Question

A mirror and a source of light are situated at the origin $O$ and at a point on $OX$ respectively. A ray of light from the source strikes the mirror and is reflected. If the DRs of the normal to the plane are $1,-1,1$, then $DCs$ of the reflected ray are

• A. $\frac{1}{3},\frac{2}{3},\frac{2}{3}$
• B. $-\frac{1}{3},\frac{2}{3},\frac{2}{3}$
• C. $-\frac{1}{3},-\frac{2}{3},-\frac{2}{3}$
• D. $-\frac{1}{3},-\frac{2}{3},\frac{2}{3}$

Answer 1

The correct option is D $-\frac{1}{3},-\frac{2}{3},\frac{2}{3}$

Let the source of light be situated at $A(a,0,0),$ where $a\ne 0.$

Let $OA$ be the normal to the mirrior at $O.$

$\therefore \angle NOA=\angle NOB=\frac{\theta }{2}$

$D{R}^{\prime }s$ of $OA$ are $<a,0,0>$ and so its $D{C}^{\prime }S$ are $(1,0,0)$

$D{C}^{\prime }s$ of $ON$ are $\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}$

$\therefore \cos \frac{\theta }{2}=\frac{1}{\sqrt{3}}$

Let $l,m,n$ be the $D{C}^{\prime }s$ of the reflected ray $OB.$

Then as in above question

$\frac{l+1}{\frac{2\cos \theta }{2}}=\frac{1}{\sqrt{3}},\frac{m+0}{\frac{2\cos \theta }{2}}=-\frac{1}{\sqrt{3}}$

and $\frac{n+0}{\frac{2\cos \theta }{2}}=\frac{1}{\sqrt{3}}$

$\Rightarrow l=\frac{2}{3}-1,m=\frac{-2}{3},n=\frac{2}{3}$

$\Rightarrow l=-\frac{1}{3},m=-\frac{2}{3},n=\frac{2}{3}$

Hence, $D{C}^{\prime }s$ of the reflected ray are $-\frac{1}{3},-\frac{2}{3},\frac{2}{3}$