Advertisements
Advertisements
प्रश्न
Why must both the objective and the eyepiece of a compound microscope have short focal lengths?
Advertisements
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.
APPEARS IN
संबंधित प्रश्न
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.
Draw a ray diagram showing image formation in a compound microscope ?
Suggest two ways by which the resolving power of a microscope can be increased?
An object is placed at a distance u from a simple microscope of focal length f. The angular magnification obtained depends
An object is to be seen through a simple microscope of focal length 12 cm. Where should the object be placed so as to produce maximum angular magnification? The least distance for clear vision is 25 cm.
A simple microscope is rated 5 X for a normal relaxed eye. What will be its magnifying power for a relaxed farsighted eye whose near point is 40 cm?
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?
Draw a neat labelled ray diagram showing the formation of an image at the least distance of distinct vision D by a simple microscope. When the final image is at D, derive an expression for its magnifying power at D.
Define the magnifying power of a microscope in terms of visual angle.
What is the advantage of a compound microscope over a simple 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.
On increasing the focal length of the objective, the magnifying power ______.
An angular magnification of 30X is desired using an objective of focal length 1.25 cm and an eye piece of focal length 5 cm. How will you set up the compound microscope for the final image formed at least distance of distinct vision?
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. |
The magnification due to a compound microscope does not depend upon ______.
