Question

When two thin lenses are kept in contact, prove that their combined or effective focal length F is given by $\frac{1}{F}=\frac{1}{f_1}+\frac{1}{f_2}$ where the terms have their usual meaning.

Answer 1

Let two thin lenses $L_1$ and $L_2$ of focal lengths $f_1$ and $f_2$ be placed in contact so as to have a common principal axis. It is required to find the effective focal length of this combination. Let $O$ be a point object on the principal axis. The refraction's through the two lenses are considered separately and the results are combined. While dealing with the individual lenses, the distances are to be measured from the respective optic centers: since the lenses are thin, these distances can also be measured from the center of the lens system (point of contact in the case of two lenses). Let u be the distance of $O$ from the center of the lens system. Assuming that the lens $L_1$ alone produces the refraction. Let the image be formed at $I$ at a distance $v$. Writing the lens equation in this case. we get

$\frac{1}{u}+\frac{1}{v^{\prime }}=\frac{1}{f_1}$ ...(i)

the image $l^{\prime }$ due to the first lens acts on the virtual object for the second lens. Let the final image be formed at $l$, at a distance $v$ from the center of the lens system. Writing the lens equation in this case, we get,

$\frac{1}{-v^{\prime }}+\frac{1}{v}=\frac{1}{f_2}$ ...(ii)

Adding equations (i) and (ii) we get

$\frac{1}{u}+\frac{1}{v^{\prime }}-\frac{1}{v^{\prime }}+\frac{1}{v}=\frac{1}{f_1}+\frac{1}{f_2}$

i.e. $\frac{1}{u}+\frac{1}{v}=\frac{1}{f_1}+\frac{1}{f_2}$ ...(iii)

Let the two lenses be replaced by a single lens which can produce the same effect as the two lenses put together produce, i.e., for an object $O$ placed at a distance $u$ from it, the image $i$ must be formed at a distance $v$. such a lens is called an equivalent lens and its focal length is called the equivalent focal length.

Writing the lens equation in this case, we get

$\frac{1}{u}+\frac{1}{v}=\frac{1}{f}$ ...(iv)

Comparing equations (iii) and (iv) we get

$\frac{1}{f}=\frac{1}{f_1}+\frac{1}{f_2}$

Hence, when thin lenses are combined, the reciprocal of the ir effective focal length will be equal to the sum of the reciprocals of the individual focal lengths.