Geometric optics, microscope

In summary, the conversation discusses the use of a microscope with specific components and measurements. The main focus is on determining the distance between the object and objective, the diameter of the observation field, and the overall magnification in the direct observation path. The lens formula is used to calculate the distance between the object and objective, and it is determined that the object is in the object space. The remaining questions involve obtaining more information about the eyepiece L2, such as its magnification and diameter in the observation field. The conversation concludes with a caveat that the speaker may not be an expert, but they have given their best explanation based on their understanding and experience.
  • #1
Grufey
30
0

Homework Statement


First, thanks in advace.

Let us consider a microscope where the objective L1 has f1=20mm and magnification 10x. In the image plane is located a diafragm M with diameter 19mm (see fig). The size of the CCD is 4,8mm (vertical) x 5,6mm (horizontal). 20mm before of the CCD camera a len L3 is introduced in order to obtain the best compatibility of the field in the both paths of the microscope. Assumins that the focal length of the eyepiece L2 is 25 mm, determine:

a) Distance Object-Objective
b) Diameter of the observation field
c) Overal magnification in the direct observation (eye) path
scheme optic system.png

Homework Equations


General geometric optics:
Magnification: m=-y'/y where y and y' are the height of the object and image respectively
Lens formula: 1/s+1/s'=1/f where s and s' are object and image distance respectively and f the focal length

It should not need more formula, I guess...

The Attempt at a Solution



My attempt of solution begins with the lens formula applied to the Len L1

[itex]\frac{1}{s}+\frac{1}{s'}=\frac{1}{f}[/itex] then, multiplying by s', we get, [itex]m=\frac{s'}{s}=\frac{s'-f}{f}[/itex]

Thus substituting m=10,f=20, we get s'=220 mm, now applying again, the lens formula, we get s=-22mm. The minus sign denotes, that the object is in the object space. Ok until here everything is undertood, isn't it?

My problen is the b) and c) part. I think I need more information about L2. How can I get the magnification and the diameter in the observation field?

On the magnification, I think it means size magnification, or it is angular magnification?

Thanks in adavance, any help will be useful
 
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  • #2
(b)
My understanding is that they want to know the diameter of the field on the object table that can be observed through the diaphragm. Not so hard if you did the calculation for the magnification factor object plane - image plane.

(c)
L2 has a focal length of 25 mm. Optical power 40 diopter. If I can normally focus to 25 cm, its magnification is ... (however, one less if you want to look with an unaccommodated eye, i.e. with the ocular focused on the image plane)
For a microscope the total magnification is ...

Caveat: I'm no expert, but I wear glasses and do possesses a microscope. Gave it my best shot, but it's your exercise, so if you don't understand or don't agree then don't take my word for it :)

Re your last question: I would stick to linear magnification. But Wiki (on the same page here ) leans the other way:
By convention, for https://www.physicsforums.com/wiki/Magnifying_glass and optical https://www.physicsforums.com/wiki/Microscope , where the size of the object is a linear dimension and the apparent size is an angle, the magnification is the ratio between the apparent (angular) size as seen in the eyepiece and the angular size of the object when placed at the conventional closest distance of distinct vision: 25 cm from the eye.
 
Last edited by a moderator:

1. What is the difference between geometric optics and physical optics?

Geometric optics deals with the behavior of light as it travels through transparent media, while physical optics focuses on the wave-like properties of light and its interactions with matter.

2. How does a microscope use geometric optics to magnify images?

A microscope uses a combination of lenses and mirrors to focus and magnify an image. The objective lens collects and magnifies the light from the specimen, while the eyepiece lens further magnifies the image for the viewer.

3. What is the difference between a compound microscope and a stereo microscope?

A compound microscope uses two sets of lenses to magnify the image, while a stereo microscope has two separate optical paths for each eye, allowing for a three-dimensional view of the specimen.

4. How does the numerical aperture affect the resolution of a microscope?

Numerical aperture is a measure of a microscope's ability to gather and resolve fine details of a specimen. A higher numerical aperture results in better resolution and a clearer image.

5. Can a microscope be used to view objects that are too small to see with the naked eye?

Yes, a microscope can magnify objects up to 1000 times their original size, allowing for the viewing of objects that are too small to see with the naked eye. This is especially useful in fields such as biology, where microscopic organisms and structures can be studied in detail.

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