Optics Problem (Curved Mirror + Lens)

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Homework Statement



A convex mirror with a focal length of -20 cm is 18 cm from a convex lens whose focal length is 6.0 cm. An object is located between them, 10 cm from the mirror and 8.0 cm from the lens. The object will produce two images, one from the light which just goes through the lens and the other from light which first hits the mirror and then goes through the lens.
(a) Where is the image from the light which just goes through the lens?
(b) Where is the image from the light which hits the mirror and then the lens? (c) If the height of the object is 1.0 cm where what is the size of both images?


Homework Equations


(1/di)+(1/do)=(1/f)
m=(hi)/(ho)=-(di)/(do)


The Attempt at a Solution


I do not have the slightest idea on how to approach this problem. Had it been two lenses opposed to a lens and a mirror I may have had a chance at producing a solution. Does anyone have an idea how to begin this problem?

Thank you!
 

Answers and Replies

  • #2
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May someone please take a stab at this!? My final's tomorrow!!! Gahhhhhhhaskldfjasd;f
 
  • #3
TSny
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Hopefully, you can answer (a). If not, show us what you tried so we can see where you're having difficulty.

For part (b), can you find the image that would be produced by the mirror alone? If so, you can use that image as the object for the lens.
 
  • #4
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For part (a) I found the image distance di to be 24 cm to the right of the lens.

(b) Image produced by mirror alone:
1/do + 1/di= 1/f
-> (-1/10cm) + 1/di)=(-1/20cm) ----> di= + 20 cm or 2cm to the right of the lens

Taking the mirror's image +2cm=do2

1/di=1/6-1/2cm

-> di= -3.03 so three cm to the left of the lens?

I thank you tremendously for your time TSny
 
  • #5
TSny
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Good for part (a)!

For part (b) the object is a real object for both the lens and the mirror. So, the object distance should be +10 cm for the mirror. You'll need to recalculate this part. Otherwise, look's like you have the right approach.

The best way to tell if an object is real (+ object distance) is to use the following rule: If the rays of light from the object are spreading apart (diverging) as the rays approach the mirror or lens, then the object is real for that mirror or lens. If the rays are converging as they approach the mirror or lens, then the object is virtual (- object distance).

If the image of the mirror does occur to the right of the lens in this problem, then the rays of light approaching the lens from the mirror would be converging. So, you would use a negative object distance for the lens. However, I think you will find that after you correct the calculation for the mirror, you will find that the image of the mirror comes out to the left of the lens. So, rays of light from the image of the mirror will be diverging as they approach the lens. So, the image of the mirror will be a real object for the lens.
 
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  • #6
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Thank you very much, TSny. That is a fantastic thing to know that i'm sure I will find helpful for times to come.

Thank you very, very much for this.

Happy Holidays

Ps
Is the answer 13cm to the right of the lens?
 
  • #7
TSny
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Didn't get that answer. What did you get for the location of the image of the mirror? Is that image behind the mirror or in front of the mirror?
 
  • #8
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Err..really? The image from the mirror was -6.67 cm --> A result of (1/10cm)+1/di=-(1/20)

This must mean it is 6.7cm to the right of the mirror, and 11.3 cm to the left of the lens.

Taking 11.3cm as our lens' new object distance do, (1/11.3cm)+(1/di)=(1/6)

1/di=.08cm -> di= 12.8cm and subsequently 12.8cm to the right of our lens.

Where did I wrong?!
 
  • #9
TSny
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You got -6.67 cm for the image distance of the mirror. Good. But note the negative sign. That means the image is virtual (on the back side of the mirror). So, the distance of the image of the mirror to the lens must be greater than 18 cm.
 

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