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The separation between the objective and the eyepiece of a compound microscope can be adjusted between 9.8 cm to 11.8 cm. If the focal lengths of the objective and the eyepiece are 1.0 cm and 6 cm respectively, find the range of the magnifying power if the image is always needed at 24 cm from the eye.A. `31.16`B. `32.16`C. `33.16`D. `34.16` |
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Answer» Correct Answer - A For eye-piece, it is given that `f_(e )=6 cm, v_(e )=-24 cm` `therefore " " (1)/(-24)-(1)/(-u_(e ))=(1)/(6)` `therefore " " u_(e )=4.8 cm` When, L = 9.8 cm, then `v_(o)=(9.8-4.8)cm = 5.0 cm` `therefore " " (1)/(5.0)-(1)/(-u_(o))=(1)/(1.0)rArr u_(o)=1.25 cm` or Magnifying power, `M = (v_(o))/(u_(o))(D)/(u_(e ))=((5.0)/(1.25))((25.0)/(4.8))=20.83` When, L = 11.8 cm, then `v_(o)=(11.8-4.8)cm = 7.0 cm` `therefore " " (1)/(7.0)-(1)/(-u_(o))=(1)/(1.0)` or `u_(o)=1.17 cm` `therefore " " M=(v_(o))/(u_(o)).(D)/(f_(e ))=((7.0)/(1.17))((25.0)/(4.8))=31.16` Therefore, range of magnifying power is from 20.83 to 31.16. |
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