Two thin lenses of focal lengths ${\mathit{f}}_{\mathbf{1}}$and${\mathit{f}}_{\mathbf{2}}$ are in contact and share the same central axis. Show that, in image formation, they are equivalent to a single thin lens for which the focal length is $\mathit{f}\mathbf{}\mathbf{=}\frac{{\mathbf{f}}_{\mathbf{1}}{\mathbf{f}}_{\mathbf{2}}}{\mathbf{(}{\mathbf{f}}_{\mathbf{1}}\mathbf{+}\mathbf{}{\mathbf{f}}_{\mathbf{2}}\mathbf{)}}$.

Two thin lenses of focal lengths $f_1\text{and}{f}_2$are in contact and share the same central axis.

Two thin lenses are in contact with each other with their focal length having the same central axis will form an image. The final image is produced by two lenses, with the image of the first lens being the object for the second. To know the image formation is equivalent to the single thin lens, we will use the concept images from the thin lenses.

Formula:

The lens formula,

$\frac{\mathbf{1}}{\mathit{f}}\mathbf{=}\frac{\mathbf{1}}{\mathit{p}}\mathbf{+}\frac{\mathbf{1}}{\mathit{i}}$ ...(i)

The focal length of the first lens can be calculated by usingequation (i) as follows:

$\frac{1}{f_1}=\frac{1}{p_1}+\frac{1}{i_1}$

Where, $p_1$is the object distance, $i_1$is the image distance.

Since$p_1=\infty$, the image is at focal point from the above equation,$i_1=f_1$for the first lens.

The focal length applied to the second lens can be calculated using equation (i) as follows:

$\frac{1}{f_2}=\frac{1}{p_2}+\frac{1}{i_2}$ ...(a)

If the thickness of the lenses can be ignored, the object distance for second lens is $p_2=-i_1\left(or{f}_1\right)$. The negative sign indicates that the image formed by the first lens is beyond the second lens. It means the second image is virtual and the object distance is negative. Thus, the thin lens equation in terms of focal lengths can be written as (for getting the focal length of the combined lens, we replace $i_2=f$in equation (a)):

$$\begin{gathered} -\frac{1}{f_1}+\frac{1}{f}=\frac{1}{f_2} \\ \frac{1}{f}=\frac{1}{f_1}+\frac{1}{f_2} \\ \frac{1}{f}=\frac{f_2+f_1}{f_1{f}_2} \\ f=\frac{f_1{f}_2}{f_1+f_2} \end{gathered}$$

Hence, the required focal length is $\frac{f_1{f}_2}{f_1+f_2}$.