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Question
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.
Solution
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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| 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. |
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| 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. |
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