# Homework Help: Lens Related Question

1. May 3, 2017

### gspsaku

1. The problem statement, all variables and given/known data
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?

2. Relevant equations
1/f = 1/di + 1/do

3. 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

2. May 3, 2017

### collinsmark

'Looks good to me.

The question, as phrased, is ambiguous regarding whether it wants the answer relative to the objective, eyepiece, or the original object.

I would specify the location relative to the eyepiece, and to be safe, specify that your answer is relative to the eyepiece.

You'll have to look that one up in your textbook. This relationship is something you'll want to memorize anyway.

The sign of the distance is related to whether the image is real or virtual. And it's also related to whether the image is on the same side of the lens as the object, or on the opposite side (the specifics of this relationship is a little different with lenses than for mirrors, by the way).

You can alternatively use the first part of this problem to help you figure this out. Was the first image, created by the objective, real or virtual? Was it on the same side of the lens as the object or the other side? Was the sign of that image distance positive or negative?

So the image created by the eyepiece has a negative distance. What does that mean then about the type of image and location?

3. May 5, 2017

### rude man

Draw a picture with the objective on the left and the eyepiece on the right (without loss of generality, WLOG)!
Obviously, the object goes to the left of the objective, so where is the image due to the objective?
Then, that image is 7cm from the eyepiece so solve the lens equation again, with the object being the image made by the objective. Careful with the units.

4. May 5, 2017

### collinsmark

[Emphasis mine]

I think you may have miss-typed the number. The (lens-to-lens) separation of the lenses is 7 cm. The distance from the eyepiece to the eyepiece's object (the eyepiece's object is the image created by the objective) is not 7 cm.

5. May 6, 2017

### rude man

You're absolutely right, i miscalculated the distance of the objective's image. It's closer to the eyepiece than to the objective in fact! My very bad.