Light travels from point A to B point via reflection at point O on the surface of a mirror. Without using calculus, show that length AOB is a minimum when the angle of incidence $\theta$is equal to the angle of reflection $\varphi$.

Light travels from point$A$ to point$B$ via reflection at point role="math" localid="1662978767434" $O$on the surface of a mirror.

Reflection is the change in direction of a wavefront at an interface between two different media so that the wavefront returns to the medium from which it originated. Using the concept of reflection, the ray direction of the light can be traced as per the reflection through the media surfaces.

Hint: Consider the image of $A$in the mirror.

Consider the ray diagram as shown below:

From the ray diagram, we can get that

$$\begin{gathered} \theta +y=\varphi +y(:\theta +y=\frac{\mathrm{\pi }}{2}=\varphi +y) \\ \theta =\varphi \end{gathered}$$

The angle of incidence is equal to angle of reflection.

Consider an incident ray$AO$with reflected ray$O\text{'}B$wheretheangle of incidence is not equal totheangle of reflection.

From the above figure, we get that

$\begin{array}{l}AO\text{'}B=AO\text{'}+O\text{'}B\\ AO\text{'}B=A\text{'}O\text{'}B\\ AO\text{'}+O\text{'}B=A\text{'}O+O\text{'}B\end{array}$

But

localid="1662980270688" $A\text{'}{O}^{\text{'}}+O^{\text{'}}B>A^{\text{'}}B(\because "Twosidesofthesumofthetwosidesofatriangle>hypotenuse")$

And

$\begin{array}{l}A\text{'}B=A\text{'}O+OB\\ A\text{'}B=AO+OB\\ AO+OB=AOB\end{array}$

Thus, from all above options, we get that the shortest path as:

$\begin{array}{l}AO\text{'}B=AO\text{'}+O\text{'}B\\ =A\text{'}O+O\text{'}B\\ >A\text{'}B\\ =A\text{'}O+OB\\ =AO+OB\\ =AOB\end{array}$

Thus, the length $AOB$of is less than $A^{\text{'}}O’B$ and $AO’B$.

Hence, the length of $AOB$is minimum.