Thin lense combined with a concave mirror

[hmsmatthew]

Homework Statement

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 !

 

[Doc Al]

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

Realize that for a diverging lens the focal length is negative.

 

[hmsmatthew]

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 ?

 

[Doc Al]

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.

 

[hmsmatthew]

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 ?

:)

 

[Doc Al]

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.