A real , `4/5` size of the object is formed 18 cm from a lens. Calculate the focal length of the lens.

Explanation

Since the image is real and diminished , the lens must be convex and the object must be placed beyond 2F.
Given : v = + 18 cm
`(h.)/(h) = (-4)/(5) " " [ because m = (h.)/h ` is negative for real image`]`
f = ?
We have, `4u = - 5 xx 18 cm therefore u = -22.5 cm`
Also focal length is given by
`1/f = 1/v - 1/u`
`= 1/(18) - (1)/(-22.5) = 1/(18) + 1/(22.5) = (9)/(90)`
f = 10 cm