In: Math
Thi sproblem involves two steps and can be calculated by considering the pairs of media In first pair of air and glass using the formula \(\frac{\mu_{2}}{\mu_{1}}=\frac{\sin \mathrm{i}}{\sin \mathrm{r}}\)
Where \(\mu_{2}\) is the refractive index of the medium in which light is reflected and \(\mu_{1}\) is the refractive index of the medium in whcih light is incident. Given \(\mu_{2}=\) refractive index of glass \(=1.52\) \(\mu_{1}=\) refractive index of air \(=1\) Angle of incidence \(=i=36^{\circ}\) Hence using the formula \(\frac{1.52}{1}=\frac{\sin 36}{\sin r}\)
Gives \(\sin r=\frac{\sin 36}{1.52}=\frac{0.5877}{1.52}=0.3866\)
Hence the angle of refraction in glass \(=r=\sin ^{-1}(0.3866)=22.74^{\circ}\) This angle of refraction acts as angle of incidence for glass water media. So again using the same formula with changing the values, \(\mu_{1}^{1}=\) refractive index of glass \(=1.52\) \(\mu_{2}^{1}=\) refractive index of water \(=1.33\) Angle of incidence \(=i^{1}=22.74^{\circ}\) Hence using snell's law \(\frac{\mu_{2}^{1}}{\mu_{1}^{1}}=\frac{\sin i^{1}}{\sin r^{\top}}\)
$$ \begin{array}{c} \frac{1.33}{1.52}=\frac{\sin \left(22.74^{0}\right)}{\sin r^{1}} \\ \text { so } \sin r^{1}=\frac{1.52}{1.33}(0.3866)=0.4418 \end{array} $$
so the angle with which the ray enters into water \(=\mathrm{r}^{1}=\sin ^{-1}(0.4418)=26.218^{\circ}\) \(\approx 26.22^{\circ}\)