Question

A plane mirror is placed at the origin so that the direction ratios of its normal are $(1,-1,1).$ A ray of light, coming along the positive direction of the $x-$axis strikes the mirror. Then the direction cosines 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 $(l,m,n)$ be the d.c.'s of reflected ray,
$(1,0,0)$ are the d.c.'s of incident ray $(x-\text{axis})$
D.c.'s of normal are $(\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}})$
Let $(l_1,m_1,n_1)=(l,m,n),(l_2,m_2,n_2)=(1,0,0)$
If $\theta$ is the angle between the normal to the plane and incident ray, then $\cos \theta =\frac{1}{\sqrt{3}}$
Where,
$(\frac{l_1+l_2}{2\cos \theta },\frac{m_1+m_2}{2\cos \theta },\frac{n_1+n_2}{2\cos \theta })=(\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}})$
$\frac{l+1}{2\cos \theta }=\frac{1}{\sqrt{3}}\Rightarrow l=-\frac{1}{3}$
$\frac{m-0}{2\cos \theta }=-\frac{1}{\sqrt{3}}\Rightarrow m=-\frac{2}{3}$
$\frac{n+0}{2\cos \theta }=\frac{1}{\sqrt{3}}\Rightarrow n=\frac{2}{3}$