Question

A microscope has a net magnification of -200. An object is .001 m from the objective. If the objective has a focal length of 1x10^-4 m, what must the focal length of the eyepiece be?

Answer 1

As we know that the magnification of compound is given by

where m_o is the linear magnification of the objective les

and m_i is the linear magnification of the eyepiece lens

for m_o we want to know v(image distance)

as we know that thin lens equation

Now taking the sign consideration and rearrangin terms

as given f(objective)=m

and object distanceism

Now as object is on the left side of the objective lens it will be taken as negative as according to sign convention

adn focal length is on positive it should be taken as positive according to sign conention

by calculating we get

we get v=

Now as we know that

by calculation we get we get magnification of objective lens

now as previous equation states that

So by calculation we get m_e=1800

Now as we know that image that will be seen will be at 25cm i.e the Distant vision of a eye

it implies that the v_e(image distance for eyepiece)=25cm=

as we know that magnification interms of focal length and image distance is given by

now putting the value of f and v we get the value of magnification to be