Derive the expression for the power of two thin lenses placed in contact with each other.

Explanation

Let `f_(1)` and `f_(2)` are the focal length of two thin convex lenses placed in contact with each other. Let a point object is placed at the point O on their common optic axis. The first lens of focal length `f_(1)` will form the image at `I_(1)`. Therefore according to lens formula

`1/(f_(1))=1/v-1/u`
Now `v=CI_(1)=` Distance of image
`u=-CO=` Distance of object
Putting these values in the above relation
`1/(f_(1))=1/(CI_(1))-1/(-CO)`.......(1)
Due ot refractive at the second lens, the final image will be formed at I i.e. for the second lens of focal length `f_(2)` the image `I_(1)` will act as object whose image will be formed at I.
Again according to lens formula `1/(f_(2))=1/(v_(1))-1/(u_(1))`
here `v_(1)=` distance of image `=CI`
`u_(1)=` distance of object `=CI_(1)`
Putting these values in the above relation, we get
`1/(f_(2))=1/(CI)-1/(CI_(1))`............2
Adding both sides of eqns (1) and (2) we get
`1/(f_(2))+1/(f_(2))=1/(CI_(1))-I/(-CO)+1/(CI)-1/(CI_(1))=1/(CI)-1/(-CO)`.............3
If f is the total focal length of combination then
`1/f=1/v-1/u`.............4
where `v=CI=` distance of image
`u=-CO=` distance of object
Putting these values in eqn (4) we get
`1/f=1/(CI-1/(-CO)`.........5
Comparing eqns (3) and (5) we get
`1/f=1/(f_(1))+1/(f_(2))`
Hence reciprocal of the total focal length is equal to the sum of reciprocals of focal length of two lens forming the combination.