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Light problem with a mirror and a lens

  1. Mar 8, 2013 #1
    1. The problem statement, all variables and given/known data

    A +15 cm focal length lens and a flat mirror are 45 cm apart (the mirror is to the right of the lens and can be treated as a concave mirror with an infinite focus length). There is an intermediate image 30 cm to the right of the lens (15 cm to the left of the mirror) with a height of -2mm.

    A) Find the height and location of the original object.

    B) Find the location of the FINAL image produced by your object in part a.

    2. Relevant equations

    1/do + 1/di = 1/f
    m=di/do

    3. The attempt at a solution

    So the solutions (in the pdf) say that there are two ways to do this. I chose solution one, but either way i choose to do this...I get stuck at the same part. So I understand that this intermediate image can first be just from the lens so I do the lens equation to find the do = 30 cm...meaning it's to the left of the lens (positive). So in the second part I first reflect the image off of the mirror and get -15cm (to the right of the mirror). NOW is the part I get confused on...this IMAGE is now the OBJECT for the lens...it is 60 cm away from the lens BUT IT IS A VIRTUAL OBJECT AND IS TO THE RIGHT OF THE LENS...therefore I believe it should be -60 cm...so I did the lens equation with -1/60 + 1/di = 1/15...which results in a different answer than the one in the solutions. I'm not sure why my professor uses a positive object distance for both solutions in the last part...it's confusing me.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     

    Attached Files:

  2. jcsd
  3. Mar 8, 2013 #2

    Doc Al

    User Avatar

    Staff: Mentor

    I would not consider it a virtual object, since the light is really coming from that point. And since the light from that "object" is moving from right to left, you would reverse the sign convention when analyzing its path through the lens. (The usual sign conventions are based on the direction the light is actually traveling.)
     
  4. Mar 8, 2013 #3
    Okay, it makes sense that the light is coming from that point. But why is this not a virtual object. The original object was to the left of the lens...the "2nd" object is between the lens and the mirror (which is a real object). Now the mirror reflects this real object into a virtual image on the opposite site of the mirror, which is being used as the object for the lens again. Light is coming off of that virtual object in the direction which is why we do the opposite of convention (because the "light" is going from right to left)??? Is this a good way to look at it? Thanks for the help Doc Al.
     
  5. Mar 8, 2013 #4

    Doc Al

    User Avatar

    Staff: Mentor

    Yes, that's a good way to look at it. And you're right, I suppose you can call it a virtual object as long as you take account of the direction of the light and don't blindly assume that the object distance will be negative.

    A more typical case of virtual object would be if you had two lenses in a row. With light going from left to right, say the image from the first lens ends up to the right of the second lens. Then to analyze the effect of the second lens that image would be treated as a virtual object, with a negative object distance. (I'm sure you are well aware of this!)
     
  6. Mar 8, 2013 #5
    very helpful response...thanks a lot!
     
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