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Reflection from a Plane Surface


There are two laws of reflection

  1. Incident ray, reflected ray, and normal lie in the same plane
  2. The angle of incidence $i$ is equal to the angle of reflection $r$ i.e., $\angle i=\angle r$


Reflection from a Plane Mirror

A virtual image of a real object is formed in a plane mirror. The image and the object are equidistant from the plane mirror.

reflection from a plane mirror

Problems from IIT JEE

Problem (IIT JEE 2013): A ray of light travelling in the direction $\frac{1}{2}\left(\hat{\imath}+\sqrt{3}\,\hat{\jmath}\right)$ is incident on a plane mirror. After reflection, it travels along $\frac{1}{2}\left(\hat{\imath}-\sqrt{3}\,\hat{\jmath}\right)$. The angle of incidence is,

  1. 30 degree
  2. 45 degree
  3. 60 degree
  4. 75 degree

Solution: Let $\hat{v}_i=\frac{1}{2}\left(\imath+\sqrt{3}\jmath\right)$ and $\hat{v}_r=\frac{1}{2}\left(\imath-\sqrt{3}\jmath\right)$ be the unit vectors along the incident and the reflected rays.


From laws of reflection, angle of incidence (say $\theta$) is equal to the angle of reflection. Thus, angle between $-\hat{v}_i$ and $\hat{v}_r$ is $2\theta$ (see figure). Using, dot products of vectors, we get, \begin{align} \cos(2\theta)=- \hat{v}_i\cdot \hat{v}_r={1}/{2}. \nonumber \end{align} Hence, $\theta={30}\;\mathrm{degree}$. The readers can see that $\hat{v}_i=\cos 60\,\hat{\imath}+\sin 60\,\hat{\jmath}$ and $\hat{v}_r=\cos (-60)\,\hat{\imath}+\sin (-60)\,\hat{\jmath}$. Drawing these vectors gives desired result.


  1. Verification of Laws of Reflection
  2. Multiple images with plane mirrors
  3. Lateral Inversion
  4. Reflection from Spherical Mirrors
JEE Physics Solved Problems in Mechanics