An equiconvex lens of focal length F is cut in to two equal parts along the vertical axis. The focal length of each part will be

Using the lens maker's formula for the initial equiconvex lens:

$\dfrac{1}{F} = (\mu - 1) \left(\dfrac{1}{R} - \dfrac{1}{-R}\right) = \dfrac{2(\mu - 1)}{R}$

When the lens is cut into two equal parts along the vertical axis, each part becomes a plano-convex lens with radii of curvature $R$ and $\infty$.

The focal length $F'$ of each part is given by:

$\dfrac{1}{F'} = (\mu - 1) \left(\dfrac{1}{R} - \dfrac{1}{\infty}\right) = \dfrac{\mu - 1}{R}$

Comparing the two equations:

$\dfrac{1}{F'} = \dfrac{1}{2} \left( \dfrac{2(\mu - 1)}{R} \right) = \dfrac{1}{2F}$

$F' = 2F$

Answer: $2F$