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.

For two thin lenses kept in contact with each other, show that:

`1/F = 1/f_1  + 1/f_2`

where the terms have their usual meaning.

Answer 1

(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`    ...(i)

(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 as a virtual object for lens L₂; it is assumed to be kept on the left, we can write for L₂.

`1/v + 1/(-v') = 1/f_2`

∴ `1/v - 1/(v') = 1/f_2`    ...(ii)

(3) Adding equations (i) and (ii), 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`   ....(iii)

(4) If a single lens of focal length (F) of an equivalent lens is used for object A and a corresponding image (I) is formed, then we have,

`1/v + 1/(-u) = 1/F`     ...(iv)

(5) From equations (iii) and (iv),

`1/F = 1/f_1  + 1/f_2`    ...(v)

Answer 2

Combined focal length of two thin lenses in contact, both lenses are convex: Suppose two thin convex lenses L₁ and L₂ of focal lengths f₁ and f₂ are placed in contact in air, having a common principal axis. A point-object O is placed on the principal axis at a distance u from the first lens L₁. Its image would be formed by the lens L₁ alone at I₁ distant v' (say) from L₁. Then, from the lens formula, we have:

`1/(v') - 1/u = 1/f_1`    ...(i)

I' serves as a virtual object for the second lens L₂ which forms a final image I at a distance v (say) from it.

Then, we have:

`1/v - 1/(v') = 1/f_2`    ...(ii)

Adding equations (i) and (ii), we get,

`1/v - 1/u = 1/f_1 + 1/f_2`    ...(iii)

If we replace these two lenses with a single lens which forms the image of an object distant u from it at a distance v, then the focal length F of this equivalent lens would be given by,

`1/v - 1/u = 1/F`    ...(iv)

From equations (iii) and (iv), we get,

`1/F = 1/f_1 + 1/f_2`    ...(v)