Thin lense combined with a concave mirror

AI Thread Summary
The discussion focuses on the application of the lens and mirror equations to solve a physics problem involving a thin lens and a concave mirror. The user initially struggles with the calculations, particularly regarding the negative focal length of a diverging lens. After some back-and-forth, they arrive at the correct equation, confirming that the image distance is -0.6f, which aligns with their ray diagram. The user contemplates how to treat the virtual image produced by the lens as an object for further calculations. The final consensus is that it is acceptable to treat the virtual image as an object for subsequent reflections and refractions.
hmsmatthew
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Homework Statement



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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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