Optics biconvex and plano convex What is the resultant focal length If a biconvex lens is cut into half and then the resulting 2 plano-convex lens arranged in such a way that the plane surface of one faces the curved surface of the other plano-convex lens considering the focal length of the plano-convex lens to be f.
 A: If the cut is perpendicular to the axis, then the combined focus length of the 2 (co-axial) lens will be the same as the original lens, whichever way you line them up. 
A: Let f' be the focal length of the biconvex lens initially & let 'f' be the focal length of the two plano convex lens each.
Then, by Gullstrand's Equation,
P'=P1+P2-P1P2d
or, in terms of focal length,
$$1/feq=1/f1+1/f2-d/f1f2$$ (for separated lens) 
or, $$1/feq=1/f1+1/f2-d/nf1f2$$ (for thick lens)
Here, P' is the power of the biconvex lens initially and P1 and P2 are the powers of the two plano convex lens.
Then, according to the question, the planoconvex lens are arranged so that the curved surface of one touches the plane surface of the other. Therefore, the effective distance between them is zero.
So, P'=P1+P2.   (Since d=0)
Since the plano convex lens are of equal focal length, they also have equal power. 
=> 1/f' = 1/f + 1/f
=> 1/f' = 2/f
Therefore, f=2f'
which implies that, the focal length of the resultant plano-convex lens is twice the focal length of the biconvex lens.

The Gullstrand's Equation used here can be easily derived from the Lens' Maker's formula also.
Using the Lens-Maker's equation:
$$ \frac{1}{f} = \frac{n_{lens} - n_o}{n_o}\left(\frac{1}{R_{left}}-\frac{1}{R_{right}}\right)$$
And because we have an equiconvex lens, $R_{right}=-R_{left}$ (the negative sign is from the sign convention of the equation), so we have:
$$ \frac{1}{f_{equi}} = \frac{n_{lens} - n_o}{n_o}\left(\frac{1}{R_{left}}+\frac{1}{R_{left}}\right) = 2\frac{n_{lens} - n_o}{n_o}\frac{1}{R_{left}}.$$
So $f_{equi}=\frac{R_{left}n_o}{2(n_{lens} - n_o)}$ and for the planoconvex lens, $R_{right} \approx \infty$ so the equation becomes:
$$ \frac{1}{f_{plano}} = \frac{n_{lens} - n_o}{n_o}\left(\frac{1}{R_{left}}+\frac{1}{\infty}\right) = \frac{n_{lens} - n_o}{n_o}\frac{1}{R_{left}}.$$
So when you cut the lens in half, the focal length is actually doubled when you cut the lens, since $f_{plano}=\frac{R_{left}n_o}{n_{lens} - n_o}=2 f_{equi}$.
For more info on the Lens-Maker's equation, you can look here for an explanation.
