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InterviewSolution
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A person with a normal near point `(25 cm)` using a compound microscope with an objective of focal length `8.0 mm` and eye piece of focal length `2.5 cm` can bring an object placed `9.0 cm` from the objective in sharp focus. What is the separation between the two lenses ? Calculate the magnifying power of the microscope ? |
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Answer» Here, `d = 25 cm, f_0 = 8.0 mm = 0.8 cm, f_e = 2.5 cm, u_0 = -9.0 mm = -0.9 cm` As `(1)/(v_e)-(1)/(u_e)=(1)/(f_e) :. (1)/(u_e)=(1)/(v_e)-(1)/(f_e)` `(v_e = - d = -25 cm)` =`(1)/(-25) - (1)/(2.5) = (-1 - 10)/(25) = -(11)/(25)` `u_e = -(25)/(11) = -2.27 cm` Again `(1)/(v_0)-(1)/(u_0) = (1)/(f_0)` `(1)/(v_0)=(1)/(f_0)+(1)/(u_0)=(1)/(0.8)+(1)/(-0.9)=(0.9 - 0.8)/(0.72) = (0.1)/(0.72)` `v_0 = (0.72)/(0.1) = 7.2 cm` `:.` Separation between the two lenses `=|u_e|+ v_0 = 2.27 + 7.2 = 9.47 cm` Magnifying power `= (v_0)/(|u_0|)(1 + (d)/(f_e)) = (7.2)/(0.9)(1 + (25)/(2.5)) = 88`. |
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