Suppose there is an equi-convex lens made of glass which has a focal length ($f$) of 30cm. Then, can we not say that the radius of curvature, $R$ of the lens is twice the focal length, i.e. $R = 60cm$?

Why do we need to use the Lens Maker's Formula for the same, which in fact gives a different result :

$$ \frac{1}{f}= \frac{\mu_2 - \mu_1}{\mu_1}\left[\frac{1}{R_1}-\frac{1}{R_2}\right] $$

For $f=30 cm$, $\mu_1 = 1$, $\mu_2 = 1.5$, $R_1 = R$ and $R_2 = -R$, we get :

$$\frac{1}{30}= 0.5\times\frac{2}{R}$$ Or, $R=30cm$

How can one explain this? Also, how can the focal length be equal to the radius of curvature?

  • $\begingroup$ I'm voting to close this question as off-topic because of insufficient prior research effort. (OP: google what is the relation between radius of curvature and focal lenght, for example) $\endgroup$ Jan 10, 2016 at 15:39
  • 2
    $\begingroup$ @AccidentalFourierTransform: OP has found the formula, applied it correctly, and finds the result counterintuitive. I think it is a fine question. $\endgroup$ Jan 10, 2016 at 15:53
  • $\begingroup$ @RossMillikan that's why flags work by voting: if you desagree with me, its perfectly fine. Upvote the question. If I'm the only one who thinks that it should be closed, it won't. (I still think the question shows insufficient prior research; if OP had researched more, it wouldn't be that counterintuitive). $\endgroup$ Jan 10, 2016 at 16:02
  • $\begingroup$ @AccidentalFourierTransform I did Google exactly what you've suggested before posting the question here. I did not get the answer as to why we cannot use the 2R method instead of the Lens Maker's Formula method. $\endgroup$
    – agdhruv
    Jan 10, 2016 at 18:29
  • 1
    $\begingroup$ @AccidentalFourierTransform Except insufficient prior research is a reason for downvoting, not for closure. $\endgroup$
    – user10851
    Jan 10, 2016 at 20:52

1 Answer 1


Intuitively, the radius of curvature has to depend on the index of refraction of the glass. If the index were $1$, the lens would have no effect at all. If the index were very high, say $10$, it would not take much curvature to get a given f'ocal length. Clearly we cannot just say the radius of curvature is twice the focal length.

You have applied the lens maker's formula correctly to your problem. There is no problem with $R=f$, in fact that is always true for $\mu_2=1.5\mu_1$. We are assuming a thin lens in this formula, so the diameter of the lens must be small compared with $R$. If the focal length is $30$ cm and the diameter of the lens is $1$ cm the thickness is twice the height of a circular segment. Given $R=30, c=1$ we have $h=R-\sqrt{R^2-(\frac c4)^2}=30-\sqrt{900-\frac 14}\approx 0.004$ so the lens is about $1$ mm thick

  • $\begingroup$ Is f=R/2 always true for a mirror, then? $\endgroup$
    – agdhruv
    Jan 10, 2016 at 17:07
  • $\begingroup$ Yes. That is a purely geometric property of the surface. As the light does not pass through the mirror, the focal length does not depend on the index of refraction. $\endgroup$ Jan 10, 2016 at 17:44

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