The focal lengths of the objective and eye`-` lens of a microscope are `1cm` and `5 cm` respectively. If the magnifying power for the relaxed eye is `45`, then the length of the tube is

Options

A. `6 cm`
B. `9 cm`
C. `12 cm`
D. `15 cm`

Answer

D

Explanation

Magnifying power for relaxed eye is
`m=(v_(o))/(u_(o))(D)/(f_(e))`
Here, `M=-45,` `f_(o)=1 cm`
`f_(e)=5 cm,` `D=25 cm`
`:. -45 =-(v_(o))/(u_(o)). (25)/(5),` or `(v_(o))/(u_(o))=9,`
For objective image is real. Therefore,
`u_(o)=v_(o)`
`u_(o)=-(v_(o))/(9),f_(o)=+1cm`
Substituting in `(1)/(v_(o))-(1)/(u_(o))=(1)/(f_(o)),`
we get `(1)/(v_(o))+(9)/(v_(o))=1` or`v_(o)=10cm`
`:.` Length of tube
`L=v_(o)+f_(e)=10+5=15cm`