Look through a clear glass or plastic bottle and describe what you see. Now fill the bottle with water and describe what you see. Use the water bottle as a lens to produce the image of a bright object and estimate the focal length of the water bottle lens. How is the focal length a function of the depth of water in the bottle?
Section summary
The microscope is a multiple-element system having more than a single lens or mirror.
Many optical devices contain more than a single lens or mirror. These are analysed by considering each element sequentially. The image formed by the first is the object for the second, and so on. The same ray tracing and thin lens techniques apply to each lens element.
The overall magnification of a multiple-element system is the product of the magnifications of its individual elements. For a two-element system with an objective and an eyepiece, this is
$m={m}_{\text{o}}{m}_{\text{e}}\text{,}$
where
${m}_{\text{o}}$ is the magnification of the objective and
${m}_{\text{e}}$ is the magnification of the eyepiece, such as for a microscope.
Microscopes are instruments for allowing us to see detail we would not be able to see with the unaided eye and consist of a range of components.
The eyepiece and objective contribute to the magnification. The numerical aperture
$(\text{NA})$ of an objective is given by
where
$n$ is the refractive index and
$\alpha $ the angle of acceptance.
Immersion techniques are often used to improve the light gathering ability of microscopes. The specimen is illuminated by transmitted, scattered or reflected light though a condenser.
The
$f/\#$ describes the light gathering ability of a lens. It is given by
$f/\#=\frac{f}{D}\approx \frac{1}{2\mathrm{NA}}.$
Conceptual questions
Geometric optics describes the interaction of light with macroscopic objects. Why, then, is it correct to use geometric optics to analyse a microscope’s image?
The image produced by the microscope in
[link] cannot be projected. Could extra lenses or mirrors project it? Explain.
Why not have the objective of a microscope form a case 2 image with a large magnification? (Hint: Consider the location of that image and the difficulty that would pose for using the eyepiece as a magnifier.)
What advantages do oil immersion objectives offer?
How does the
$\text{NA}$ of a microscope compare with the
$\text{NA}$ of an optical fiber?
Problem exercises
A microscope with an overall magnification of 800 has an objective that magnifies by 200. (a) What is the magnification of the eyepiece? (b) If there are two other objectives that can be used, having magnifications of 100 and 400, what other total magnifications are possible?
(a) 4.00
(b) 1600
(a) What magnification is produced by a 0.150 cm focal length microscope objective that is 0.155 cm from the object being viewed? (b) What is the overall magnification if an
$\mathrm{8\times}$ eyepiece (one that produces a magnification of 8.00) is used?
(a) Where does an object need to be placed relative to a microscope for its 0.500 cm focal length objective to produce a magnification of
$\mathrm{\u2013400}$ ? (b) Where should the 5.00 cm focal length eyepiece be placed to produce a further fourfold (4.00) magnification?
(a) 0.501 cm
(b) Eyepiece should be 204 cm behind the objective lens.
You switch from a
$1.40\text{NA}\phantom{\rule{0.25em}{0ex}}\text{60}\times $ oil immersion objective to a
$1.40\text{NA}\phantom{\rule{0.25em}{0ex}}\text{60}\times $ oil immersion objective. What are the acceptance angles for each? Compare and comment on the values. Which would you use first to locate the target area on your specimen?
An amoeba is 0.305 cm away from the 0.300 cm focal length objective lens of a microscope. (a) Where is the image formed by the objective lens? (b) What is this image’s magnification? (c) An eyepiece with a 2.00 cm focal length is placed 20.0 cm from the objective. Where is the final image? (d) What magnification is produced by the eyepiece? (e) What is the overall magnification? (See
[link] .)
(a) +18.3 cm (on the eyepiece side of the objective lens)
(b) -60.0
(c) -11.3 cm (on the objective side of the eyepiece)
(d) +6.67
(e) -400
You are using a standard microscope with a
$0.10\text{NA}\phantom{\rule{0.25em}{0ex}}\text{4}\times $ objective and switch to a
$0.65\text{NA}\phantom{\rule{0.25em}{0ex}}\text{40}\times $ objective. What are the acceptance angles for each? Compare and comment on the values. Which would you use first to locate the target area on of your specimen? (See
[link] .)
Unreasonable Results
Your friends show you an image through a microscope. They tell you that the microscope has an objective with a 0.500 cm focal length and an eyepiece with a 5.00 cm focal length. The resulting overall magnification is 250,000. Are these viable values for a microscope?
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