Given, length of mirror, m=50cm=50×10−2m Distance of source from mirror, d=60cm =60×10−2m Distance of man from mirror, dm=1.2m By using the concept of ray diagram of plane mirror shown below
Now, using the concept of similar triangle, ∆HAl∼△GAE and ∆BAl∼∆CAE ∴
AI
AE
=
HI
EG
⇒
0.60
1.8
=
0.25
EG
⇒EG=0.25×
1.8
0.6
=0.25×3=0.75m As, CG=2EG ⇒CG=0.75×2=1.50m
AI
AE
=
HI
EG
⇒
0.60
1.8
=
0.25
EG
⇒EG=0.25×
1.8
0.6
=0.25×3=0.75m As, CG=2EG ⇒CG=0.75×2=1.50m Hence, distance between the extreme points, where he can see image of light source in mirror is 150cm.