Geometric Optics: Solving for Position & Nature of Image

[Darth Frodo]

Homework Statement

An object is placed 400 mm in front of a convex lens of focal length 80 mm. Find the position of the image formed. State the nature of this image.

A second convex lens of magnifying power X8 is placed 125 mm behind the first convex lens.
What is the focal length of this lens?
Find the position of the final image formed and state its nature.

The Attempt at a Solution

So the first part is pretty easy,

$\frac{1}{U} + \frac{1}{V} = \frac{1}{F}$

$\frac{1}{V} = \frac{1}{80} - \frac{1}{400}$

$V = 100 mm$


Next part is where I'm unsure. Here's my attempt,

$m = \frac{-V}{U}$

$8 = \frac{-V}{U}$

$8 = \frac{-V}{25}$

$V = -200mm$

$\frac{1}{U} + \frac{1}{V} = \frac{1}{F}$

$\frac{1}{F} = \frac{1}{25} - \frac{1}{200}$

$F = 28.6 mm$

Is this the correct method?

Thanks.

 

[rude man]

The second lens magnifying power is defined (most of the time) by 1/4 diopters + 1. It is not a function of where the object is placed. So 8 = 1/4 d + 1, solve for d, then f = 1/d in meters. But now you have 2 lenses with a distance between them so your job now is to come up with the effective focal length of the two lenses with their separation accounted for.

 

[Darth Frodo]

Hmm, that's strange. What if I put this into context and said I was a freshman and had never encountered this formula before?

 

[rude man]

Hmm, that's strange. What if I put this into context and said I was a freshman and had never encountered this formula before?

If it's any consolation, I had to look up the meaning of a " ... lens with magnification power of x" also. And I got my degree in 1962!

And it's always possible the question had a different intent, like maybe the magnification of the object was 8 after inserting the second lens ...