For the derivation of the focal length of a planoconvex lens with one surface silvered, we consider the phenomenon taking place in three steps:

  1. Refraction through first surface
  2. Reflection through silvered surface
  3. Refraction through first surface again

So the inverse of the resultant focal length can be obtained as the summation of twice the inverse of the focal length of the planoconvex lens and the inverse of the focal length of the silvered surface.

My question here is:

Why do we consider the focal length of the planoconvex lens (while taking into consideration the refracting surface) and why not just the focal length of that particular refracting surface?


2 Answers 2


We have to consider the individual refractive surfaces only and not the planoconvex lens as a whole, but it so happens, that the final answer can be simply expressed in terms of the focal length the planoconvex lens would have without the silvering.

For a general derivation, consider a lens with radius ($R_1,R_2$), with $R_2$ silvered. Cartesian sign convention are used here.
For an object distance (from the first surface, but for a thin lens, the distance is same for both the surfaces) $u$, we get for the first surface:- $$\frac{\mu}{v_1}=\frac{\mu-1}{R_1}+\frac{1}{u}$$ (where $\mu$ is the refractive index of lens and that of air is $1$)
Now for surface two, a mirror, $v_1$ is the object:- $$\frac1{v_2}=\frac2{R_2}-\frac{1}{v_1}$$ Now the second image $v_2$ acts as the object for first surface again, but now with inverted sign conventions. $$-\frac 1v=\frac{1-\mu}{-R_1}+\frac{\mu}{v_2}$$
using the value of $v_1$ and $v_2$ from preceding equations you come up with $$\frac 1v+\frac 1u=\frac{2\mu}{R_2}-\frac{2(\mu-1)}{R_1}$$ This is similar to a mirror with focal length given by the left hand side of the equation. It can be rearranged to $$\frac 1f=-\bigg(\frac{2(\mu-1)}{R_1}-\frac{2(\mu-1)}{R_2}\bigg)+\frac 2{R_2}=\frac{-2}{f_{\text{lens}}}+\frac{1}{f_{\text{mirror}}}$$ which is the required equation.


Basically the real net equation for the final focal length of mirror is based on net power:

$$P_{\rm net} \mathrm{(equivalent\ mirror)} = 2P_{\rm lens} + P_{\rm mirror}$$


$$-\frac{1}{f_{\rm eq(mirror)}} = \frac{2}{f_{\rm lens}} + -\frac{1}{f_{\rm mirror}}$$

  • $\begingroup$ What should we do if we have two convex lens touching each other and the Second one is silvered a side ??? $\endgroup$
    – jeevan s
    Feb 18, 2018 at 16:32
  • $\begingroup$ According to the derived formula, it should be (-)2P(lens), how can it be justified? $\endgroup$
    – Harshil
    Oct 8, 2022 at 5:38

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