By giving sign-conventions, derive the lens formula relating object distance, image distance and focal length for a thin convex lens. Draw a ray diagram to show the formation of image of an object placed between optical centre and focus of a convex lens.

Explanation

New cartesian sign conventions :
(i) All the distances are measured from the optical centre of the lens.
(ii) All the distances measures in the direction of incident ray are taken as positive and opposite to direction of incident ray are taken as negative.
(iii) All the distances measured vertically upwards from the principal axis are taken as positive and vertically downwards as negative.

Consider a convex lens of focal length f. Let AB an object placed normally on the principal axis of the lens. The ray of light from the object AB after refracting through the convex lens meet at point B. So A.B. is the real image of the object AB.
`Delta.s` ABC and A.B.C are similar
`:.(AB)/(AB)=(CA)/(CA)`....i
Similarly, `Delta.s CDF` and A.B. F are similar
`(CD)/(AB)=(CF)/(FA)`
but `CD=AB:.(AB)/(AB)=(CF)/(FA)`.........ii
From eqn i and ii we have
`(CA)/(CA)=(CF)/(FA)`...........iii
Now `FA.=CA.-CF`
( `:.` all distances are measured from the optical centre of the lens)
`:.` Eqn (iii) becomes `(CA)/(CA)=(CF)/(CA-CF)`...........iv
Applying sign conventions we have
`CA=-u` (Distance of the object from the lens)
`CA.=+v` (Distance of the image from the lens)
`CF=+f` (Focal length of lens)
Hence eq iv becoms
`-u/v=f/(v-f)` or `-uv+uf=uf`
Dividing both sides but uvf, we get
`-1/f+1/v=1/u` or `-1/u+1/v=1/f`........v
Which is the lens formula.