Figures $(a)$ and $(b)$ show refraction of a ray in air incident at $60^{\circ}$ with the normal to a glass-air and water-air interface,respectively. Predict the angle of refraction in glass when the angle of incidence in water is $45^{\circ}$ with the normal to a water-glass interface [Figure $(c)$].

(N/A) For the glass-air interface (Figure $a$):
Angle of incidence,$i = 60^{\circ}$
Angle of refraction,$r = 35^{\circ}$
Using Snell's law,the refractive index of glass with respect to air is:
$\mu_{g}^{a} = \frac{\sin i}{\sin r} = \frac{\sin 60^{\circ}}{\sin 35^{\circ}} = \frac{0.8660}{0.5736} \approx 1.51$
For the air-water interface (Figure $b$):
Angle of incidence,$i = 60^{\circ}$
Angle of refraction,$r = 47^{\circ}$
Using Snell's law,the refractive index of water with respect to air is:
$\mu_{w}^{a} = \frac{\sin i}{\sin r} = \frac{\sin 60^{\circ}}{\sin 47^{\circ}} = \frac{0.8660}{0.7314} \approx 1.184$
The relative refractive index of glass with respect to water is:
$\mu_{g}^{w} = \frac{\mu_{g}^{a}}{\mu_{w}^{a}} = \frac{1.51}{1.184} \approx 1.275$
For the water-glass interface (Figure $c$):
Angle of incidence,$i = 45^{\circ}$
Let the angle of refraction be $r$.
Using Snell's law: $\mu_{w} \sin i = \mu_{g} \sin r$,which implies $\frac{\sin i}{\sin r} = \frac{\mu_{g}}{\mu_{w}} = \mu_{g}^{w}$.
$\sin r = \frac{\sin 45^{\circ}}{\mu_{g}^{w}} = \frac{0.7071}{1.275} \approx 0.5546$
$r = \sin^{-1}(0.5546) \approx 33.68^{\circ}$