Question

Unit vector along a ray of light that is incident on a plane mirror is 1/√3(i+j+k) and unit vector along the
normal to the mirror at the point of incidence is1/√2(i+j)
Unit vector along the reflected ray can be expressed
as:​

Answer 1

Answer:

Unit vector along a Roy of light is incident 1√2+2357=2315787

Answer 2

Given:

Unit vector along incident ray $=\frac{1}{\sqrt{3} }(i+j+k)$

Unit vector along reflected ray $=\frac{1}{\sqrt{2} } (i+j)$

To find:

Unit vector along reflected ray.

Solution:

We have been given that the incident ray is falling on the plane mirror.

• We know, that, in a plane mirror there is no dissipation of any incident energy and is similar to elastic collision where the ray is reflected back along the angle similar to the angle of incidence.

• When considered the rays in a two dimensional plane, or x-y plane, the incident ray after falling on the mirror gets reflected in the opposite plane or the direction of the rays is reversed.

• According to the given question, the incident ray $(i+j+k)$ when reflected in another x-y plane, the sign of the coordinates of x and y gets changed.

Hence,

The reflected ray will be $(-i-j+k)$

Therefore,

The unit vector will be $=\frac{-i-j+k}{\sqrt{(-1)^{2}+ (-1)^{2}+ (1)^{2} } }$

                                     $=\frac{1}{\sqrt{3} }(-i-j+k)$

Final answer:

Hence, the unit vector along reflected ray will be $\frac{1}{\sqrt{3} }(-i-j+k)$