# Problem about compound optics in microscopes

• help I have 12 hours
In summary: This is because the eyepiece magnifies the image, and the real image is located very close to the lens.
help I have 12 hours
Member has been reminded to show some effort from his side.
Homework Statement
The distance between the object and the eyepiece of a compound microscope is 25.0 cm. The focal length of its objective lens is 0.200 cm and the eyepiece has a focal length of 2.60 cm. A person with a near point of 25.0 cm and a far point at infinity is using the microscope.
(b) What is the total magnification of the microscope when used by the person of normal eyesight?
Relevant Equations
(1/s)+(1/s')=1/f
m=-(s')/s
angular magnification=near point/f
total magnification=(angular magnification)*linear magnification
angular magnification=9.62 part a

1/s+1/s'=5 1/(25-s-s')+1/s"=1/2.6 -s"/s=m

can't find m, don't know how to use the info that a person of normal eyesight is looking into the microscope.

When the near point is 25 cm., the angular spread ## \Delta \theta ## (radians) of the object of size ##h ## will be ## \Delta \theta=h/(25 \, cm) ## when viewed without a microscope. When viewed with a microscope, that same object of size ##h ## will make an image at ##+\infty ## where ## \Delta \theta_m ## can be readily computed. The magnification ## M=\frac{\Delta \theta_m}{\Delta \theta} ##.

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what is
Δθm
thank you

I looked over the numbers you have for the problem, and they were presented in a way that is somewhat confusing. Usually, (if I understand it correctly), the object, when viewed with a compound microscope, sits just outside the focal point of the objective. This makes for an image length inside the microscope that can be rather large, and thus a relatively large image. This image occurs at the focal point of the eyepiece lens. What they didn't give you when they wrote up the problem is the distance from the object to the objective, or the distance between the objective and the eyepiece. They leave it to the reader to figure that out. ## \\ ## Given an image size from the objective of ## h' ##, you can compute ## \Delta \theta_m=h'/f_{eyepiece} ##. You just need to compute ## h' ## from the numbers they give you. First compute the image distance for the objective's image, (simple arithmetic from the numbers they give you), and then you can compute/estimate ## h' ##, (as some number times the object size ## h ##). The ## \Delta \theta_m ## you then compute is the angular span of the final image (which is at ## +\infty ##), when viewed through the eyepiece. ## \\ ## Edit note: I changed the ##b's ## to ## h's ##. The letter ## b ## in optics is used for object distance. It should not be used for object size.

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First compute the image distance for the objective's image,

How do I find the image distance of the objective if I don't know where the objective is?
Thanks

The object distance ## b ## of the object from the objective lens can be taken as ## b \approx f_{objective}=.2 ## cm, even though it will be just slightly larger than ## f_{objective} ##.
Meanwhile, the image from the objective lens is located at the focal point of the eyepiece lens. This means the image distance ## m ## from the objective is ## m=25.0-.2-2.6=22.2 ## cm. (simple arithmetic). (I would do well to draw a diagram, but I think you can follow where the image is located).
If the object is ## h ## units high, what is the image height ## h' ## of this intermediate image?(there is a formula to compute it, which is often written as ## h_i/h_o=s_i/s_o ##. I'm using ## b ## for ## s_o ##, and ## m ## for ## s_i ##, etc.)
This intermediate image is then viewed with the eyepiece lens. The final image forms at infinity, and it can be shown that the angular spread ## \Delta \theta_m=h'/f_{eyepiece} ## for the final image which results from viewing something (the image from the objective lens) that sits on the focal plane of the eyepiece lens.

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if the distance from the object to the objective lens is .2 cm. How does 1/s+1/s'=1/f?
1/.2+1/22.2=1/.2 ?

You can compute the distance ## b ## from ##1/b+1/m=1/f ##, to get ## b=fm/(m-f)=.20(22.2)/22 =.202 \approx .20 ##, as mentioned above. (See first sentence of post 7). ## \\ ## In general, if you want to get a large real image from a convex lens, (the real image will be at a fairly large distance on the opposite side of the lens), you place the object just outside the focal point of the lens. ## \\ ## Note: The compound microscope can take a little effort to get things well focused=it is necessary to get the real image from the objective lens to be near the focal point of the eyepiece lens, etc. When you turn the dial on the microscope, it moves the microscope up and down, and it can be somewhat fussy how far the objective lens is from the object, as we can see from this calculation. In any case, using ## b=.20 ## is good enough to compute the number for the magnification ## M ## that we need. ## \\ ## It should also be noted that while the objective lens makes a real image near the focal point of the eyepiece lens, the eyepiece lens works like a simple magnifying glass to further magnify the real image from the objective lens.

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thank you
I'm very busy and confused, can you tell me the answer?

or get me very close to the answer, cus I'm not getting it and i don't have time to get it.
thank you

The homework helpers are not allowed to give out answers according to PF rules. The homework help is intended for students who are willing to put in the necessary time to work through the problems and learn something in the process. Meanwhile, with all the equations I have provided, I think I probably already gave out more than a homework helper should. It should be a simple matter to compute the answer from what I already gave. The students are supposed to show some effort.

scottdave, jedishrfu and BvU
Sadly we cannot provide the answer to you. Our mission is to help students with problems in STEM courses by providing hints provided we see some attempt to solve it. We find that our longtime students appreciate this approach.

Perhaps if look for resources at your school or in your book or on Khan Academy you will find a faster way to understanding your problem.

## 1. What is the purpose of compound optics in microscopes?

Compound optics in microscopes are used to magnify small objects and allow for better visualization of their details. They also help to improve the resolution and clarity of the image.

## 2. How do compound optics work in microscopes?

Compound optics in microscopes work by using a combination of lenses to magnify the image of the object being viewed. The objective lens collects and focuses the light from the object, and the eyepiece lens further magnifies the image for the viewer.

## 3. What are the main components of compound optics in microscopes?

The main components of compound optics in microscopes include the objective lens, eyepiece lens, and the body tube which holds these lenses in place. Some microscopes may also have additional lenses, such as a condenser lens, to improve the quality of the image.

## 4. How does the quality of compound optics affect the performance of a microscope?

The quality of compound optics is crucial in determining the performance of a microscope. High-quality lenses with precise curvature and coatings can greatly improve the resolution and clarity of the image, while lower quality lenses may result in a distorted or blurry image.

## 5. Are there any common problems with compound optics in microscopes?

One common problem with compound optics in microscopes is chromatic aberration, which occurs when different colors of light are focused at slightly different points, resulting in a blurry image. This can be corrected by using achromatic or apochromatic lenses. Another issue is spherical aberration, which causes the edges of the image to appear blurry and can be reduced by using aspheric lenses.

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