हिंदी
कर्नाटक बोर्ड पी.यू.सी.पीयूसी विज्ञान कक्षा ११

A Compound Microscope Consists of an Objective of Focal Length 1 Cm and an Eyepiece of Focal Length 5 Cm. - Physics

Advertisements
Advertisements

प्रश्न

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?

संक्षेप में उत्तर
Advertisements

उत्तर

For the compound microscope, we have:
Focal length of the objective, f0 =1.0 cm Focal length of the eyepiece, fe = 5 cm Distance of the object from the objective, u0 =0.5 cm

Distance of the image from the eyepiece, ve =30 cm

The lens formula for the objective lens is given by

`1/v_0 -1/u_0 = 1/f_0`

`=> 1/v_0 +1/0.5 =1/1`

`=>1/v_0 =1 -10/5 =-1`

⇒ v0= -1 cm

The objective will form a virtual image at the side same as that of the object at a distance of 1 cm from the objective lens. The image formed by the objective will act as a virtual object for the eyepiece.  

The lens formula for the eyepiece is given by

`1/v_e -1/u_e =1/f_e`

`=> 1/30 -1/u_e =1/5`

`=> -1/u_e =1/5 -1/30 =(6-1)/30 =1/6`

⇒ ue = -6 cm 

∴ Separation between the objective and the eyepiece = 6 − 1 = 5 cm

 

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 19: Optical Instruments - Exercise [पृष्ठ ४३२]

APPEARS IN

एचसी वर्मा Concepts of Physics Vol. 1 [English] Class 11 and 12
अध्याय 19 Optical Instruments
Exercise | Q 10 | पृष्ठ ४३२

संबंधित 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.


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.


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?


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

Does the magnifying power of a microscope depend on the colour of the light used? Justify your answer.


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


A compound microscope forms an inverted image of an object. In which of the following cases it it likely to create difficulties? 


Consider the following two statements :-

(A) Line spectra contain information about atoms.

(B) Band spectra contain information about molecules.


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. 


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.


A microscope is focussed on a mark on a piece of paper and then a slab of glass of thickness 3 cm and refractive index 1.5 is placed over the mark. How should the microscope be moved to get the mark in focus again?


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 ______.


A thin converging lens of focal length 5cm is used as a simple microscope. Calculate its magnifying power when image formed lies at:

  1. Infinity.
  2. Least distance of distinct vision (D = 25 cm).

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:


What is meant by a microscope in normal use?


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×