A pole is vertically submerged in swimming pool, such that it gives a length of shadow 2.15 m within water when sunlight is incident at an angle of 30° with the surface of water. If swimming pool is filled to a height of 1.5 m, then the height of the pole above the water surface in centimeters is $$(n_w = \dfrac{4}{3})$$ ______.

The angle of incidence $$i$$ with the normal is calculated from the angle made with the water surface:  $$i = 90^\circ - 30^\circ = 60^\circ$$

Applying Snell's law at the air-water interface to find the angle of refraction $$r$$:

$$\sin i = n_w \sin r$$

$$\sin 60^\circ = \frac{4}{3} \sin r \implies \frac{\sqrt{3}}{2} = \frac{4}{3} \sin r \implies \sin r = \frac{3\sqrt{3}}{8} \approx 0.65$$

$$\tan r = \frac{\sin r}{\sqrt{1 - \sin^2 r}} = \frac{\frac{3\sqrt{3}}{8}}{\sqrt{1 - \left(\frac{3\sqrt{3}}{8}\right)^2}} = \frac{3\sqrt{3}}{\sqrt{64 - 27}} = \frac{3\sqrt{3}}{\sqrt{37}} \approx 0.854$$

The total shadow length within the water is composed of the shadow component from the air section ($$x_1$$) and the water section ($$x_2$$):

$$x_{\text{total}} = x_1 + x_2$$ $$\implies$$ $$2.15 = h_a \tan 60^\circ + h_w \tan r$$

$$2.15 = h_a \sqrt{3} + 1.5 \times 0.854$$ $$\implies$$ $$2.15 = h_a \sqrt{3} + 1.281$$ $$\implies$$ $$h_a \sqrt{3} = 2.15 - 1.281 = 0.869$$

$$h_a = \frac{0.869}{\sqrt{3}} = \frac{0.869}{1.732} \approx 0.50\ \text{m}$$

$$h_a = 0.50 \times 100 = 50\ \text{cm}$$