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| 1. |
A compound microscope uses an objective lens of focal length 4 cm and eyepiece lens of focal length 10 cm. An object is placed at 6 cm from the objective lens. Calculate the magnifying power of the compound microscope. Also calculate the length of the microscope |
| Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> : It is given that focal length of objective lens `f_(0)= + 4 cm`focal length of eyepiece lens `f_e = +10 cm` and `u_(0) = - 6 cm` <br/> Let the final image is formed at the least distance of distinct vision (D = 25 cm). <br/> For object lens `1/v_(0) -1/u_(0) =1/f_(0) rArr 1/v_(0) + 1/f_(0) = -1/6 + 1/4 =(-2+3)/12 = 1/12` <br/> `therefore v_(0) = + 12 cm` <br/> Again for eyepiece lens `v_e = - D = - 25 cm`, <a href="https://interviewquestions.tuteehub.com/tag/hence-484344" style="font-weight:bold;" target="_blank" title="Click to know more about HENCE">HENCE</a> <br/> `1/v_(<a href="https://interviewquestions.tuteehub.com/tag/e-444102" style="font-weight:bold;" target="_blank" title="Click to know more about E">E</a>) -1/u_(e) =1/f_(e) rArr 1/u_(e) -1/_(f_(e)) =1/(-25) -1/(10) = (-2-5)/50 = -7/50` <br/> or `u_(e) = -50/7 =-7.4 cm` <br/> Length of the microscope tube `L = |v_(0)| + |u_(e)| = 12 + 7.4 = 19.4 cm` <br/> and magnifying the power `m=m_(0) xx m_(e) =(v_(0)/u_(0)) xx(v_(e)/v_(f)) =12/(-6) xx (-25)/(-7.4) = -7`</body></html> | |