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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.
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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Magnifying power of a simple microscope is inversely proportional to the focal length of the lens. What then stops us from using a convex lens of smaller and smaller focal length and achieving greater and greater magnifying power?
When are two objects just resolved? Explain.
How can the resolving power of a compound microscope be increased? Use relevant formula to support your answer.
Draw a ray diagram showing the image formation by a compound microscope. Hence obtained expression for total magnification when the image is formed at infinity.
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?
| 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. |
Which of the following is not correct in the context of a compound microscope?
| 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. |
A compound microscope consists of an objective of 10X and an eye-piece of 20X. The magnification due to the microscope would be:
