Question

Suppose a man stands in front of a mirror as shown below. His eyes are 1.38 m above the floor and the top of his head is 0.284 m higher. Find the height above the floor of the top of the smallest mirror in which he can see both the top of his head and his feet.

answer to 3 decimal places.

Answer 1

Answer :_

some points

law of reflection $\theta _i=\theta_r$

diagram must be like this

For the Top of the mirror :-

From the figure , it is obvious that the top of the mirror is equal to GD, which is equal to

GD=GA+AD

Similarly as before we can find the distance GA , where thew triangle AGE and FGE shares the same adjecent side , and have the same angle as before due to law of reflection where the incident angle is equal to the reflected angle , and hence

$\theta_i=\theta_r$

$tan(\theta_i)=tan(\theta_r)\\ \frac{GA}{GE}=\frac{GF}{GE}\\ GA=GF$

And , from the givens we know that

AF=0.284

=GA+GF

=2GA

=2GF

thus,

$GA=\frac{0.284}{2}=0.142 \ m$

thus, height of the top mirror which is equal to GD is

GD=GA+AD

= 0.142+1.38

= $\boxed{1.522 m}$ answer