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in Physics by Rising star (572 points)
A compound microscope consists of an objective of focal length 1cm and eye piece of focal length 5cm separated by 12.2cm. (a) At what distance from the objective should an object be placed so that the final image is formed at least distance of distinct vision? (b) Calculate the angular magnification in this case

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1 Answer

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by Expert (1.4k points)

given, f_0=1cm,f_e=5cm  and L = 12.2 cm

so, final image seen at least division of distinct vision (25cm) , 

v_e=-25cm

now using formula, \frac{1}{f_e}=\frac{1}{v_e}-\frac{1}{u_e}

or, 1/5 = 1/-25 - 1/u_e

or, 1/5 + 1/25 = -1/u_e

or, -6/25 = 1/u_e

or, u_e = -25/6 = -4.2 cm

given length of tube , L  = |v_0|+|u_e| = 12.2 cm

or, |v_0|+4.2=12.2\implies |v_0|=8cm

again using lens maker formula,

\frac{1}{f_0}=\frac{1}{v_0}-\frac{1}{u_0}

or, 1/1 = 1/8 - 1/u_0

u_0 = -8/7 = -1.1 cm

now, angular magnification, m = \frac{v_0}{-|u_0|}\left(1+\frac{D}{f_e}\right)

= 8/-(-1.1)[ 1 + 25/5]

= (8/1.1) × 6

= 48/1.1

= 43.6

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