Question

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 1.5m, then the height of the pole above the water surface in centimeters is ($n_w$ = 4/3)____[JEE MAIN-2023]

Answer 1

50 cm

Answer 2

Solution:

Let the pole’s total length be 1.5 m (underwater, fixed to the pool bottom) plus an additional height h above water. A ray of sunlight that grazes the top of the pole is incident from air (where it makes 30° with the water surface, i.e. 60° with the vertical) and then refracts into water.

1. In air (from top to water surface):

   • Vertical drop = h
   • Since the ray makes 30° with the horizontal, its vertical component is sin 30° = ½ and horizontal component is cos 30° = (√3)/2.
   • The distance traveled in air = h/sin30° = 2h, so the horizontal displacement = 2h·cos30° = 2h·(√3/2) = h√3.

2. At the water surface:

   • The incident angle (with the vertical) in air = 60°.

   • By Snell’s law, with n_air ≃ 1 and n_water = 4/3:

     sinθ_water = sin60°/(4/3) = (√3/2)·(3/4) = (3√3)/8.

   • Thus, θ_water ≃ arcsin(3√3/8) ≃ 40.6°.

   • In water, the ray makes an angle 40.6° with the vertical. Over the water depth of 1.5 m (from surface to pool bottom), the horizontal shift = 1.5·tan(40.6°) ≈ 1.5·0.855 = 1.2825 m.

3. Total horizontal displacement (shadow length) on the pool bottom:

   It is the sum of the air and water contributions:

   Shadow = h√3 + 1.2825.

   Given that the shadow length is 2.15 m,

   h√3 = 2.15 – 1.2825 = 0.8675,

   h = 0.8675/√3 = 0.8675/1.732 ≈ 0.5 m.

4. Conversion:

   Height above water = 0.5 m = 50 centimeters.

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Minimal explanation:

• Horizontal displacement in air = h √3; in water = 1.5·tan(θ_water) with sinθ_water = (sin60°)/(4/3) ≃ 0.6495, so tanθ_water ≃ 0.855.
• Total displacement: h√3 + 1.2825 = 2.15; solving gives h ≃ 0.5 m = 50 cm.