Combination of lenses and mirrors.

[Timballisto]

Homework Statement

The lens and mirror in the figure below have focal lengths of +74.0 cm and -59.1 cm, respectively. An object is placed 1.00 m to the left of the lens as shown.

http://www.jmlproductions.net/jmlproductions/p36-54.gif

I need to find the final image location.

Homework Equations

This problem uses the lens equation:

1/p + 1/q = 1/f

For verification, I use the equations M = -q/p and M_f = M₁M₂ as the second part asks for final magnification.

The Attempt at a Solution

I have tried this a few times. Calculate the image position for the first lens:

1/100 + 1/q = 1/74

1/((1/74)-(1/100)) = q

So q = 284.62 cm, so the distance of p for the mirror then is -q+100, which means that it's behind the mirror with p = -184.62 cm. Using the lens equation again:

1/-184.62 + 1/q = 1/-59.1

1/((1/-59.1)+(1/184.62)) = q

q = -86.93, so the image is formed in front of the mirror (? convex mirror? Should this happen?)

To convert q to p for the first lens again, add 100 and multiply by -1.

p = -13.07

Then find the final image position.

1/-13.07 + 1/q = 1/74

1/((1/74)+(1/13.7)) = q

q = 11.1

So add this to 100 to get final position at 111.1 cm

I have not actually submitted this answer, but using those numbers to calculate the final magnification, it is also incorrect as the magnification is wrong. What I do know is that the final image is inverted. Many thanks to those that help.

 

[rl.bhat]

Hi Timballisto, welcome to PF.
When the converging beam falls on a convex mirror, if the image falls on the center of curvature C of the mirror, the rays will retrace their path.
If the image is formed within C, the rays will converge and form an image in front of the mirror.
If the image is formed beyond C, the rays will diverge and image will be formed behind the mirror.
Check what happens in the given problem.

 

[Timballisto]

Hi, thanks for the reply.

Since the image distance for the first lens is > 100, the image forms behind the mirror, so the object for the mirror is virtual. I don't know how that changes my solution though other than to make the object position for the mirror -184.62. Should it be a different number?

 

[rl.bhat]

C for mirror is 118.2 cm. Its distance form the lens is 218.2 cm. Image distance form the lens is 284.62 cm. So the image is beyond C. Draw the ray diagram and check whether the rays from the lens converge or diverge after reflection from the mirror.