The top of a swimming pool is at ground level. If the pool is 3.00 m deep, how far below ground level does the bottom of the pool appear to be located for the following conditions? (The index of refraction of water is 1.333.)

(a) The pool is completely filled with water.
m below ground level

(b) The pool is filled halfway with water.
m below ground level

a) 2.25 m

b) 2.625 m

Explanation:

Refraction is the name given to the phenomenon of the speed of light changing the the boundary when it moves from one physical medium to the other.

Refractive index is the ratio of the speed of light in empty vacuum (air is an appropriate substitution) to the speed of light in the medium under consideration.

In terms of real and apparent depth, the refractive index is given by

η = (real depth)/(apparent depth)

a) Real depth = 3.00 m

Apparent depth = ?

Refractive index, η = 1.333

1.333 = 3/(apparent depth)

Apparent depth = 3/1.3333 = 2.25 m.

Hence the bottom of the pool appears to be 2.25 m below the ground level.

b) Real depth = 1.5 m

Apparent depth = ?

Refractive index, η = 1.333

1.3333 = 1.5/(apparent depth)

Apparent depth = 1.5/1.3333 = 1.125 m

But the pool is half filled with water, there is a 1.5 m depth on top of the pool before refraction starts.

So, apparent depth of the pool = 1.5 + 1.125 = 2.625 m below the ground level

we have been given a quadratic function and we need to restrict the domain such that it becomes a one to one function.

we know that vertex of this quadratic function occurs at (5,2).

further, we know that range of this function is .

if we restrict the domain of this function to either or , it will become one to one function.

let us know find its inverse.

upon interchanging x and y, we get:

let us now solve this function for y.

hence, the inverse function would be if we restrict the domain of original function to and the inverse function would be if we restrict the domain to .