Derive the equation for thin lens and obtain its magnification .

Explanation

(i) Consider an object OO' of height `h_1` placed on the principal axis with its height perpendicular to the principal axis. (ii) The ray OP passing through the pole of the lens goes undeviated. The inverted real image Il' formed has a height `h_2`. (iii) 'The lateral or transverse magnification m is defined as the ratio of the height of the mage to that of the object.
` m = (II')/(OO') " " ...(1)`
From the two similar triangles `trianglePOO'` and `triangle PII' `, we can write,
` (II')/(OO') = (PI)/(PO) " " ....(2)`
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`
(iv) 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.
(v) 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)`