Reflection from a plane mirror

[JJBladester]

Homework Statement

You need to place a mirror on the left wall of the figure so that the reflected light from the bulb exactly fills the right wall.

A) What is the proper height of the mirror?
B) How far below the ceiling should the top edge of the mirror be?

Homework Equations

\theta_i=\theta_r

The Attempt at a Solution

This is the image I created below. I'm not sure if "height" of the mirror is the height of it from its top to its bottom or if the height is the distance from the floor to the top of the mirror. How do I get theta?

 

[ehild]

Draw the reflected rays as if they came straight out of the mirror image of the lamp.

ehild

 

[JJBladester]

ehild,

Thanks for your response. I have redrawn the problem with the ray coming from the image one meter to the left of the mirror. I think I have the setup correct now. Am I trying to find theta_r so I can get the mirror's height? If I get theta_r, how does that help me get the height of the mirror?

 

[ehild]

Draw the other ray from the image to the bottom of the wall. The mirror extends between those points of the wall where these rays cross it. But still there will be a shadow of the screen in the middle of the wall.

ehild

 

[gneill]

What is the vertical position of the bulb? Is it centered between floor and ceiling?

 

[JJBladester]

What is the vertical position of the bulb? Is it centered between floor and ceiling?

From the picture, it appears so, but can we assume that... That's another question? If we assume the bulb's vertical position is 1.5m from the floor, then:

\theta_r=tan^{-1}\left (\frac{1.5}{5}\right )=16.7deg

But where do I go from there?

I am trying to calulate the height of the mirror which I believe would be between the two red marks I put in this diagram of the situation:

 

[JJBladester]

Nvm... Got the height! It was 0.600m which I got from 2*[1*tan(theta_r)]. Thanks all for your help. I guess it's the geometry that got me on this one.

 

[ehild]

Nvm... Got the height! It was 0.600m which I got from 2*[1*tan(theta_r)]. Thanks all for your help. I guess it's the geometry that got me on this one.

The laws of Geometry are valid even for Physics:) .

Remember this method of image.

ehild