Troubleshooting Multiple Optics Problems: Finding the Final Image Location

[asd1249jf]

Homework Statement

A plane mirror is located at the origin. A converging lens with focal length 5.00m is located at x=1m. An object is placed at 31m.

What is the location of the final image, as seen by an observer looking toward the mirror through the lens?

Homework Equations

1/f = 1/s + 1/s'

The Attempt at a Solution

First, we need to find the image created by the lens as if the mirror was not there.

so 1/(5m) = 1/(30) + 1/s'

s' = 6m

In terms of the coordinate system used

x = -5m, since the image is projected on the negative side.

Now, we need to find the image produced by the mirror due to the reflection from the image produced by the lens. Since angle of incidence = angle of reflection, we immidiately know that the projected distance is the same but on the opposite, therefore, the image distance created by the mirror is x = 5m.

Now the image created by the mirror is the image for lens. We apply the equation again so

1/(5m) = 1/(4m) + 1/s' <- Note the 4m is the object distance.

s' = -20m

So in terms of the coordinate system, we have x = -19m, which turned out to be wrong.


What did I do wrong in the steps? Can someone give me any pointers?

 

[asd1249jf]

Anyone have any idea?

 

[andrevdh]

The mirror folds the beams back towards the lens. The lens will therefore "see" the virtual object 4 meter on its (now) output side. This means that you have to insert it as - 4.0 m into the equation.