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प्रश्न
Why must both the objective and the eyepiece of a compound microscope have short focal lengths?
उत्तर
The objective of the magnification of a microscope is `v_o/|u_o| = 1/((|u_o|/f_o - 1))`. It is clear from this that to increase this magnification, the value of |uo| should be slightly more than fo. But a microscope is used for nearby objects that are kept near the objective.
Hence, for these objects the value of |uo| is less, hence the value of fo has to be kept even less.
The magnification of the eyepiece is `(1 + "D"/"f"_"e")`; hence, it is clear that to increase it, the value of fe is kept less.
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संबंधित प्रश्न
A giant refracting telescope has an objective lens of focal length 15 m. If an eye piece of focal length 1.0 cm is used, what is the angular magnification of the telescope ?
A compound microscope consists of an objective lens of focal length 2.0 cm and an eyepiece of focal length 6.25 cm separated by a distance of 15 cm. How far from the objective should an object be placed in order to obtain the final image at
- the least distance of distinct vision (25 cm), and
- infinity?
What is the magnifying power of the microscope in each case?
A person with a normal near point (25 cm) using a compound microscope with the objective of focal length 8.0 mm and an eyepiece of focal length 2.5 cm can bring an object placed at 9.0 mm from the objective in sharp focus. What is the separation between the two lenses? Calculate the magnifying power of the microscope.
You are given the following three lenses. Which two lenses will you use as an eyepiece and as an objective to construct a compound microscope?
| Lenses | Power (D) | Aperture (cm) |
| L1 | 3 | 8 |
| L2 | 6 | 1 |
| L3 | 10 | 1 |
Define the magnifying power of a compound microscope when the final image is formed at infinity. Why must both the objective and the eyepiece of a compound microscope has short focal lengths? Explain.
How can the resolving power of a compound microscope be increased? Use relevant formula to support your answer.
Draw a ray diagram to show the working of a compound microscope. Deduce an expression for the total magnification when the final image is formed at the near point.
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, estimate the magnifying power of the microscope.
Draw the labelled ray diagram for the formation of image by a compound microscope.
Derive the expression for the total magnification of a compound microscope. Explain why both the objective and the eyepiece of a compound microscope must have short focal lengths.
In which of the following the final image is erect?
(a) Simple microscope
(b) Compound microscope
(c) Astronomical telescope
(d) Galilean telescope
A compound microscope forms an inverted image of an object. In which of the following cases it it likely to create difficulties?
Consider the following two statements :-
(A) Line spectra contain information about atoms.
(B) Band spectra contain information about molecules.
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
A compound microscope consists of an objective of focal length 1 cm and an eyepiece of focal length 5 cm. An object is placed at a distance of 0.5 cm from the objective. What should be the separation between the lenses so that the microscope projects an inverted real image of the object on a screen 30 cm behind the eyepiece?
A lady uses + 1.5 D glasses to have normal vision from 25 cm onwards. She uses a 20 D lens as a simple microscope to see an object. Find the maximum magnifying power if she uses the microscope (a) together with her glass (b) without the glass. Do the answers suggest that an object can be more clearly seen through a microscope without using the correcting glasses?
compound microscope consists of two convex lenses of focal length 2 cm and 5 cm. When an object is kept at a distance of 2.1 cm from the objective, a virtual and magnified image is fonned 25 cm from the eye piece. Calculate the magnifying power of the microscope.
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 thin converging lens of focal length 5cm is used as a simple microscope. Calculate its magnifying power when image formed lies at:
- Infinity.
- Least distance of distinct vision (D = 25 cm).
The near vision of an average person is 25 cm. To view an object with an angular magnification of 10, what should be the power of the 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. |
Which of the following is not correct in the context of a compound microscope?
What is meant by a microscope in normal use?
