In the figure shown here, what is the equivalent focal length of...

The focal length $\left({f}_{L}\right)$ of the left plano-concave lens can be calculated as follows-

$\begin{cases}\dfrac{1}{{f}_{L}} & = & \left(n - 1\right)\left(\dfrac{1}{{R}_{1}} - \dfrac{1}{{R}_{2}}\right) \\ & = & \left(1 . 6 - 1\right)\left(\dfrac{1}{\infty} - \dfrac{1}{20cm}\right) \\ & = & - \dfrac{0 . 6}{20cm} . . . \left(1\right)\end{cases}$

The focal length \left({f}_{M}\right)$of the middle bi-convex lens can be calculated as follows-</p><p>$\begin{cases}\dfrac{1}{{f}_{M}} & = & \left(n - 1\right)\left(\dfrac{1}{{R}_{1}} - \dfrac{1}{{R}_{2}}\right) \\ & = & \left(1 . 5 - 1\right)\left(\dfrac{1}{20cm} - \dfrac{1}{- 20cm}\right) \\ & = & \dfrac{0 . 5}{10cm} . . . \left(2\right)\end{cases}$</p><p>The focal length \left({f}_{R}\right)$ of the right plano-concave lens can be calculated as follows-

$\begin{cases}\dfrac{1}{{f}_{R}} & = & \left(n - 1\right)\left(\dfrac{1}{{R}_{1}} - \dfrac{1}{{R}_{2}}\right) \\ & = & \left(1 . 6 - 1\right)\left(\dfrac{1}{- 20cm} - \dfrac{1}{\infty}\right) \\ & = & - \dfrac{0 . 6}{20cm} . . . \left(3\right)\end{cases}$

Hence, the effective focal length \left(f\right)$of the combination of lenses can be calculated as follows-</p><p>$\begin{cases}\dfrac{1}{f} & = & \dfrac{1}{{f}_{L}} + \dfrac{1}{{f}_{M}} + \dfrac{1}{{f}_{R}} \\ & = & - \dfrac{0 . 6}{20cm} + \dfrac{0 . 5}{10cm} - \dfrac{0 . 6}{20cm} \\ & = & - \dfrac{0 . 1}{10cm} \\ & ⇒ & f = - 100cm\end{cases}