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(a) Draw a ray diagram to show the working of a compound microscope. Deduce an expression for the total magnification when the final image is formed at the near point. (b) 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 eyepiece has a focal length of 5 cm and the final image is formed at the near point, estimate the magnifying power of the microscope |
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Answer» Solution :(a) A ray diagram showing the WORKING of a compound microscope has been shown in Fig. 9.90. Total ANGULAR magnification (Magnifying POWER) of a compound microscope is defined as the ratio of the angle subtended at the eye by the final image to the angle subtended at the eye by the object as viewed directly by placing it at near point of eye. In figure `C_2B" = D`= least distance of distinct vision. Let us imagine the object AB to be shifted to `A_(1)B..`so that it is also at a distance D from the eye.  If `angleAC_(2)B.. = beta` and `angleA_(1)C_(2)B.. =alpha,` then by definition `m=beta/alpha = (tan beta)/(tanalpha)` In `triangleA..B..C_(2), tan beta =(A.B.)/(C_(2)B.)` and in `triangleA_(1)B..C_(2), tan alpha =(A_(1)B.)/(C_(2)B.)` `therefore m_(e) xx m_(0)` Applying lens formula for eyelens, we have `therefore m = m_(0) xx m_(e) =-v_(0)/u_(0)(1+D/f_(e))` As a first approximation `v_(0)=L` , distance between the OBJECTIVE and eye lens or the length of the microscope tube and`u_(0)=f_(0)`, the focal length of objective lens. Hence, `m=-L/f_(0)(1+D/f_(e))` Therefore, to increase the magnifying power of a compound microscope, focal lengths of both objective as well as eye lens should be as small as possible. (b) As per question `u_0 = 1.5 cm, f_(0) = 1.25 cm, f_e = 5 cm` and final image is formed at near point of eye i.e., at least distance of distinct vision D = 25 cm. `1/v_(0) =1/1.25-1/1.5 = 1/7.5` or `v_(0) = 7.5 cm` `therefore` magnifying power `|m| = v_(0)/u_(0) (1+D/f_(e)) = 7.5/1.5(1+25/5) = 30` |
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