You are given two converging lenses of focal lengths 1.25 cm and 5cm to design a compaound microscope.if it is desired to have a magnification of 30, find out the separation between the object and the eyepiece.
or
A small telescope has an objective lens of focal length 150 cm and eyepiece of focal length 5 cm. What is the magnifying of the telescope for viewing distant objects on normal adjustment ?
If this telescope is used to view a 100 m tall tower 3 km away, what is the height of the image of the tower formed 3km away, what is the height of the image of the tower formed by the objective lens ?

Explanation

It is given that `f_(e)= 5cm " " f_(o 1.25 cm m= 30 L=? `
Now, `m= L/(f_(0)) (1+ D/f_(e)) Rightarrow 30 = L/1.25 (1+25/5) Rightarrow (Lxx6)/(1.25) Rightarrow L = ( 30 xx 1.25)/6`
Given ` f_(o) = 150`
`f_(e) = 5 cm`
When final image is at infinity , magnifying power,
` M= (-f_(o))/(f_(e)) = ( -150)/(5.0) = -30`
Negative sign shows that the image is inverted.
Let AB be tower and A'B' its image, then
`H/u = (h_(1))/u`
height of image formed by objective
`h_(1) = H/u v`
Here, H= 100 m,
u = 3km = 3000 m
For distant object `v~~ f_(o) = 150 cm = 1.50 m= h_(1) = H/u f_(0) `
` = (100m)/(3000m) xx 1.40 m = 4.7 xx 10^(-2) m = 4.7 cm `