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, $d = 25 \,cm$ $f_0 = 8.0 \,mm, f_e = 2.5\, cm$, $u_0 = -9.0\, mm = -0.9 \,cm$ Now, $\frac{1}{v_{e}}-\frac{1}{u_{e}}=\frac{1}{f_{e}}$ $\therefore \frac{1}{u_{e}} = \frac{1}{v_{e}} -\frac{1}{f_{e}} = \frac{1}{-25}-\frac{1}{2.5}=\frac{-11}{25}$ $\left(\because v_{e} = -d = -25 \,cm\right)$ $u_{e} = \frac{-25}{11}=2.27 \,cm$ Again, $\frac{1}{v_{0}} - \frac{1}{u_{0}} = \frac{1}{f_{0}}$ $\frac{1}{v_{0}} = \frac{1}{f_{0}} +\frac{1}{u_{0}} = \frac{1}{0.8} +\frac{1}{-0.9}$ $\frac{0.9-0.8}{0.72} = \frac{0.1}{0.72}$ $v_{0} = \frac{0.72}{0.1} = 7.2\, cm$ Therefore, separation between two lenses $= u_{e} + v_{0} = 2.27 +7.2 = 9.47 \,cm$ Magnifying power, $m = \frac{v_{0}}{u_{0}}\left(1+\frac{d}{f_{e}}\right)$ $= \frac{7.2}{0.9}\left(1+\frac{25}{2.5}\right) = 88$