In case of thin lenses in contact the effective power of the combination is given as: $P=P_1+P_2+P_3$..... For a lens - mirror in contact, like a lens with silvered surface (where lens and mirror are seperated by virtually $0$ distance) can we say that the effective power is the sum of power of mirror and lens.If so how?


1 Answer 1


For a silvered lens, the power formula would be $P_{eqv.}=P_{lens}+P_{mirror}+P_{lens}$.
(the lens power is doubled as rays travel through it twice).

eg. for a biconvex lens ($\eta,R$) with one side mirrored:
1. $f_{lens}=\frac{R}{2(\eta-1)}$
2. $f_{mirror}=R/2$
3. $P_{eqv.}=\frac{2(2\eta-1)}{R}$
4. in other words $P_{eqv.}=(2+\frac{1}{k})P_{lens}$ where $k=f_m/f_l$

The power formula doesn't seem to care what the underlying optical instrument is as long as they are thin and close enough for linearity to hold. When mirrors are involved appropriate multiplicities account for multiple passage of light rays.

  • $\begingroup$ How can we prove $P_{eqv}=P_{lens}+P_{mirror}+P_{lens}$. I was able to prove $P=P_1+P_2+P_3$ for combination of only lenses but unable to prove for combination of lens and mirror $\endgroup$
    – Param_1729
    Feb 10, 2020 at 16:50
  • $\begingroup$ @Param_1729 its just linear optical instruments in series $\endgroup$
    – lineage
    Feb 11, 2020 at 13:16
  • $\begingroup$ its linear optical instruments in series but mirror formula is different from lens formula. Consider two lens of focal length $f_1 and f_2 $ if u is the coordinate of the object then $\frac{1}{v_{1}} = \frac{1}{u}+\frac{1}{f_1} $now$ v_1 $will act as object for second lens therefore $\frac{1}{v_{final}} = \frac{1}{v_1}+\frac{1}{f_2}$ by using these two equation we get $\frac{1}{f_{eqv}} = \frac{1}{f_1}+\frac{1}{f_2} = \frac{1}{v_{final}}-\frac{1}{u}$ and in the similar way we can prove formula of $P_{eqv}$for combination of more than two lenses but i am unable to prove it for lense + mirror $\endgroup$
    – Param_1729
    Feb 11, 2020 at 14:27

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