Simple Optics Problem (plane mirror and lens)

[AKJ1]

Homework Statement

A plane mirror is placed at the origin. A converging lens with a focal length of 5.00m is located at x=12.50m. A object is placed at x=22.5m

Find the final location of the image in terms of its X coordinate & magnification.

Homework Equations

1/q + 1/p = 1/f (lens)

1/q = -1/p (plane mirror; focal length is infinite)

q = image distance
p= object distance

The Attempt at a Solution

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So I first find the distance from the lens to the object (10m), then following from the equation

1/q + 1/p = 1/f

1/q +1/10 = 1/5

q = 10m

Now we consider the mirror. The image from the lens will serve as the object for the mirror.

1/q = -1/p

1/q = -1/10

q = -10m ?

However the answer says 20m. I just started studying geometric optics (I have yet to have a formal lecture on it), so I am likely missing something fundamental here. Is there another image somehow formed?

Thanks

 

[TSny]

So I first find the distance from the lens to the object (10m), then following from the equation

1/q + 1/p = 1/f

1/q +1/10 = 1/5

q = 10m

Looks good so far.

Now we consider the mirror. The image from the lens will serve as the object for the mirror.

1/q = -1/p

1/q = -1/10

Yes, the image of the lens serves as the object for the mirror. But, 10 m is not what you should use for the object distance for the mirror. A diagram where you show the image of the lens should help.

 

[Simon Bridge]

The light reflected from the mirror will pass through the lens - making another image.
Try sketching a ray diagram.

 

[AKJ1]

Looks good so far.
Yes, the image of the lens serves as the object for the mirror. But, 10 m is not what you should use for the object distance for the mirror. A diagram where you show the image of the lens should help.

Okay so I sketched a diagram. The 10m I got is the distance from the lens, so that distance from the mirror is actually 2.5m

1/q = -1/p

1/q = -1/2.5

q = -2.5 m

The image is virtual and behind the mirror (Correct?)

So now the distance from the lens is (12.5-(-2.5)) = 15 m

1/p + 1/q = 1/f

1/15 + 1/q = 1/5 (Is the focal length still positive because we have a convex lens?)

q = 7.5 m, but the distance from the origin is 12.5 + 7.5 = 20m

Okay assuming I did it correctly, how do I know how many images will be formed in any system? Will I just need to take it step by step and realize that mirrors will reflect light back to the lens?

Thank you!

 

[AKJ1]

The light reflected from the mirror will pass through the lens - making another image.
Try sketching a ray diagram.

Will a mirror always reflect light back in the direction of which it originated? Or does this depend on the type of mirror?

 

[Simon Bridge]

Will a mirror always reflect light back in the direction of which it originated? Or does this depend on the type of mirror?

The mirror will always reflect light back into the space that the light came from - that is what reflect means. The alternative would be if the mirror somehow reflected light through itself ... there are objects that do that, they are called windows.

Some of the light reflected by the mirror will go through the lens.