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 ...

 

[Nickie]

I can confirm that your method is correct. You have correctly used the lens equation to find the position of the image formed by the second convex lens. The negative sign in the magnification equation indicates that the image is inverted, which is consistent with the nature of the image formed by a convex lens.

To determine the focal length of the second lens, you have correctly used the lens equation with the known values of the object distance (125 mm) and the image distance (-200 mm). This gives a focal length of 28.6 mm, which is a positive value as expected for a convex lens.

Finally, the position of the final image can be found by adding the image distance of the first lens (100 mm) and the image distance of the second lens (-200 mm). This gives a final image distance of -100 mm, indicating that the final image is located 100 mm in front of the second lens. The nature of this image would also be inverted, as it is formed by a convex lens.

Overall, your approach and calculations are correct. Well done!