Question

A person with normal near point 25 cm using a compound microscope with objective of focal length 8.0 mm and an eye piece of focal length 2.5 cm can bring an object placed at 9.0 mm from the objective in sharp focus. The separation between two lenses and magnification respectively are:

• A. 9.47 cm, 88
• B. 3.36 cm, 44
• C. 6.00 cm, 22
• D. 7.49 cm, 11

Answer 1

Answer: (A)

9.47 cm, 88

Answer 2

: Here, $d = 25cm,f_{o} = 8.0mm,f_{e} = 2.5cm,$

$u_{o} = - 9.0mm = - 0.9cm$

Now,$\frac{1}{v_{e}} - \frac{1}{u_{e}} = \frac{1}{f_{e}}\therefore\frac{1}{u_{e}} = \frac{1}{v_{e}} - \frac{1}{f_{e}} = \frac{1}{- 25} - \frac{1}{2.5}$

$= \frac{- 1 - 10}{25} = \frac{- 11}{25}(\because v_{e} = - d = - 25cm)$

$u_{e} = \frac{- 25}{11} = 2.27cm$

Again, $\frac{1}{v_{o}} - \frac{1}{u_{o}} = \frac{1}{f_{o}}$

$\frac{1}{v} = \frac{1}{f_{o}} + \frac{1}{u_{o}} = \frac{1}{0.8} + \frac{1}{- 0.9} = \frac{0.9 - 0.8}{0.72} = \frac{0.1}{0.72}$

$v_{o} = \frac{0.72}{0.1} = 7.2cm$

Therefore, separation between two lenses

$= u_{e} + v_{o} = 2.27 + 7.2 = 9.47cm$

Magnifying power,,$m = \frac{v_{o}}{u_{o}}\left( 1 + \frac{d}{f_{e}} \right) = \frac{7.2}{0.9}\left( 1 + \frac{25}{2.5} \right) = 88$