Question

Two plane mirrors $AB$ and $CD$ each of length $2$ m are arranged parallel to each other $\sqrt{3}$ m apart and a ray of light is incident at $A$ as shown in the figure. How many reflections does the ray of light undergo? What is the distance travelled by the ray of light between the two mirrors? What is the angle of deviation?

Answer 1

Step 1: Drawing a refracted ray

The angle of incidence $=$Angle of reflection

$\Rightarrow \angle PAD=\angle EAD={30}^{°}$

Let $DE=a$m

It is given that $AD=\sqrt{3}$m

In triangle $DAE$,

$\tan {30}^{°}=\frac{a}{\sqrt{3}}$

$\frac{1}{\sqrt{3}}=\frac{a}{\sqrt{3}}$

$a=1$m

Redrawing the diagram we get,

From above diagram we can see that ray of light undergo $2$ reflections.

Step 2: Finding the required parameters:

Using Pythagoras theorem in triangle $ADE$

$$\begin{gathered} A{E}^2=A{D}^2+D{E}^2 \\ A{E}^2={\left(\sqrt{3}\right)}^2+1^2 \\ A{E}^2=4 \end{gathered}$$

$AE=2$m

$\Rightarrow AE=EF=2$m

Distance traveled by the ray of light between the two mirrors is $2+2=4$m

As the angle of incidence of ray $EB$ is ${30}^{°}$ which is equal to the ray $PA$

$\Rightarrow$The angle of deviation is $0^{°}$

Hence,

1. Ray of light undergo $2$ reflections.

2. The distance traveled by the ray of light between the two mirrors is $4$m

3. Angle of deviation is $0^{°}$