# Compound Microscope: Calculating Image Location with Two Lenses

• gspsaku
In summary, the image of the object placed 3.0 mm from the objective lens is located 28 mm in from the eyepiece.
gspsaku

## 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

gspsaku said:

## 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.
'Looks good to me.

So I believe I have solved everything. Do I need to take into account the distance between the lenses (7cm or 70mm) again?
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.

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

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?

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.

rude man said:
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.
[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.

collinsmark said:
[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.
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.

## What is a lens?

A lens is a transparent object with curved surfaces that can refract or bend light, allowing us to see objects in focus.

## How does a lens work?

A lens works by bending light as it passes through it, which causes the light rays to converge or diverge. This bending of light allows us to see objects in focus.

## What are the different types of lenses?

There are two main types of lenses: convex lenses, which are thicker in the middle and converge light, and concave lenses, which are thinner in the middle and diverge light. Other types of lenses include plano-convex, plano-concave, and meniscus lenses.

## What is the focal length of a lens?

The focal length of a lens is the distance between the lens and the point where light rays converge or diverge to form an image. It is typically measured in millimeters (mm).

## How do I choose the right lens for my camera?

Choosing the right lens for your camera depends on your specific photography needs. Factors to consider include focal length, aperture, and lens type. It is best to do research and read reviews to determine which lens will best suit your needs.

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