What refractive index should be used while calculating the focal length of lens using lensmaker's formula?

Question

Suppose a lens is half dipped in liquid. Then should the refractive index of liquid with respect to air should be taken. Or should refractive index of glass wih respect to liquid be taken. Or perhaps something else.

Answer 1

In this case, there is no single focal length. There are front and rear focal lengths, as well as the effective focal length. You can calculate refractive powers first:

$\Phi_1 = \frac{n_\text{lens} - n_1}{R_1}$

$\Phi_2 = \frac{n_2 - n_\text{lens}}{R_2}$

$\Phi = \Phi_1 + \Phi_2$

Then front and rear focal lengths:

$f_F = \frac{n_1}{\Phi}$

$f_R = \frac{n_2}{\Phi}$

In this situation, the thin lens equation is:

$\frac{n_1}{f_F} = \frac{n_1}{s_1} + \frac{n_2}{s_2} = \frac{n_2}{f_R}$

Effective focal length can be used as a single number representing both front and rear ones:

$f_E = \frac{1}{\Phi}$

$\frac{n_1}{f_F} = \frac{n_2}{f_R} = \frac{1}{f_E}$

All expressions taken from the free chapter of Physics of Digital Photography by Andy Rowlands.