Chapter 4.2, Problem 75E

Solution

To identify: The what is the length PB in the given figure, a light ray shining from point A to point P on the mirror will bounce to point B in such a way that the angle of incidence a will equal the angle of reflection ß.

This is the law of reflection derived from physical experiments. Both angles are measured from the normal line, which is perpendicular to the mirror at the point of reflection P. If A is 2 m farther from the mirror than is B, and if $a=30°andAP=\text{5m}$

  

The distance PB is 2.69 meter

Given information A is 2 meter father from the mirror than is B and if $\alpha =30°$ and $AP=5m$

  

Explanation:

Consider the given information

Construction: join the normal with A at C

  

Therefore, the distance PC (distance of A from mirror) that is $\left(d_A\right)$

Recall that

  $\begin{array}{c}\cos \theta =\frac{adjacent}{hypotenuse}\\ \cos 30°=\frac{PC}{5}\\ PC=5\cos 30°\\ d_A=5\cos 30°\end{array}$

Hence the distance of B from mirror $\left(d_B\right)$ would be

  $\begin{array}{c}{d}_B=d_A-2\\ =5\cos 30°-2\end{array}$

As according to law of reflection angle of incidence $\alpha$ will be equal to the angle of reflection $\beta$ that is $\beta =30°$

Then

  $\begin{array}{c}PB=\frac{d_B}{\cos 30°}\\ =\frac{5\cos 30°-2}{\cos 30°}\\ =5-\frac{2}{\cos 30°}\end{array}$

Again, recall that $\cos 30°=\frac{\sqrt{3}}{2}$ .

Therefore

  $\begin{array}{c}PB=5-\frac{4}{\sqrt{3}}\\ =2.69\text{meter}\end{array}$

Hence the distance PB is 2.69 meter