question_answer

Show that the length of a plane mirror in which a person can see his fall image is half the height of the person.          

Answer:

                           Let\[{{M}_{1}}\text{ }{{M}_{2}}\]be the plane mirror fixed on a wall. HEO is the position of a person standing in front of the mirror. E is the position of the eye of the person. To know the height of plane mirror, the key thing is to remember that a ray of light from leg must reach his eye and similarly a ray of light from the top of his head should also reach his eye. A ray of light coming from the head of the person after reflection from mirror enters the eye of the person. This ray appears to come from H behind the mirror. Similarly, a ray from the feet of the person after reflection from the mirror enters the eye of the person. This ray appears to come from O behind the mirror. Thus, HO be the position of the full image of the person. Now, draw the perpendiculars\[{{M}_{1}}N\]and\[{{M}_{2}}M\]on the mirror. According to law of reflection, \[\angle H{{M}_{1}}N=\angle E{{M}_{1}}N\]                       \[\therefore \]\[HN=NE\] Also, \[\angle E{{M}_{2}}M=\angle O{{M}_{2}}M\] \[\therefore \] \[EM=MO\] Now, \[{{M}_{1}}{{M}_{2}}=NM=HO(HN+MO)\] Using equations, (1) and (2), we get \[{{M}_{1}}{{M}_{2}}=HO(NE+EM)=HONM=HO\] \[{{M}_{1}}{{M}_{2}}\] or \[2{{M}_{1}}{{M}_{2}}=HO\] or \[{{M}_{1}}{{M}_{2}}=\frac{1}{2}HO\] i.e., length of mirror\[=\frac{1}{2}\times \]height of the person