Question

A plate of glass with parallel faces having a refractive index of 1.52 is resting on the surface of water in a tank. A ray of light coming from above in air makes an angle of incidence 36.0 with the normal to the top surface of the glass. What angle does the ray refracted into the water make with the normal to the surface? Use 1.33 for the index of refraction of water.

Answer 1

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 \(=\mathrm{r}^{1}=\sin ^{-1}(0.4418)=26.218^{\circ}\) \(\approx 26.22^{\circ}\)