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 30o 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) ____________

Answer 1

Correct Answer: 50

Refraction and Pole Height Problem

Given: \( \sin 60^\circ = \frac{4}{3} \sin r \)

Then:

\( \sin r = \frac{3}{4} \times \frac{\sqrt{3}}{2} = \frac{3\sqrt{3}}{8} \qquad (i) \) 

\( \cos r = \sqrt{1 - \frac{27}{64}} = \sqrt{\frac{37}{64}} = \frac{\sqrt{37}}{8} \approx 0.75 \)

\( \tan r = \sqrt{\frac{27}{37}} \)

\( \frac{x}{1.5} = 0.85 \)

\( x = 0.85 \times 1.5 = 1.275 \text{ m} \)

\( \tan 30^\circ = \frac{y}{2.15 - 1.275} = 0.50 \)

\( y = \frac{0.875}{1.732} = 0.50 \)

Conclusion:

So, the length of the pole above the water surface = \( \mathbf{0.50 \text{ m} = 50 \text{ cm}} \).

Answer 2

Snell’s Law and Pole Height Calculation

Step 1: Snell’s Law  

Snell’s Law relates the angle of incidence (\(i\)) and angle of refraction (\(r\)) for light passing through two different media with refractive indices \(n_1\) and \(n_2\):

\( n_1 \sin i = n_2 \sin r \)

In this case: - \( n_1 = 1 \) (air), - \( i = 60^\circ \) (since the angle with the surface is 30°), - \( n_2 = \frac{4}{3} \) (water).

Step 2: Calculate the Angle of Refraction

Using Snell’s law:

\( \sin 60^\circ = \frac{4}{3} \sin r \)

Solving for \( \sin r \):

\( \sin r = \frac{3}{4} \sin 60^\circ = \frac{3}{4} \times \frac{\sqrt{3}}{2} = \frac{3\sqrt{3}}{8} \)

Step 3: Calculate \( \tan r \)

We first find \( \cos r \):

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

Now we calculate \( \tan r \):

\( \tan r = \frac{\sin r}{\cos r} = \frac{3\sqrt{3}/8}{\sqrt{37}/8} = \frac{3\sqrt{3}}{\sqrt{37}} \approx 0.85 \)

Step 4: Calculate the Length of the Pole Submerged in Water

Let \( x \) be the horizontal length of the shadow. We have \( \tan r = \frac{x}{1.5 \, \text{m}} \), where 1.5 m is the depth of the water.

Thus:

\( x = 1.5 \times \tan r = 1.5 \times 0.85 = 1.275 \, \text{m} \)

Since the total length of the shadow is 2.15 m, the horizontal distance from the pole to the point where light enters the water is:

\( 2.15 - 1.275 = 0.875 \, \text{m} \)

Step 5: Calculate the Height of the Pole Above Water

Let \( y \) be the height of the pole above the water. Since the angle of incidence is 30° with the water surface, we have:

\( \tan 30^\circ = \frac{y}{0.875} \)

Solving for \( y \):

\( y = 0.875 \times \tan 30^\circ = 0.875 \times \frac{1}{\sqrt{3}} \approx 0.875 \times 0.577 = 0.50 \, \text{m} = 50 \, \text{cm} \)

Conclusion:

The height of the pole above the water surface is \( \mathbf{50 \, \text{cm}} \).