Question
Prove geometrically that when plane mirror turns through a certain angle, the reflected ray turns through twice the angle.
✓
Answer
Consider a ray of light $A B$, incident on plane mirror in position $MM ^{\prime}$, such that BC is the reflected ray and BN is the normal.
$
\begin{array}{l}
\angle ABM=\angle CBN=\angle i \\
\angle ABC=2 \angle i \ldots(i)
\end{array}
$
Let the mirror be rotated through an angle ' 0 ' about point $B$, such that $M_1 M_1$, is the new position of the mirror and $BN _1$ is the new position of normal. As the position of the incident ray remains same, therefore new angle of the incidence is $\angle A B N$, whose magnitude is $(i+\theta)$. Let BD be the reflected ray, such that $\angle DNB { }_1$ is the new angle of reflection.
$
\begin{aligned}
\angle ABD & =\angle ABN_1+\angle DBN_1 \\
& =\angle(i+\theta)+\angle(i+\theta) \\
& =2 \angle i+2 \angle \theta\quad \quad \ldots \ldots(ii)
\end{aligned}
$
Subtracting (i) from (ii),
$
\begin{array}{l}
\angle ABD-\angle ABC=2 \angle i+2 \angle \theta-2 \angle i \\
\therefore \angle CBD=2 \angle \theta
\end{array}
$
Thus, for a given incident ray, if the plane mirror is rotated through a certain angle, then the reflected ray rotates through twice the angle.
$
\begin{array}{l}
\angle ABM=\angle CBN=\angle i \\
\angle ABC=2 \angle i \ldots(i)
\end{array}
$
Let the mirror be rotated through an angle ' 0 ' about point $B$, such that $M_1 M_1$, is the new position of the mirror and $BN _1$ is the new position of normal. As the position of the incident ray remains same, therefore new angle of the incidence is $\angle A B N$, whose magnitude is $(i+\theta)$. Let BD be the reflected ray, such that $\angle DNB { }_1$ is the new angle of reflection.
$
\begin{aligned}
\angle ABD & =\angle ABN_1+\angle DBN_1 \\
& =\angle(i+\theta)+\angle(i+\theta) \\
& =2 \angle i+2 \angle \theta\quad \quad \ldots \ldots(ii)
\end{aligned}
$
Subtracting (i) from (ii),
$
\begin{array}{l}
\angle ABD-\angle ABC=2 \angle i+2 \angle \theta-2 \angle i \\
\therefore \angle CBD=2 \angle \theta
\end{array}
$
Thus, for a given incident ray, if the plane mirror is rotated through a certain angle, then the reflected ray rotates through twice the angle.
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