Question

A pole of length 1 m is half dipped in a swimming pool with water level 50.0 cm higher than bed. The refractive index of water is 1.33 and sunlight is coming at an angle of 45 degree with the vertical. Find the length of shadow of the pole on bed.

Answer 1

The shadow length on the bed is approximately 0.81 m.

Answer 2

We start by noting that the pole is 1 m long and “half‐dipped” so that 0.5 m lies above the water and 0.5 m is underwater. The pool depth is 0.5 m (i.e. the water–bed gap).

Because the sunlight makes an angle of 45° with the vertical in air, a ray traveling in air falls with a horizontal displacement equal to the vertical drop (since tan 45° = 1). However, once the ray reaches the water surface it refracts. Its direction in water is given by Snell’s law.

1. At the water–air boundary, let the incident angle be

   θ_air = 45°

   Snell’s law gives

   n_air·sin(θ_air) = n_water·sin(θ_water)

   Since n_air ≃ 1 and n_water = 1.33,

   sin(θ_water) = sin(45°)/1.33 = 0.7071/1.33 ≃ 0.531

   Thus,

   θ_water ≃ arcsin(0.531) ≃ 32°

   and

   tan(θ_water) ≃ tan(32°) ≃ 0.625.

2. Consider the ray that just grazes the top of the pole.

   (a) In air (from the top of the pole to the water surface):

   Vertical distance = 0.5 m

   Horizontal shift = 0.5·tan45 = 0.5·1 = 0.5 m.

   (b) In water (from the water surface to the bed):

   Vertical distance = 0.5 m

   Horizontal shift = 0.5·tan(32°) ≃ 0.5·0.625 = 0.3125 m.

3. Thus the total horizontal displacement (from the vertical under the pole) of the ray that just grazes the top is

   0.5 + 0.3125 = 0.8125 m (approximately).

Since the lower end of the pole touches the bed (because the underwater 0.5 m exactly meets the 0.5 m water depth), the shadow on the bed extends from the base of the pole to the point reached by this ray. Therefore, the length of the shadow is about 0.81 m.

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Explanation (minimal):

1. The top of the pole is 0.5 m above the water; its ray in air moves horizontally 0.5 m (tan 45° = 1).

2. At the water surface, Snell’s law gives θ_water where sinθ_water = (sin45°)/1.33 ≃ 0.531; hence tanθ_water ≃ 0.625.

3. The ray in water (0.5 m deep) shifts horizontally by 0.5 × 0.625 = 0.3125 m.

4. Total shadow = 0.5 + 0.3125 ≃ 0.8125 m.