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प्रश्न
Draw a labelled ray diagram showing the formation of image by a compound microscope in normal adjustment. Derive the expression for its magnifying power.
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उत्तर
A simple microscope has a limited maximum magnification (≤ 9) for realistic focal lengths. For much larger magnifications, one uses two lenses, one compounding the effect of the other. This is known as a compound microscope. A schematic diagram of a compound microscope is shown in Fig. The lens nearest the object, called the objective, forms a real, inverted, magnified image of the object. This serves as the object for the second lens, the eyepiece, which functions essentially like a simple microscope or magnifier, produces the final image, which is enlarged and virtual. The first inverted image is thus near (at or within) the focal plane of the eyepiece, at a distance appropriate for final image formation at infinity, or a little closer for image formation at the near point. Clearly, the final image is inverted with respect to the original object. We now obtain the magnification due to a compound microscope. The ray diagram of Figure shows that the (linear) magnification due to the objective, namely h'/h, equals
`"m"_o = "h'"/"h"= "L"/f_o`
where we have used the result
`tan beta = ("h"/f_o ) = ("h'"/"L")`
Here h′ is the size of the first image, the object size being h and fo being the focal length of the objective. The first image is formed near the focal point of the eyepiece. The distance L, i.e., the distance between the second focal point of the objective and the first focal point of the eyepiece (focal length fe) is called the tube length of the compound microscope. As the first inverted image is near the focal point of the eyepiece, we use the result for the simple microscope to obtain the (angular) magnification me due to it equation m = `(1 + "D"/f)`, when the final image is formed at the near point, is
`"m"_"e" = (1 + "D"/f_"e")`
When the final image is formed at infinity, the angular magnification due to the eyepiece is
`"m"_"e"` = (D/fe)
Thus, the total magnification, when the image is formed at infinity, is
m = `"m"_o"m"_"e" = ("L"/f_o)("D"/f_e)`
Clearly, to achieve a large magnification of a small object (hence the name microscope), the objective and eyepiece should have small focal lengths.
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संबंधित प्रश्न
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An object is placed at a distance u from a simple microscope of focal length f. The angular magnification obtained depends
In which of the following the final image is erect?
(a) Simple microscope
(b) Compound microscope
(c) Astronomical telescope
(d) Galilean telescope
Can the image formed by a simple microscope be projected on a screen without using any additional lens or mirror?
The separation between the objective and the eyepiece of a compound microscope can be adjusted between 9.8 cm to 11.8 cm. If the focal lengths of the objective and the eyepiece are 1.0 cm and 6 cm respectively, find the range of the magnifying power if the image is always needed at 24 cm from the eye
How does the resolving power of a microscope change when
(i) the diameter of the objective lens is decreased?
(ii) the wavelength of the incident light is increased ?
Justify your answer in each case.
| A compound microscope consists of two converging lenses. One of them, of smaller aperture and smaller focal length, is called objective and the other of slightly larger aperture and slightly larger focal length is called eye-piece. Both lenses are fitted in a tube with an arrangement to vary the distance between them. A tiny object is placed in front of the objective at a distance slightly greater than its focal length. The objective produces the image of the object which acts as an object for the eye-piece. The eye-piece, in turn, produces the final magnified image. |
The focal lengths of the objective and eye-piece of a compound microscope are 1.2 cm and 3.0 cm respectively. The object is placed at a distance of 1.25 cm from the objective. If the final image is formed at infinity, the magnifying power of the microscope would be:
In a compound microscope an object is placed at a distance of 1.5 cm from the objective of focal length 1.25 cm. If the eye-piece has a focal length of 5 cm and the final image is formed at the near point, find the magnifying power of the microscope.
