1. Given data

• Length of shadow inside water: $L = 2.15\,\text{m}$
• Water depth: $h_w = 1.5\,\text{m}$
• Refractive index of water: $n_w = \dfrac{4}{3}$
• Sunlight is incident at angle $30^\circ$ with the surface of water.

So the angle with the normal in air is $i = 90^\circ - 30^\circ = 60^\circ.$

Let the height of the pole above water surface be $h$ m.

2. Refraction at water surface

Using Snell's law: $n_{air}\sin i = n_w \sin r$ $1\cdot \sin 60^\circ = \frac{4}{3}\sin r$ $\sin r = \frac{3}{4}\cdot \frac{\sqrt{3}}{2} = \frac{3\sqrt{3}}{8}.$

Hence, $\tan r = \frac{\sin r}{\sqrt{1-\sin^2 r}}.$

Now, $\sin^2 r = \left(\frac{3\sqrt{3}}{8}\right)^2 = \frac{27}{64}$ so $\cos r = \sqrt{1-\frac{27}{64}} = \sqrt{\frac{37}{64}} = \frac{\sqrt{37}}{8}.$

Therefore, $\tan r = \frac{\frac{3\sqrt{3}}{8}}{\frac{\sqrt{37}}{8}} = \frac{3\sqrt{3}}{\sqrt{37}}.$

Numerically, $\tan r \approx 0.854.$

3. Shadow formation geometry

The pole has two parts:

• height above water surface = $h$
• submerged part = $1.5\,\text{m}$

The sunlight travels:

• in air at angle $30^\circ$ with the surface, i.e. $60^\circ$ with normal
• in water at angle $r$ with normal

So horizontal shift due to the part in air: $x_1 = h\tan 30^\circ = \frac{h}{\sqrt{3}}$ because the ray makes $30^\circ$ with vertical? Let us carefully use the correct angle.

Since the ray makes $30^\circ$ with the surface (horizontal), it makes $60^\circ$ with the vertical. Therefore horizontal displacement in air is $x_1 = h\tan 60^\circ = h\sqrt{3}.$

Horizontal shift in water: $x_2 = 1.5\tan r.$

Thus total shadow length inside water is $L = x_1 + x_2$ $2.15 = h\sqrt{3} + 1.5\tan r.$

Substitute $\tan r = \dfrac{3\sqrt{3}}{\sqrt{37}} \approx 0.854$: $2.15 = h\sqrt{3} + 1.5(0.854)$ $2.15 = h\sqrt{3} + 1.281$ $h\sqrt{3} = 2.15 - 1.281 = 0.869$ $h = \frac{0.869}{\sqrt{3}} \approx \frac{0.869}{1.732} \approx 0.502\,\text{m}.$

So height above water surface is approximately $h \approx 0.50\,\text{m} = 50\,\text{cm}.$

4. Final integer answer

$\boxed{50}$