91Ó°ÊÓ

Corner reflector in which three mirrors are fastened together to form corners of the cube

From the laws of reflection, we know that incident and reflected rays lie in the same plane, and the angle of reflection is the same as the angle of incidence, but the direction of propagation is reversed.

Formula:

$\mathbf{∠}\mathbf{i}\mathbf{=}\mathbf{∠}\mathbf{r}$

3D vector,$\stackrel{\mathbf{→}}{\mathbf{R}}\mathbf{=}\mathbf{a}\hat{\mathbf{i}}\mathbf{+}\mathbf{b}\hat{\mathbf{j}}\mathbf{+}\mathbf{c}\hat{\mathbf{k}}$

Consider a right-handed $xyz$coordinate system and assume that$\mathrm{xy}\mathrm{plane},\mathrm{yz}\mathrm{plane}$and$\mathrm{zx}\mathrm{plane}$form the mirror surfaces. Suppose an incident ray R strikes the mirror in the xy plane; then the direction of R is given by the unit vector.

$\stackrel{→}{R}=cos\left(\mathrm{Î\pm }\right)\hat{i}+cos\left(\mathrm{β}\right)\hat{j}+cos\left(\mathrm{γ}\right)\hat{k}$

Where, $\mathrm{Î\pm },\mathrm{γ},\mathrm{β}$are the angles made $\stackrel{→}{R}$with the three axes.

After reflection, x and y components of $\stackrel{→}{R}$don’t change their direction, but z component changes the sign as shown in the figure.

After reflection, the unit vector becomes

$\stackrel{→}{R}=cos\left(\mathrm{Î\pm }\right)\hat{i}+cos\left(\mathrm{β}\right)\hat{j}-cos\left(\mathrm{γ}\right)\hat{k}$

Next, it strikes the mirror in the zx plane.Then, following the same logic as described above, we get

$\stackrel{→}{R}=cos\left(\mathrm{Î\pm }\right)\hat{i}-cos\left(\mathrm{β}\right)\hat{j}-cos\left(\mathrm{γ}\right)\hat{k}$

When the ray strikes the mirror in the yz plane,we have the final unit vector as follows:

$\stackrel{→}{R}=\cos \left(\mathrm{Î\pm }\right)\hat{i}-cos\left(\mathrm{β}\right)\hat{j}-cos\left(\mathrm{γ}\right)\hat{k}$

This is the unit vector in the direction of the reflected ray.

So $\stackrel{→}{R}$is exactly the same but opposite in direction.