Question

If we need a magnification of $375$ from a compound microscope of tube length $150 \,mm$ and an objective of focal length $5\, mm$, the focal length of the eye-piece, should be close to :

• A. $22\, mm$
• B. $12 \,mm$
• C. $33\, mm$
• D. $2 \,mm$

Answer 1

Answer: (A)

$22\, mm$

Answer 2

Given $M =375, L =150 \,mm , f _{0}=5 \, mm , fe =?, d =25 \, cm .$
$M$ for a compound microscope is given by
$M =\frac{ v _{0}}{+ u _{0}}\left(1+\frac{ d }{ fe }\right)$
focal length of objective lens is small $u_{0} \simeq f_{0}$.
Alos as focal length of eye is small $v_{0} \simeq L$
$\therefore M =\frac{ L }{+ f _{0}}\left(1+\frac{ d }{ f _{ e }}\right)$
$\Rightarrow 375=\frac{150}{5}\left(1+\frac{250}{ f _{ e }}\right)$
$\Rightarrow \frac{375}{30}=\left(1+\frac{250}{ f _{ e }}\right)$
$\Rightarrow \frac{375}{30}-1=\frac{250}{ f _{ e }}$
$\Rightarrow \frac{345}{30}=\frac{250}{ f _{ e }}$
$\Rightarrow f _{ e }=2.2\, cm$
$F_e = 22\,mm$