A ray of light travels along the line $x=1$ and gets reflected by a mirror on $x+y=1$. Find the equation of the reflected ray.


I am to solve this problem using only $\tan\theta=|\dfrac{m_1-m_2}{1+m_1m_2}|$ where $m_1$ and $m_2$ are the slopes of the lines between which the angle $\theta$ is subtended.$$$$

I know one way of solving it is to find any general point on $x=1$, find its image about $x+y=1$ and find the equation of the reflected ray using the 2 point form(once we have the point of intersection of $x=1$ and $x+y=1$. However, this method is not allowed by my teacher.

$$$$I would be grateful for any assistance. Many thanks.

  • $\begingroup$ you should define $\theta, m_1, m_2$ above (especially since it means something geometrical). Then try to find what they could correspond to in your case. Then convert the geometric property of mirror to this. ;-) $\endgroup$ – Fabrice NEYRET Nov 1 '15 at 9:54
  • $\begingroup$ Sorry. I'll do that now. Could you please show me how to find what they correspond to? $\endgroup$ – Better World Nov 1 '15 at 9:55
  • $\begingroup$ @FabriceNEYRET Assuming that $\theta$ is the angle subtended between the normal and the incident ray, I get the slope of the normal as 1, and the slope of the incident ray as $\infty$. $\endgroup$ – Better World Nov 1 '15 at 9:58
  • $\begingroup$ In geometry, you know things about the angle of the reflected ray as compare to the angle of the incident ray. Your formula is about $tan(\theta_1-\theta_2)$. It is used to work in slope space instead of angles space, i.e., to translate relations about angles into relations about slopes. $\endgroup$ – Fabrice NEYRET Nov 1 '15 at 10:01
  • $\begingroup$ Sir, I know that the angle of incidence is equal to the angle of refraction. However I can't understand how to use it correctly in this question. $\endgroup$ – Better World Nov 1 '15 at 10:18

Depending on the direction of the ray we get $(-\infty,1]$ or $[1,\infty)$:

enter image description here

  • $\begingroup$ Sir, I couldn't get your answer. I have to find the equation of the reflected ray. $\endgroup$ – Better World Nov 1 '15 at 10:18
  • $\begingroup$ The equation of the red horizontal line is $y=0$ if $x\le 1$ and it is not defined if $x>1$. The equation of the blue horizontal line is $y=0$ if $x\ge 1$ and not defined otherwise. The red and blue horizontal lines can be described as $y=0$. But the reflection depends on the direction of the ray coming along $x=1$. $\endgroup$ – zoli Nov 1 '15 at 10:21


Not quite sure if this is the answer you're looking for, but perhaps you could approach it this way. Intuitively we know that the reflection would be the perpendicular of the line $x=1$. So we shall aim to show that algebraically.

For convenience, we shall rewrite $x+y=1$ as $y=-x+1$.

Let $\theta$ be the angle between $y=-x+1$ and $x=1$, and let $m_1, m_2$ be the slopes of the lines $y=-x+1$ and $x=1$ respectively. Now we define a slope $m_3$, which will become the slope of the reflected line. Since angle of incidence equals angle of reflection, we have the following equation necessarily.

\begin{align} \tan \theta &= \left \lvert \frac{m_1 - m_2}{1+ m_1 m_2} \right \rvert = \left \lvert \frac{m_1 - m_3}{1+ m_1 m_3} \right \rvert \end{align}

Now with $m_1 = -1$, we equate the following: \begin{align} \left \lvert \frac{(-1)- m_2}{1- m_2} \right \rvert = \left \lvert \frac{(-1)- m_3}{1-m_3} \right \rvert \end{align}

After some algebraic manipulation, we have $m_2 \cdot m_3 = -1$, verifying that the reflected line is the perpendicular of $x=1$. Since it passes through the point $(1,0)$, the equation of the reflected line must be $y=0$.


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