Object distance Image distance do di Eye Object Virtual image Plane mirror (b) Figure . Formation of image in plane mirror for (a) point and (b) extended objects - - - - Unit ray optics Table . Image by inclined mirrors θ The position of object placed Number of images n Even Symmetrical n = q Unsymmetrical n = q Odd Symmetrical n = q Unsymmetrical n = q Figure . Images formed by inclined mirrors EXAMPLE .
What is the height of the mirror needed for a person to see his/her image fully on the mirror? Solution Let us assume a person of height h is standing in front of a vertical plane mirror. The person could see his/her head when light from the head falls on the mirror and gets reflected to the eyes. Same way, light from the feet falls on the mirror and gets reflected to the eyes.
A light ray AO from the point object is incident on the mirror and it is reflected along OB . The normal is ON . The angle of incidence ∠ AON = angle of reflection ∠ BON Another ray AD incident normally on the mirror at D is reflected back along DA . When BO and AD are extended backwards, they meet at a point A' .
Thus, the rays appear to come from a point A' which is behind the plane mirror. The object and its image are at equal perpendicular distances from the plane mirror which can be shown by the following explanation. Angle ∠ AON = angle ∠ DAO [Since they are alternate angles] Angle ∠ BON = angle ∠ O ¢ A D [Since they are corresponding angles] Hence, it follows that angle, ∠ DAO = ∠ O ¢ A D The triangles ∆ ODA and ∆ OD ¢ A are congruent ∴ AD = ¢ A D This shows that the image distance d i inside the plane mirror is equal