Advertisements
Advertisements
प्रश्न
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.
Advertisements
उत्तर
Focal length of the objective lens, fo = 8 mm = 0.8 cm
Focal length of the eyepiece, fe = 2.5 cm
Object distance for the objective lens, uo = −9.0 mm = −0.9 cm
Least distance of distant vision, d = 25 cm
Image distance for the eyepiece, ve = −d = −25 cm
Object distance for the eyepiece = ue
Using the lens formula, we can obtain the value of ue as:
`1/v_e - 1/u_e = 1/f_e`
`1/u_e = 1/v_e - 1/f_e`
`1/u_e = 1/-25 - 1/2.5`
`1/u_e = (-1 - 10)/25`
`1/u_e = (-11)/25`
ue = `-25/11`
∴ ue = −2.27 cm
We can also obtain the value of the image distance for the objective lens (vo) using the lens formula.
`1/v_o = 1/u_o = 1/f_o`
`1/v_o = 1/f_o + 1/u_o`
`1/v_o = 1/0.8 - 1/0.9`
`1/v_o = (0.9 - 0.8)/0.72`
`1/v_o = 0.1/0.72`
∴ vo = 7.2 cm
The distance between the objective lens and the eyepiece = |ue| + vo
= 2.27 + 7.2
= 9.47 cm
The magnifying power of the microscope is calculated as:
`v_o/|u_o| (1 + d/f_e)`
= `7.2/0.9 (1 + 25/2.5)`
= 8 × (1 + 10)
= 88
Hence, the magnifying power of the microscope is 88.
APPEARS IN
संबंधित प्रश्न
Explain the basic differences between the construction and working of a telescope and a microscope
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?
An angular magnification (magnifying power) of 30X is desired using an objective of focal length 1.25 cm and an eyepiece of focal length 5 cm. How will you set up the compound microscope?
Suggest two ways by which the resolving power of a microscope can be increased?
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.
A compound microscope uses an objective lens of focal length 4 cm and eyepiece lens of focal length 10 cm. An object is placed at 6 cm from the objective lens. Calculate the magnifying power of the compound microscope. Also calculate the length of the microscope.
A man is looking at a small object placed at his near point. Without altering the position of his eye or the object, he puts a simple microscope of magnifying power 5 X before his eyes. The angular magnification achieved is
Consider the following two statements :-
(A) Line spectra contain information about atoms.
(B) Band spectra contain information about molecules.
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?
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.
A convex lens of a focal length 5 cm is used as a simple microscope. Where should an object be placed so that the image formed by it lies at the least distance of distinct vision (D = 25 cm)?
In the case of a regular prism, in minimum deviation position, the angle made by the refracted ray (inside the prism) with the normal drawn to the refracting surface is ______.
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?
| 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 focal lengths of the objective and eye-piece of a compound microscope are 1.2 cm and 3.0 cm respectively. The object is placed at a distance of 1.25 cm from the objective. If the final image is formed at infinity, the magnifying power of the microscope would be:
