Question

Four identical mirrors are made to stand vertically to form a square arrangement as shown in a top view. A ray starts from the midpoint $M$ of mirror $AD$ and after two reflections reaches corner $D$. Then, angle $\theta$ must be

• A. $ta{n}^{-1}\left(0.75\right)$
• B. $co{t}^{-1}\left(0.75\right)$
• C. $si{n}^{-1}\left(0.75\right)$
• D. $co{s}^{-1}\left(0.75\right)$

Answer 1

Answer: (B)

The ray starting from point $M$ at an angle $\theta$ reaches the corner $D$ at the right along a parallel path. Let $a$ be the length of the side.

From figure,
$tan\theta =\frac{x}{\left(a/2\right)}...\left(i\right)$
$tan\theta =\frac{a-x}{y}...\left(ii\right)$
$tan\theta =\frac{a}{a-y}...\left(iii\right)$
From $\left(i\right)$ and $\left(ii\right)$, we get
$\frac{2x}{a}=\frac{a-x}{y}$ or $2xy=a^2-xa...\left(iv\right)$
From $\left(ii\right)$ and $\left(iii\right)$, we get
$\frac{a-x}{y}=\frac{a}{a-y}$
$\Rightarrow 3xy=2ay$ (Using $(iv)$)
$x=\frac{2a}{3}$
Substituting this value of $x$ in equation $(i)$, we get
$tan\theta =\frac{\left(2a/3\right)}{\left(a/2\right)}=\frac{4}{3}$
$\therefore cot\theta =\frac{1}{tan\theta }=\frac{3}{4}$
or $\theta =co{t}^{-1}\left(0.75\right)$