# Is focal length of a planoconvex lens twice to the focal length of equiconvex lens from which the planoconvex lens was made?

If an equiconvex lens is divided equally, we will obtain a planoconvex lens. Let assume focal length of our equiconvex lens is 'f'. I would like to know what will be the focal length of planoconvex lens that we obtained after dividing equiconvex lense. Will it be twice of 'f'? Or will it be 0.5 of 'f'? How to determine it?

• Just think about the refraction and the resulting deviation of light at each face of the two types of lenses. – Farcher Aug 25 '16 at 14:22

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 $f_{plano}=\frac{R_{left}n_o}{n_{lens} - n_o}=2 f_{equi}$.