A ray of light is reflected by a plane mirror | vedclass

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(B) The incident ray vector is $\hat{e}_0$ and the reflected ray vector is $\hat{e}$. The normal vector is $\hat{n}$.From the law of reflection, the a

Vedclass · Accepted Answer

$\hat{e}_0 - 2 ( \hat{e}_0 \cdot \hat{n} ) \hat{n}$

Vedclass · Answer

(B) The incident ray vector is $\hat{e}_0$ and the reflected ray vector is $\hat{e}$. The normal vector is $\hat{n}$.From the law of reflection, the angle of incidence equals the angle of reflection, and all three vectors lie in the same plane.Let $	heta$ be the angle between the incident ray and the normal. Since $\hat{e}_0$ is directed towards the mirror, the angle between $\hat{e}_0$ and $\hat{n}$ is $(180^\circ - 	heta)$.Thus, $\hat{e}_0 \cdot \hat{n} = |\hat{e}_0| |\hat{n}| \cos(180^\circ - 	heta) = -\cos 	heta$.We can resolve the incident ray $\hat{e}_0$ into components parallel and perpendicular to the normal $\hat{n}$:$\hat{e}_0 = \hat{e}_{0\perp} + \hat{e}_{0\parallel} = (\hat{e}_0 \cdot \hat{n}) \hat{n} + (\hat{e}_0 - (\hat{e}_0 \cdot \hat{n}) \hat{n})$.The reflected ray $\hat{e}$ has the same component parallel to the mirror surface but the component along the normal is reversed:$\hat{e} = -(\hat{e}_0 \cdot \hat{n}) \hat{n} + (\hat{e}_0 - (\hat{e}_0 \cdot \hat{n}) \hat{n})$.$\hat{e} = \hat{e}_0 - 2(\hat{e}_0 \cdot \hat{n}) \hat{n}$.Therefore, the correct option is $(B)$.

$A$ ray of light is reflected by a plane mirror. $\hat{e}_0$, $\hat{e}$ and $\hat{n}$ be the unit vectors along the incident ray, reflected ray and the normal to the reflecting surface respectively. Which of the following gives an expression for $\hat{e}$?

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