Obtain the equation for lateral magnification for thin lens.

Explanation

Let us consider an object O O. of heigh `h_(1)` placed on the principal aixs with its height perpendicular to the principal axis. The ray OP passing through the pole of the lens goes undeviated. The inverted real image II. formed has a height `h_(2)`.
`m = (II.)/(OO.)`
From the two similar triangles `Delta POO. and Delta PII.` we can write
`(H.)/(OO.)=(PI)/(PO)`
Applying sign convention,
`(-h_(2))/(h_(1))=(v)/(-u)`
Substituting this in the equation (1) for magnification,
`m=(-h_(2))/(h_(1))=(v)/(-u)`
After rearranging.
`m=(h_(2))/(h_(1))=(v)/(-u)`
The magnification is negative for real image and positive for virtual image. In the case of a concave lens, the magnification is always positive and lens than one. We can also have the equation for magnification by combining the lens equation with the formula for magnificaiton as,
`m=(h_(2))/(h_(1))=(f)/(f+u)(or)m=(h_(2))/(h_(1))=(f-v)/(f)`