Advertisements
Advertisements
Question
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?
Solution
For the compound microscope, we have:
Power of the objective lens = 20 D
The focal length of the objective lens is given by
`f_0 = 1/(20D) = 0.05 text/"m"= 5 text/ "cm"`
Power of the eyepiece = 10 D
The focal length of the eyepiece is given by
`f_e =1/(10D) = 0.1 text/ "m" = 10 text/" cm"`
Least distance for clear vision, D = 25 cm,
To distinguish the two points having minimum separation, the magnifying power should be maximum.
For the eyepiece, we have:
Image distance, ve= -25 cm
Focal length, Fe = 10 cm
The lens formula is given by
`1/u_e = 1/v_e -1/f_e`
= `1/-25 -1/10`
=`(-2-5)/50=-7/50`
Separation between the objective and the eyepiece = 20 cm
So, the image distance for the objective lens (v0) can be obtained as:
`v_0 =20 -50/7 =90/7 text/"cm"`
The lens formula for the objective lens is given by
`1/u_0 =1/v_0-1/f_0`
=`7/90 -1/5 =(7-18)/90 = -11/90`
`=> u_0 =-90/11text/" cm"`
Magnification of the compound microscope:
`m = v_0/u_0 (1+D/f_e)`
=`-(90/7)/(-90/11) (1+25/10)`
=`11/7 × (3.5) = 5.5`
`therefore `Minimum separation that the eye can distinguish =`0.22/5.5 text/"mm" =0.04 text/ "mm"`
APPEARS IN
RELATED QUESTIONS
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.
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.
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.
An object is placed at a distance u from a simple microscope of focal length f. The angular magnification obtained depends
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
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.
Can the image formed by a simple microscope be projected on a screen without using any additional lens or mirror?
A simple microscope using a single lens often shows coloured image of a white source. Why?
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.
Draw a labelled ray diagram showing the formation of image by a compound microscope in normal adjustment. Derive the expression for its magnifying power.
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.
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. |
A compound microscope consists of an objective of 10X and an eye-piece of 20X. The magnification due to the microscope would be:
The focal lengths of the objective and the eye-piece of a compound microscope are 1.0 cm and 2.5 cm respectively. Find the tube length of the microscope for obtaining a magnification of 300.
