Chapter 6, Problem 1PS

To determine

The distance that the light travel from the red mirror back to the blue mirror.

Answer to Problem 1PS

  $2.01\text{'}$ .

Explanation of Solution

Given information:

In the figure, a beam of light is directed at the blue mirror, reflected to the red mirror, and then reflected back to the blue mirror. Find $PT$ , the distance that the light travel from the red mirror back to the blue mirror.

Calculation:

From the given information,a beam of light is directed at the blue mirror, reflected to the red mirror, and then reflected back to the blue mirror.

By using law of cosines to find $PQ$ than angle alpha can be found by the law of sines. Theta angle can be defined by the exterior angle theorem, then angle PTO by the triangle sum theorem and finally $PT$ can be found by the law of sines.

  

  $\begin{array}{l}\overline{OP}=4.7\text{'},\overline{OQ}=6\text{'},\angle POT={25}^o,\overline{PT}=?\\ \overline{PQ}=\sqrt{{4.7}^2+6^2-2(4.7)(6)\cos {25}^o}\approx 2.64\text{'}\\ \frac{\overline{PQ}}{\sin {25}^o}=\frac{\overline{OP}}{\sin \alpha }\therefore \alpha =\arcsin \frac{\overline{OP}\sin {25}^o}{\overline{PQ}}\approx {48.78}^o\end{array}$

  $\begin{array}{l}\theta =\alpha +{25}^o\approx {73.78}^o\\ \angle PTO={180}^o-(\theta +{25}^o)\approx {81.22}^o\\ \frac{\overline{OP}}{\sin \angle PTO}=\frac{\overline{PT}}{\sin {25}^o}\therefore \overline{PT}=\frac{\overline{OP}\sin {25}^o}{\sin \angle PTO}\approx 2.01\text{'}\end{array}$

Thus, the distance that the light travel from the red mirror back to the blue mirror is $2.01\text{'}$ .