Advertisements
Advertisements
प्रश्न
The total magnification produced by a compound microscope is 20. The magnification produced by the eye piece is 5. The microscope is focussed on a certain object. The distance between the objective and eyepiece is observed to be 14 cm. If least distance of distinct vision is 20 cm, calculate the focal length of the objective and the eye piece.
Advertisements
उत्तर
For the least distance of clear vision, the total magnification is given by:
`m=-L/`
where, L is the separation between the eyepiece and the objective
fo is the focal length of the objective
fe is the focal length of the eyepiece
D is the least distance for clear vision
Also, the given magnification for the eyepiece:
`m_e=5=(1+D/f_e)`
`=>5=1+20/f_e`
⇒fe= 5 cm
Substituting the value of m and me in equation (1), we get:
m=mo.me
`=>m_o=m/m_e=20/5=4`
Now, we have:
`m_0=L/(|f_0|)`
`=>f_0=14/4=3.5 "cm"`
APPEARS IN
संबंधित प्रश्न
Explain the basic differences between the construction and working of a telescope and a microscope
If this telescope is used to view the moon, what is the diameter of the image of the moon formed by the objective lens ? the diameter of the moon is 3.48 × 106 m and the radius of lunar orbit is 3.8 × 108m.
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?
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?
How can the resolving power of a compound microscope be increased? Use relevant formula to support your answer.
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.
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 using a single lens often shows coloured image of a white source. Why?
| 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:
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.
