- #1

gspsaku

- 18

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## Homework Statement

A compound microscope consists of two converging lenses (the objective and the eyepiece) placed 7.0 cm apart. If the objective has a focal length of 2.8 mm and the eyepiece has a focal length of 3.3 cm, what would be the location of the final image of an object placed 3.0 mm from the objective lens?

## Homework Equations

1/f = 1/di + 1/do

## The Attempt at a Solution

Alright. So first I solved for the image relative to the objective lens.

1/2.8 = 1/di + 1/3 => di = 42mm

Since there is a 7cm, or (70mm) gap between the two lenses, I take that into account therefore it is 70mm-42mm= 28mm. So the image is 28mm in from the eye piece.

Now using the same formula above, 1/f = 1/di + 1/do using mm again to be consistent

1/33 = 1/di + 1/28 => di = -184.8 mm.

So I believe I have solved everything. Do I need to take into account the distance between the lenses (7cm or 70mm) again? Lastly, since it is -184.8mm, does that mean it is in front or behind of the eye piece? This is where I get confused

Thanks