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Question
Derive the equation for thin lens and obtain its magnification.
Solution
- Let us consider an object OO′ of height h1 placed on the principal axis with its height perpendicular to the principal axis.
- The ray OP passing through the pole of the lens goes undefeated.
- The inverted real image II’ formed has a height h2.
- 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'")`
- From the two similar triangles ∆ POO’ and PII’, we can write,
`("II'")/("OO'") = "PI"/"PO"`
Applying sign convention,
`(-"h"_2)/"h"_1 = "v"/(- "u")` - Substituting this in equation 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.
- 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"`
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