Question

Open in App

Updated on : 2022-09-05

Solution

Verified by Toppr

Correct option is B)

Given,

Separation between lense,$L=21.7cm,$

Focal length of eye piece, $L=21.7cm,f_{e}=2.5cm$

Focal length of objective lens,$f_{0}=1.6cm,v_{0}=19.2cm,u_{0}=?$

$L=v_{0}+f_{e}$

Where $L$ is the length of the microscope and $f_{e}$ e the focal length of the objective

Substituting we get:

$21.7=v_{0}+2.5$

$βv_{0}=21.7β2.5=19.2cm$

According to lens formula we know that:

$f_{0}1β=v_{0}1ββu_{0}1β$

$v_{0}=19.2cm,u_{0}=?$

$1.61β=19.21ββu_{0}1β$

$βu_{0}1β=1.61ββ19.21β$

$βu_{0}1β=1610ββ19210β$

$βu_{0}1β=192120β10β=192110β$

$u_{0}=β110192βcm=β1.75cm$

Magnification, $m=uβvβ=1.75β19.2β=10.97=11$

Hence, linear magnification is $11$Β

Solve any question of Ray Optics and Optical Instrument with:-

Was this answer helpful?

0

0