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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.

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  • $\begingroup$ Take a look at Fresnel's formulas. $\endgroup$
    – my2cts
    Jul 22, 2018 at 9:17
  • $\begingroup$ By "half dipped", do you mean that the axis of the lens is parallel to the surface of the liquid, or perpendicular to the surface of the liquid? $\endgroup$
    – S. McGrew
    Jul 22, 2018 at 12:57
  • $\begingroup$ Axis is perp to liquid layer. $\endgroup$
    – theDude823
    Jul 22, 2018 at 13:50

1 Answer 1

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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.

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  • $\begingroup$ Note: all of this assumes thin lens. The quoted book deals with thick lens as well. $\endgroup$ Jul 23, 2018 at 8:03

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