MHB Find Distance of Object From Pole Given Angle/Ilumination on a Pole

[leprofece]

on a pole, two reflectors located on the heights a and b, focus the same object on the ground. at what distance from the pole will the object be When is the angle forming light rays maximum?

Answer sqrt (ab)

maybe the angle is tang ^-1a/b

or sin (h/d²)?

 

[soroban]

Hello, leprofece!

On a pole, two reflectors located at heights a and b
focus on the same object on the ground.
At what distance from the pole will the object be
when the angle forming light rays is a maximum?

Answer: sqrt (ab)

First, we need an accurate diagram.

      -   *
      :   |*
      :   | *
      :   |  *
      :   |   *
      a - *    *
      : : | *   *
      : : |   * θ*
      : b |     * *
      : : |     α **
      - - * - - - - *
               x

We want to find the distance x
so that angle \theta is maximum.

We have: .\tan\alpha \,=\,\frac{b}{x} \quad\Rightarrow\quad x \,=\,\frac{b}{\tan\alpha}\;\;[1]

And: .\tan(\theta + \alpha) \,=\,\frac{a}{x} \quad\Rightarrow\quad x \,=\,\frac{a}{\tan(\theta+\alpha)}\;\;[2]

Equate [1] and [2]: .\frac{b}{\tan\alpha} \:=\:\frac{a}{\tan(\theta+\alpha)}

. . . . . .b\tan(\theta+\alpha) \:=\:a\tan\alpha

. . b\frac{\tan\theta+\tan\alpha}{1-\tan\alpha\tan\theta} \:=\:a\tan\alpha

Solve for \tan\theta\!:\;\;\tan\theta \:=\:\frac{(a-b)\tan\alpha}{b - a\tan^2!\alpha}

Differentiate and simplify:
. . \sec^2\!\theta\frac{d\theta}{d\alpha} \:=\:\frac{(a-b)\sec^2\!\alpha\,(b-a\tan^2\!\alpha)}{(b+a\tan^2\!\alpha)^2}

Then: .b-a\tan^2\!\alpha \:=\:0 \quad\Rightarrow\quad \tan^2\!\alpha \,=\,\frac{b}{a}
Hence: .\tan\alpha \,=\,\sqrt{\frac{b}{a}}
Substitute into [1]: .x \:=\:\frac{b}{\tan\alpha} \:=\:\frac{b}{\sqrt{\frac{b}{a}}} \:=\:b\cdot\frac{\sqrt{a}}{\sqrt{b}}
Therefore: .x \;=\;\sqrt{ab}