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The focal lengths of the objective and the eye-piece of a compound microscope are 2.0 cm and 3.0 cm respectively. The distance between the objective and the eye-piece is 15.0 cm. The final image formed by the eye-piece is at infinity. The two lenses are thin. The distances in cm of the object and the image produced by the objective measured from the objective lens are respectively

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a

2.4 and 15.0

b

2.0 and 3.0

c

2.0 and 12.0

d

2.4 and 12.0

answer is A.

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Detailed Solution

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When final image is formed at infinity then Length of the tube

$\begin{array}{l} = v_{o} + f_{o} = v_{o} + f_{o} \Rightarrow 15 = v_{o} + 3 \Rightarrow v_{o} = 12 cm \\ For objective lens \frac{1}{f} = \frac{1}{v_{o}} - \frac{1}{u_{o}} \Rightarrow \frac{1}{(+ 2)} = \frac{1}{(+ 12)} - \frac{1}{u_{o}} \Rightarrow u_{o} = - 2 .4 cm \end{array}$

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The focal lengths of the objective and the eye-piece of a compound microscope are 2.0 cm and 3.0 cm respectively. The distance between the objective and the eye-piece is 15.0 cm. The final image formed by the eye-piece is at infinity. The two lenses are thin. The distances in cm of the object and the image produced by the objective measured from the objective lens are respectively