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InterviewSolution
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A compound microscope has a magnifying power of 100 when the image is formed at infinity. The objective has a focal length of0.5 cm and the tube length is 6.5 cm. What isthe focal length of the eyepiece. |
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Answer» Solution :GIVEN data : Magnifying power of compound microscope = m = 100 The focal LENGTH of objective `f_0 = 0.5 cm` (or) ` 5 xx 10^(-3) m` Tube length = 6.5 cm To find :The focal length of eye PIECE `f_e` = ? Since the image is formed at infinity, the real image produced by the objective lens should lie on the focus of the eye piece. So ` v_0 + f_e = 6.5 = 6.5cm "" ....(1)`  Formula : Magnifying power = `(v_0)/(u_0) xx (D)/(F_e)`(for normal adjustment) ` rArrm = - [ 1 - (v_0)/(f_0) ] (D)/(F_e) [ therefore (v_0)/(u_0) =1 - (v_0)/( f_0)]` ``rArr 100= - [ 1 - (v_0)/(0.5) ] xx (25)/(f_e) ` (Taking D = 25 cm) ` rArr 100 f_e = - 1 ( 1-2 v_0) xx 25` ` rArr 2v_0 -4f_e = 1 "" ...(2)` Solving (1) & (2) ` v_0 =4.5 cm ` ` f_e = 2 cm ` The focal length of eye piece = 2 cm |
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