How to derive this equation from the geometry of the attached figure?

Question

How can I derive this equation $ \displaystyle{x+x'-l=\sqrt{(h_t+h_r)^2+d^2}-\sqrt{(h_t-h_r)^2+d^2}} $ from the geometry of the attached figure?

Answer 1

Assuming the picture shows a light reflection, $x$ and $x'$ have the same angle to the surface axis.

In triangle $ABC$, apply Pythagoras Theorem we get

$$l=\sqrt{(h_t-h_r)^2+d^2}$$

Reflect $x'$ over the surface axis, let $E$ be the reflected point of $C$ over the surface, then since the angle $\theta$ is also reflected we get $AE$ is a straight line. Let $D$ be the vertical extension of $AB$ such that $DE$ is parallel to the surface axis.

In triangle $ADE$, apply Pythagoras Theorem we get

$$x+x'=\sqrt{(h_t+h_r)^2+d^2}$$

Subtract the two equations and the result follows.

Answer 2

It is nice assuming the picture shows a light reflection as did @cr001 in his answer. However the statement is not true in general as shown in the attached figure in which we have $$AB=x=13,BC=x'=5,AC=l=8\sqrt2\\d=8,h_t=12\text { and }h_r=4$$ In this case of shape analogue to the given one we have $$x+x'-l=18-8\sqrt2\\\sqrt{(h_t+h_r)^2+d^2}-\sqrt{(h_t-h_r)^2+d^2}=8\sqrt5-8\sqrt2$$

Answer 3

Useful in geometric optics for light or microwave antenna propagation, T is the transmit tower and R the receiver tower.

In order to find the point of incidence I on the mirror (or ground) surface the reflection of R as R' is used e.g., in Fermat principle by finding minimum total length or minimum total time.

$$ \sin \theta= \frac{h_t+h_r}{x+x'}$$ or $$ \tan \theta= \frac{h_t+h_r}{d}$$

Optical path difference can be computed as:

$$ =x+x'-l= TR'-TR =\sqrt{(h_t+h_r)^2+d^2}-\sqrt{(h_t-h_r)^2+d^2}.$$ which is difference of two hypotenuses in triangle using Pythagoras thm.(Triangles drawn have sum and difference of tower vertical heights with horizontal separation $d$ ).