Question

A child’s toy sits on the bottom of a swimming pool in which the water depth is d=1.8 m. To a child standing at the pool edge, the toy appears to be a horizontal distance of 4.2 m away from the sidewall. Assume the child’s eyes are 3.5m above the bottom of the pool. What is the actual horizontal distance from the sidewall to the toy?

Answer 1

Answer:

Horizontal distance is 1.28 + 2.04 = 3.32 m

Explanation:

Given data:

from  below figure

Applying Pythagoras theorem

from Snell's law

$\frac{sin i}{sin r} = \frac{n_2}{n_1} = \frac{\frac{EF}{AB}}{\frac{DE}{BD}}$

$1.33 =\frac{ \frac{2.04}{\sqrt{(3.5 -1.8)^2 + 2.04^2}}}{\frac{x}{\sqrt{1.8^2 + x^2}}}$

solving for x we have

$x^2 = 1.626$

x = 1.28

Horizontal distance is 1.28 + 2.04 = 3.32 m