(a) and (b) show refraction of a ray in air incident at 60° 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 45º with the normal to a water - glass interface.
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As per the given figure, for the glass - air interface: Angle of incidence, i = 60° Angle of refraction, r= 35° The relative refractive index of glass with respect to air is given by Snell's law as: $=\text{Mh}_2=6\times4.7=28.2 \ \text{cm}$ $\mu^\circ_\text{g}=\frac{\sin\text{i}}{\sin\text{r}}$ $\frac{\sin 60^\circ}{\sin35^\circ}=\frac{0.8660}{0.5736}=1.51\dots(1)$ As per the given figure, for the air - water interface: Angle of incidence, i = 60° Angle of refraction, r= 47° The relative refractive index of water with respect to air is given by Snell's law as: $\mu^\circ_\text{g}=\frac{\sin\text{i}}{\sin\text{r}}$ $\frac{\sin 60^\circ}{\sin47^\circ}=\frac{0.8660}{0.5736}=1.184\dots(2)$ Using (1) and (2), the relative refractive index of glass with respect to water can be obtained as: $\mu^\text{w}_\text{g}=\frac{\mu^\text{a}_\text{g}}{\mu^\text{a}_\text{w}}$ $=\frac{1.51}{1.184}=1.275$ The following figure shows the situation involving the glass - water interface. 
Angle of incidence, i= 45° Angle of refraction = r From Snell's law, rcan be calculated as: $\frac{\sin\text{i}}{\sin\text{r}}=\mu^\text{w}_\text{g}$ $\frac{\sin45^\circ}{\sin\text{r}}=1.275$ $\sin\text{r}=\frac{\frac{1}{\sqrt2}}{1.275}=0.5546$ $\therefore \ \text{r}=\sin^{-1}(0.5546)=38.67^\circ$ Hence, the angle of refraction at the water - glass interface is 38.68°.
Angle of incidence, i= 45° Angle of refraction = r From Snell's law, rcan be calculated as: $\frac{\sin\text{i}}{\sin\text{r}}=\mu^\text{w}_\text{g}$ $\frac{\sin45^\circ}{\sin\text{r}}=1.275$ $\sin\text{r}=\frac{\frac{1}{\sqrt2}}{1.275}=0.5546$ $\therefore \ \text{r}=\sin^{-1}(0.5546)=38.67^\circ$ Hence, the angle of refraction at the water - glass interface is 38.68°.
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