Interesting Physics/Calculus Problem

Question

Tim and John want to find the width of body of water. They stand on opposite sides. John holds the mirror above his head. Tim holds a flashlight above his head and aims it at a spot between the water so that it reflects off into Tim's mirror. They know the following properties of light:

• 1) Light travels on the path that minimizes time
• 2) The shortest path between two points is a line that connects them
• 3) The speed of light = $v$

They know their heights, and Tim knows the angle he is aiming is flashlight at.

Find the width of the body of water.

Note: do not use angle of reflection = angle of incidence.

Hint: If you do not use calculus and all 3 properties, you are missing something.

Answer 1

Let $T$ be the point where Tim holds the flashlight. Let $J$ be John's mirror. Suppose the light reflects on the water at $P$. Let $T$ be a height $h$ above the ground and let $T′$ be the same distance below the ground (so that $TT'$ is perpendicular to the ground which it meets at $F$, its midpoint).

By (1) $TP+PJ$ must be minimised. But $TP=T′P$, so $T′P+PJ$ must be minimised and hence by 2) $P$ must lie on the straight line $T′J$.

[By definition $TF=FT′$ and $\angle PFT=\angle PFT′=90^\circ$. So the triangles $PFT,PFT′$ are congruent. So $TP=T′P$.]

So if Tim is aiming his flashlight at an angle $\alpha$ below the horizontal, we have that the width is $(h+h')$ cot $\alpha$, where $h'$ is the height of John's mirror above the ground.

The minimisation approach is known as Fermat's Principle. There is a good discussion of it in Feynman's Lectures on Physics.

Answer 2

Minimizing the time is the same as minimizing the length of the path. So the question is: among all paths from the light source to the point of reflection and from there to the observer, which one is shortest?

The answer can be seen by considering the reflection of the observer in the plane of the lake: If his eye is five feet above the plane, then its reflection is five feet below. Now consider this important fact:

The distance from any point on the lake to the observer's eye (five feet above the plane of the water's surface) is the same as the distance from that point to the eye of his reflection (five feet below the plane of the water's surface).

This becomes obvious if you draw the picture.

Therefore the following two paths have equal lengths:

• From the light source to the point where the light hits the water, and from there to the observer's eye.
• From the light source to the point where the light hits the water, and from there to the reflection of the observer's eye.

Consequently the point on the surface of the lake that minimizes the total length of the first of these paths is the same as the point the minimizes the second one.

However, the point that minimizes the second one is just the point where the straight line from the light source to the reflection of the observer's eye.

As a corollary, you get that the angle of incidence equals the angle of reflection. So we are not using that as an axiom, but we are deducing it from what was given.

Thus however many degrees below horizontal the flashlight is aimed, the observer will see the light coming from that same number of degrees below the horizontal.

The cosine of that angle multiplied by the height of the observer's eye above the water's surface, plus the cosine multiplied by the height of the light source above the water's surface, therefore is the horizontal distance from the light source to the observer.