Optics and height of mirror problem

[DanicaK]

Homework Statement

A man is 160 cm tall and hiss eyes are 150 cm above the floor. He looks at his image formed by a plane mirror pleaced on a wall.

Homework Equations

a) In order to see his feet, what should the distance beetwen the bottom edge of the mirror and the floor be?
b)To see his image completely, what should the height of the mirror be?

The Attempt at a Solution

 

[rl.bhat]

Hi, DanicaK, welcome to PF.
Can you draw the ray diagram. What is your attempt?

 

[DanicaK]

Here it is.

 

[rl.bhat]

Here it is.

I can't see the picture.

 

[DanicaK]

hm ... It's attachement and it is drown in Paint.

 

[essecks]

That diagram seems to have it.

Now just think about how you'd be able to figure out the height, and the measurements of that mirror?

:)

[edit] also, the picture obviously worked for me too :)

 

[mgb_phys]

the diagram is correct, all you have to remember is that the angle between the incoming ray and the mirror ( actually the normal to the mirror ie. the horiontal) is the same as the reflected ray.

ps. Attachments take a few minutes to show up - they are checked by the staff to make sure they aren't spam/porn.

 

[DanicaK]

I know the things you said, but ...

 

[rl.bhat]

I know the things you said, but ...

Can you state the laws of reflection?

 

[DanicaK]

The angle between the incoming ray and the normal of the mirror is the same as the angle between the normal and the reflected ray.

 

[DanicaK]

The solutions are:
a) 75 cm
b) 80 cm.

But how to find them?

 

[rl.bhat]

Draw a perpendicular from the lower edge of the mirror on the line joining foot of the person and his eye. It bisects the line. So what will be the distance between the lower edge of the mirror and the floor?
Repeat the same thing for the line joining the top of the head and eye.

 

[DanicaK]

I can't solve it yet.

 

[rl.bhat]

I can't solve it yet.

Distance between foot to eye is 150 cm.To see the foot, lower edge of the mirror should be at a height equal to half of this distance.

 

[essecks]

I can't solve it yet.

The angle between the incoming ray and the normal of the mirror is the same as the angle between the normal and the reflected ray.

That is a key part of the solution. If the angles are the same, what does that say about the vertical distances (with regard to each other)?

 

[DanicaK]

O, OK :) Thanks a lot!