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 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=4 / 3) ____________.

Solution

Physics sir · 4 steps

Split the shadow in two: the part cast by the pole above water, and the bend the ray takes inside the water.

1. Angle of incidence. The sunlight makes 30° with the surface, so it makes i = 60° with the normal. Snell's law gives sin r = (sin 60°)/(4/3) = 3√3/8, so cos r = √37/8 and tan r = 3√3/√37 ≈ 0.854.
2. Two pieces of shadow. The ray grazing the top of the pole first travels through air at 30° to the surface. It meets the water at a horizontal distance h cot 30° = √3 h from the pole. It then bends and covers another 1.5 tan r horizontally before reaching the floor.

   √3 h + 1.5 tan r = 2.15

3. Put in numbers. 1.5 × 0.854 = 1.281 m, so √3 h = 2.15 - 1.281 = 0.869 m. Then h = 0.869/1.732 ≈ 0.50 m.
4. So. The pole stands about 0.5 m above the water, which is 50 cm.

The idea to keep. Always convert an angle given with the surface into an angle with the normal before using Snell's law. A shadow in water is the shadow in air plus the extra stretch from the refracted ray.

Answer from the NTA final answer key. Solution by PrepWiser.