Derive the expression for effective focal length of two thin lenses kept in contact.

Explanation

Correct ray diagram with Indication of arrows}
Refraction through first lens in
the absence of second, equation `(1)/(v_(1))-(1)/(u)=(1)/(f_(1))`}
Refraction through second lens
In the absence of first, equation `(1)/(v)-(1)/(v_(1))=(1)/(f_(2))` }
Adding equation `(1)/(v)-(1)/(u)=(1)/(f_(1))+(1)/(f_(2).)(1)/(v)-(1)/(u)=(1)/(f)` }
Arriving `(1)/(f)=(1)/(f_(1))+(1)/(f_(2))`.

Consider two thin lenses of focal lengths `f_(1) and f_(2)` respectively placed in contact with each other.
Let .O. be the point object placed on the principle axis of the lense.
If second lens is not present than the first lens forms an image `I_(1)` of the object .O. at a distance `v_(1)` from it.
`therefore -(1)/(u)+(1)/(v_(1))=(1)/(f_(1))` . . (1) Since second lens is in contact with the first, So `I_(1)` acts as an object to the second lens which forms the Image I at a distance v from it
`therefore -(1)/(v_(2))+(1)/(v)=(1)/(f_(2))` . . (2)
adding (1) and (2) we get
`-(1)/(u)+(1)/(v)=(1)/(f_(1))+(1)/(f_(2))`
Or `-(1)/(u)+(1)/(v)=(1)/(f)`
where `(1)/(f)=(1)/(f_(1))+(1)/(f_(2))`
Thus, the two thin lenses in contact behave as a single lens of focal length f. this single lens is known as equivalent lens and its focal length f is called equivalent focal length.