by:

$\sin \theta_c = \dfrac{\mu_w}{\mu_g} = \dfrac{4/3}{3/2} = \dfrac{8}{9}$

$\theta_c = \sin^{-1}$\left(\dfrac{8}{9}\right)

When the angle of incidence $i$ is less than the critical angle ($0 \le i < \theta_c$), refraction occurs. The angle of deviation $\delta$ is given by $\delta = r - i$. As $i$ increases from $0$ to $\theta_c$, $r$ increases from $0$ to $\pi/2$. The deviation $\delta$ increases from $0$ to a maximum value of:

$\delta_1 = \dfrac{\pi}{2} - \theta_c$

When the angle of incidence $i$ is greater than or equal to the critical angle ($\theta_c \le i \le \pi/2$), total internal reflection (TIR) occurs. The angle of deviation is given by $\delta = \pi - 2i$. As $i$ increases from $\theta_c$ to $\pi/2$, the deviation $\delta$ decreases from a maximum of $\pi - 2\theta_c$ to $0$.

The range of deviation for refraction is $[0, \pi/2 - \theta_c)$ and for TIR is $[0, \pi - 2\theta_c]$. Since $\theta_c < \pi/2$, we have $\pi/2 - \theta_c < \pi - 2\theta_c$.

For a deviation $\delta$ to have two corresponding angles of incidence (one from refraction and one from TIR), it must lie in the overlapping range of both cases. This overlapping range is from $0$ to $\delta_1$.

The upper limit of this range is:

$\delta_1 = \dfrac{\pi}{2} - \sin^{-1}$\left(\dfrac{8}{9}\right) = \cos^{-1}\left(\dfrac{8}{9}\right)

Thus, the range of the angle of deviation for which there are two angles of incidence is $0$ to $\cos^{-1} \left(\dfrac{8}{9}\right).$