Chapter 23, Problem 74GP

To determine

We must find the depth of the pool.

Answer to Problem 74GP

Solution:

Depth of pool can be determined by using laws of refraction.

Explanation of Solution

Since the ray from the bottom of pool is being refracted from water to air. So by the law of refraction we will find the depth of pool by finding the refracted angle and then applying Pythagoras theorem.

Given:

that width of pool= 6∙50 m. Visible angle from horizontal for bottom of the pool is 13∙0^°.

Formula used:

(1)Value of a right angle is ${90}^{\circ }$.

(2)As the ray is being refracted from water to air, here Snell’s law would be used to find refracted angle say r.

Snell’s law is given by n₁ sin θ₁= n₂ sin θ₂

Where, n₁ and n₂ are refractive index of two different mediums and Ϙ₁ and Ϙ₂ are incident and refracted angles.

(3)Refractive index of water = 1∙33 and refractive index of air = 1∙00

(4)In a right-angle triangle,

tan Ϙ = perpendicular/ base.

Calculation:

Let’s assume that incident angle of the ray from the water is i (say). And refraction angle of refracted ray is r (say).

Since angle r and 13^° is making right angle. So, value of angle

$\begin{array}{l}r={90}^o-{13}^o={77}^o\\ \text{Snell's}\text{law}\\ n_w\sin i=n_a\sin r\\ 1.33\sin i=1\times \sin {77}^o\\ \sin i=\sin \frac{{77}^o}{{1.33}^o}\\ i={\sin }^{-1}\sin \frac{{77}^o}{{1.33}^o}={47.138}^o\\ \tan i=\frac{\text{breadth}\text{of}\text{pool}}{\text{depth}\text{of}\text{pool}}\\ \tan i=\frac{6.5\text{m}}{D}\\ \tan {47.138}^o=6.50\text{m/}D\\ D=\left(6.50/1.0776\right)\text{m}\text{=}6.03\text{m}\end{array}$