A) 3d
B) 2d
C) d
D) \[\frac{d}{2}\]
Correct Answer: A
Solution :
In the given figure \[\Delta AED\]and \[\Delta ABC\]are similar triangles. So,\[\frac{BC}{ED}=\frac{AC}{AD}\Rightarrow \frac{BC}{ED}=\frac{2L}{L}\]\[\Rightarrow BC=2ED\] ?(i) Also, \[\Delta AED\]and \[\Delta ASD\]are congruent triangles. So, ED = DS ...(ii) Using (i) and (ii), BC = d So, the distance over which the man can see the image of the light source in the mirror is \[d+d+d=3d\]You need to login to perform this action.
You will be redirected in
3 sec