rays). Still the laws of reflection are valid at every point of incidence in irregular reflection as shown in Figure . (b). (a) - - - - Unit ray optics ∠ R OR = ∠ N'OR – ∠ N'OR = ( i + θ ) – ( i – θ ) ∠ R OR = θ .
i i i–θ i+θ i+θ θ 2θ θ R R N A' B' N' O . . Image formation in plane mirror Let us consider a point object A placed in front of a plane mirror. The point of incidence is O on the mirror as shown in the Figure.
. (a). Figure . Angle of deviation due to reflection i r O C N X Y (a) d= -2i O C N X Y (b) d= EXAMPLE .
Prove that for the same incident light when a reflecting surface is tilted by an angle θ , the reflected light will be tilted by an angle θ . Solution AB is the reflecting surface as shown in the Figure. Both the incident ray IO and the reflected ray OR subtend angle i with the normal N as the angle of incidence is equal to angle of reflection. When the surface AB is tilted to ¢ ¢ A B by an angle θ , the normal N is also is tilted to ′ N by the same angle θ .
Remember that the position of the incident ray IO remains unaltered. But the reflected ray now is OR . Now, in the tilted system, the angle of incidence, ∠ N'OI = i+θ and the angle of reflection, ∠ N'OR = i+θ are the same. The angle between ON' and OR is, ∠ N'OR = i – θ .
The angle tilted on the reflected light is the angle between OR and OR which is ∠ R OR . From the geometry we can write, i i O D b b (a) Plane mirror N