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 d.c's 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

Answer: (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 incident ray and $OB$ the reflected ray.

$ON$ is the normal to the mirror at $O.$

$\therefore \angle AON=\angle NOB=\frac{\theta }{2}$ (say)

d.r's of $OA$ are $a,0,0$ and so its d.r's are $1,0,0$

d.r'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.r's of the reflected ray $OB.$

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

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's of the reflected ray are $-\frac{1}{3},-\frac{2}{3},\frac{2}{3}$