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A microscope ishaving objective of focal length 1 cm and eyepiece of focal length 6 cm of tube length is 30 cm and image is formed at the least distance of distinct vision, what is the magnification produced by the microscope. Take D = 25 cm. |
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Answer» 6 `m = (L)/(f_(0)) (1 + (D)/(f_(E)))` (where length of tube L = 30 cm, focal length of objective lens `f_(0) = 1 cm`, focal length of eye - PIECE `f_(e) = 6 cm, D = 25 cm`) `=(30)/(1)(1+(25)/(6))=30xx(6+25)/(6)` `= 5 xx 31` = ` 155 cm cong 150`. |
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