Question

2. Consider the following retroreflector (a device that reflects light back to its source), which consists of two perpendicular mirrors. Prove that any ray incident on this retroreflector is reflected backward regardless of its incident angle. 3. The numerical aperture (NA) of an optical fiber is defined as NA-sin Cnax), where

Consider the following retroreflector (a device that reflects light back to its source), which consists of two perpendicular mirrors. Prove that any ray incident on this retroreflector is reflected backward regardless of its incident angle

Answer 1

Mirror AO is perpendicular to mirror OD

i.e $\dpi{150} \bg_white \angle$ AOD= 90°

To prove any incident rays get reflected backward regardless of its angle of incidence , it is enough to prove incident rays and reflected rays become parallel to each other after reflection from retroreflector.

Proof:

Let PB is incident ray that falls at B making an angle $\dpi{150} \bg_white \Theta$ 1 with the normal BM

BC is the reflected ray that makes angle $\dpi{150} \bg_white \Theta$ 2 with normal BM

Now the reflected ray will act as incident ray for the mirror OD making an angle $\dpi{150} \bg_white \Theta$ 3 with normal CM

CQ is the reflected ray making angle $\dpi{150} \bg_white \Theta$ 4 with normal

Now,

BM is normal to mirror OA and CM is normal to OD

$\dpi{150} \bg_white \angle$ BOC = 90°

Hence closed figure BOCM is a rectangle

and $\dpi{150} \bg_white \angle$ BMC= 90°

Now coming to traingle BCM

Sum of angle of traingle is 180°

$\dpi{150} \bg_white \Theta$2 +$\dpi{150} \bg_white \Theta$3+ $\dpi{150} \bg_white \angle$ BMC = 180

$\dpi{150} \bg_white \Theta$2+ $\dpi{150} \bg_white \Theta$ 3 = 180° - 90° = 90°

....... eqn1

Now multiply eqn 1 by 2

2*($\dpi{150} \bg_white \Theta$2 +$\dpi{150} \bg_white \Theta$3) = 90* 2

(2 * $\dpi{150} \bg_white \Theta$ 2 + 2*$\dpi{150} \bg_white \Theta$3 )= 180°

($\dpi{150} \bg_white \Theta$1+$\dpi{150} \bg_white \Theta$2 )+ ($\dpi{150} \bg_white \Theta$3+$\dpi{150} \bg_white \Theta$4)= 180°

$\dpi{150} \bg_white \angle$PMC + $\dpi{150} \bg_white \angle$ BCQ = 180°

{ Note: angle of incidence and angle of reflection are equal

hence ,angle PBM = angle CBM

and angle BCQ =angle QCM ( see diagram)}

Hence sum of interior angle on same side of transversal. ( supplementary angles )= 180°

This proves that Ray PB and CQ are parallel to each other

Conclusion :

As rays of incident and final reflected rays become parallel to each other in retroreflector, this implies whatever may be the angle of incidence is! final reflected rays will always be parallel to incidence rays regardless of its angle of incidence