Draw a ray diagram to show the formation of the image of an object placed between the optical centre and focus of a convex lens. Deduce the relationship between object distance, image disatnce and focal length of lens.

Explanation

When object is placed between optical centre and principal focus.
Consider a convex lens of focal length f and optical centre C. Let an object AB be placed perpendicular to the principal axis at a point between C and F. The virtual image of this object A.B. is formed on the same side of the lens. The image is virtual, erect and is behind the object as shown in fig.

`Delta` ABC and `Delta` A.B.C are similar, so
`(A.B.)/(AB)=(CB.)/(CB)` ...(1)
`Delta` DCF and `Delta` A.B.F are similar, so
`(A.B.)/(DC)=(B.F.)/(CF)` ...(2)
But DC = AB
`:. (A.B.)/(AB)=(B.F)/(CF)`
From Eqs. (1) & (2), we have
`(CB.)/(CB)=(B.F)/(CF)`
or `(CB.)/(CB)=(CB.+CF)/(CF)` ...(3)
Using sign conventions,
Let CB. = -v, CB = -u, CF = f
So Eq. (3) becomes
`(-v)/(-u)=(-v+f)/(f)`
or vf= - uv + uf
Dividing both sides by uvf, we get
`(vf)/(uvf)=(-uv)/(uvf)+(uf)/(uvf)`
or `(1)/(u)= -(1)/(f)+(1)/(v)`
or `(1)/(f)=-(1)/(u)+(1)/(v)`
or `-(1)/(u)+(1)/(v)=(1)/(f)`
This is lens formula.