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प्रश्न
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.
उत्तर
Show L in figure
`tan beta = h/(f_0) .... (1)`
`tan beta = h'/L ..... (2)`
`h'/h =Lf_0` [Using (1) and (2)]
`m_0 = h'/h = L/f_0` [Magnification due to objective]
`m_e = 1+D/f_e`
Net magnification (m) =`m_0m_e`
`|m = (L/f_0) (D/f_e)|`
f0 and fe are in denominator.
This formula contains foand fe in denominator. Therefore, both the objective and the eyepiece of a compound microscope must have short focal lengths.
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संबंधित प्रश्न
Draw a labelled ray diagram showing the formation of a final image by a compound microscope at least distance of distinct vision
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 viewing through a compound microscope, our eyes should be positioned not on the eyepiece but a short distance away from it for best viewing. Why? How much should be that short distance between the eye and eyepiece?
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 |
How can the resolving power of a compound microscope be increased? Use relevant formula to support your answer.
The focal length of the objective of a compound microscope if fo and its distance from the eyepiece is L. The object is placed at a distance u from the objective. For proper working of the instrument,
(a) L < u
(b) L > u
(c) fo < L < 2fo
(d) L > 2fo
The magnifying power of a converging lens used as a simple microscope is `(1+D/f).` A compound microscope is a combination of two such converging lenses. Why don't we have magnifying power `(1+D/f_0)(1+D/f_0)`?In other words, why can the objective not be treated as a simple microscope but the eyepiece can?
An eye can distinguish between two points of an object if they are separated by more than 0.22 mm when the object is placed at 25 cm from the eye. The object is now seen by a compound microscope having a 20 D objective and 10 D eyepiece separated by a distance of 20 cm. The final image is formed at 25 cm from the eye. What is the minimum separation between two points of the object which can now be distinguished?
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).
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.
