Thin lense combined with a concave mirror

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Homework Help Overview

The discussion revolves around the interaction between a thin lens and a concave mirror, specifically focusing on the calculations involving image distances, object distances, and magnification. The original poster is preparing for a physics final exam and is grappling with the implications of using a diverging lens in conjunction with a mirror.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the application of the lens formula and the implications of using a diverging lens, particularly regarding the sign of the focal length. There are attempts to reconcile algebraic manipulations with the physical characteristics of the images formed, including questions about treating virtual images as objects.

Discussion Status

Participants are actively engaging with the problem, questioning assumptions about the focal length and the nature of the images produced. Some guidance has been provided regarding the treatment of virtual images, and there is a recognition of the need to consider the positioning of images relative to the lens.

Contextual Notes

The original poster mentions difficulty in drawing a ray diagram and the constraints of not being able to upload it. There is also a sense of urgency due to the upcoming exam, which may influence the depth of exploration in the discussion.

hmsmatthew
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Homework Statement



Untitled.png


Homework Equations



1/f=1/i + 1/p
m=-i/p

Where m is magnification, i is image distance, p is object distance, and f is the focal length.

The Attempt at a Solution



1/f = 1/1.5f + 1/i
i=3f

I have also attempted to draw a ray diagram to use the image of the object through the lens as a starting point for the object to be reflected by the mirror. Using a conventional ray diagram i cannot equal 3f, this does not make sense with a diverging lense. My diagram has gone nowhere and i cannot upload it with ease.

Any help at all would be greatly appreciated. Physics final exam is next week !
 
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hmsmatthew said:
1/f = 1/1.5f + 1/i
i=3f
Realize that for a diverging lens the focal length is negative.
 
Doc Al said:
Realize that for a diverging lens the focal length is negative.

yes i realize this, and the image is also virtual and negative. i thought algebraically you treat the variables as if they are unknown when solving for something ?
 
hmsmatthew said:
yes i realize this, and the image is also virtual and negative.
But your equation does not reflect that fact.
i thought algebraically you treat the variables as if they are unknown when solving for something ?
The focal length is not an unknown.
 
Doc Al said:
But your equation does not reflect that fact.

The focal length is not an unknown.

Ok i think i have the answer now. I think its safe to say -1/f = +1/1.5f + 1/i which gives i=-0.6f

This agrees with my ray diagram.

From there i think i can treat each image as if it were an object for the reflection and refraction. In this way i can get image 2 and image 3. Only trouble is, image 1 is behind the lens, so not sure if i can treat it as a object and pretend the lens isn't there hmmmm ?

:)
 
hmsmatthew said:
Ok i think i have the answer now. I think its safe to say -1/f = +1/1.5f + 1/i which gives i=-0.6f

This agrees with my ray diagram.

From there i think i can treat each image as if it were an object for the reflection and refraction. In this way i can get image 2 and image 3.
All good.
Only trouble is, image 1 is behind the lens, so not sure if i can treat it as a object and pretend the lens isn't there hmmmm ?
That's exactly what you need to do.
 

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