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Question
When two thin lenses are kept in contact, prove that their combined or effective focal length F is
given by :
`1/F = 1/f_1 + 1/f_2`
where the terms have their usual meaning.
Solution
1) The image is formed in two steps :
In the first step, the lens L1 produces the image I' of the object A. It is a real image at a distance v' from L1
Then for the lens L1 , we have,
`1/(v') + 1/(-u) = 1/f_1`
`:. 1/(v') - 1/u = 1/f_1` ... (1)
2) In the second step, the image I' acts as an object for L2 and the final image I is formed. The image I' is not observed, hence it acts a virtual object for lens L2 it is assumed to be kept on left, we can write for L2.
`1/v + 1/(-v') = 1/f_2`
`:. 1/v - 1/(v') = 1/f_2` ...(2)
3) Adding equation (1) and (2), we get
`1/(v') - 1/u + 1/v - 1/(v') = 1/f_1 + 1/f_2`
`:. 1/v - 1/u = 1/f_1 + 1/f_2` ....(3)
4) If a single lens of focal length f of equivalent lens is used for object A and corresponding image I is formed then we have,
`1/v + /(-u) = /1/f` ...(4)
5) From equations (3) and (4),
`1/f = 1/f_1 + 1/f_2` ...(5)
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