InterviewSolution
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InterviewSolution
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The total magnification produced by a compound microscope is 20. The magnification produced by the eye piece is 5. The microscope is focussed on a certain object. The distance between the objective and eye piece is observed to be 14 cm. If least distance of distinct vision is 20 cm, calculate the focal length of objective and eye piece. |
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Answer» SOLUTION :Here, `m=-20 , m_e=5 , v_e=-20` cm For EYEPIECE , `m_e=v_e/mu_e` `rArr 5=(-20)/mu_e rArr =(-20)/5`=-4 cm Using lens formula, `1/v_e-1/u_e=1/f_e` `-1/20+1/4=1/f_e` `(-1+5)/20=1/(f_e) rArr f_e=5cm` Now , TOTAL MAGNIFICATION `m=m_exxm_0` `-20=5xxm_0` `m_0=1-v_0/f_0` `-4=1-10/f_0` `-5=10/(f_0) rArr f_0` = 2cm |
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