Two concave lenses L_1 and L_2 are placed in contact with one...

When two concave lenses are placed in contact, they touch at their edges, leaving a gap in the middle that is thicker at the center and thinner at the edges. This gap has the shape of a biconvex lens.

Let the powers of the two concave lenses be $P_1$ and $P_2$. Since they are concave, their powers are negative ($P_1 < 0$, $P_2 < 0$).

The initial equivalent power of the combination (with air in the gap) is:
$P_i = P_1 + P_2$

When the gap is filled with a substance of refractive index $\mu > 1$, it acts as a convex lens. Let the power of this liquid lens be $P_3$. Since it is a convex lens, its power is positive ($P_3 > 0$).

The new equivalent power of the combination is:
$P_f = P_1 + P_2 + P_3$

Since $P_1 + P_2$ is negative and $P_3$ is positive, the addition of the positive power reduces the overall magnitude of the negative power of the combination. Thus, $|P_f| < |P_i|$.

The focal length of a lens combination is the reciprocal of its equivalent power ($F = \dfrac{1}{P}$).

Since the magnitude of the equivalent power decreases, the magnitude of the equivalent focal length increases.