Obtain the equation for lateral magnification for thin lens.

Explanation

(i) Consider an object OO' of height `h_1` placed on the principal axis 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`
(ii) The lateral or transverse magnification m is defined as the ratio of the height of the image to that of the object.
` m = (II')/(OO') " " ...(1)`
(iii) From the two similar triangles `DeltaPOO`' and `Delta PII'` we can write,
` (II')/(OO') = (PI)/(PO)`
Applying sign convention,
` (-h_2)/(h_1) = (v)/(-u)`
(iv) Substituting this in the equation (1) for magnification,
`m = (-h_2)/(h_1) = (v)/(-u)`
After rearranging,
` m = (h_2)/(h_1) = (v)/(-u)`
(v) 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 less than one.
(vi) The equations for magnification by combining the lens equation with the formula for magnification as,
` m = (h_2)/(h_1) = (f)/(f+ u) " or " m = h_2/h_1 = (f - v)/(f)`