Question

In a compound microscope, the focal lengths of two lenses are 1.5cm 6.25 cm , If an object is placed at 2 cm from objective and the final image is formed at 25 cm from eye lens, the distance between the two lenses is

• A. 6.00 cm
• B. 7.75 cm
• C. 9.25 cm
• D. 11.0 cm

Answer 1

Answer: (D)

11.0 cm

Answer 2

: Here, $f_{o} = 1.5cm,f_{e} = 6.25cm,u_{o} = - 2cm$

$v_{e} = - 25m$

For objective,$\frac{1}{v_{o}} - \frac{1}{u_{o}} = \frac{1}{f_{o}}\therefore\frac{1}{v_{o}} - \frac{1}{- 2} = \frac{1}{1.5}$

$\frac{1}{v_{o}} = \frac{1}{1.5} - \frac{1}{2}orv_{o} = 6cm$

For eye piece

$\frac{1}{v_{e}} - \frac{1}{u_{e}} = \frac{1}{f_{e}}$

$\frac{1}{- 25} - \frac{1}{u_{e}} = \frac{1}{6.25}$

$- \frac{1}{u_{e}} = \frac{1}{6.25} + \frac{1}{25}oru_{e} = - 5cm$

Distance between two lenses

$= \left| v_{o} \right| + \left| u_{e} \right| = 6cm + 5cm = 11cm$