Figure shows a biconvex lens of having radius of curvatures as R1 and R2 , in the ray depicted shouldnt the meeting point of the rays be at 2F and the object should be placed at -2F , where F is the focal length from the lens maker formula for thin lenses ? The figure depicted -2F1 going to 2F1 this is wrong isnt ? It should be F in place of F1 ,where 1/F= 1/F1 +1/F2 isnt ? And is the F1 and F2 called the focal point of lenses having one radius of curvature as infinity ?is that the definition of first and secondary focus(other than the parallel rays meeting one)? nu

  • $\begingroup$ Depending on the context (where is it from ?) it might serve to illustrate aberration experienced by off-center rays. $\endgroup$
    – tobalt
    Commented Dec 26, 2021 at 22:02
  • $\begingroup$ I think so otherwise it should have been at 2F2 only $\endgroup$
    – Orion_Pax
    Commented Dec 27, 2021 at 4:53

1 Answer 1


F$_1$ and F$_2$ are the principal foci. They are the same distance, f, (the focal length) either side of the optical centre of the lens.

I can therefore make no sense of your equation "1/F= 1/F1 +1/F2".

The rays on the diagram are roughly correct. An object in a plane at distance 2f from the optical centre of the lens should form an image on the plane at a distance 2f from the optical centre but to the other side of the lens. You can show this by putting $u=2f$ into the equation $$\frac 1u + \frac 1v = \frac 1f\ \ \ \ \ \ \ \text{[real-is-positive convention]}$$ or, equivalently, by putting $u=-2f$ into the equation $$\frac 1u + \frac 1f = \frac 1v\ \ \ \ \ \ \ \text{[cartesian convention]}$$

I assume that 2F$_1$ and 2F$_2$ are supposed to be points on the lens axis a distance 2f from the optical centre of the lens, on either side of the lens. 2F$_2$ seems to be marked too close to the lens.

When I first looked at the diagram I was misled by the labels C$_1$ and C$_2$, expecting them to stand for the centres of curvature of the two faces of the lens. But these will not be at distance 2f from the centre of the lens, as on the diagram. [It would require a refractive index of 2 for the lens material, as you can show from the lens-makers' formula.] Clearly C$_1$ and C$_2$ have nothing to do with centres of curvature!

  • $\begingroup$ Sir i clearly understand now what you mean , regarding the final note u said: 1/f = (2-1)(1/R1 +1/R2) (all terms positive) , we arrived at R1R2/(R1+R2) = f , now how u showed this to be 2f = R1 and 2f=R2 ? $\endgroup$
    – Orion_Pax
    Commented Dec 27, 2021 at 4:41
  • 1
    $\begingroup$ Sorry. I'm assuming that $R_1 = R_2 = R$ (say), so $R=2f$. $\endgroup$ Commented Dec 27, 2021 at 8:50
  • $\begingroup$ I see Sir so in general we can say foy any lens having radii of curvature R1,R2 , F1O =OF2 and 2F1O =2OF2 is not equal to R1 and R2 right? $\endgroup$
    – Orion_Pax
    Commented Dec 27, 2021 at 13:07
  • 1
    $\begingroup$ Yes. The principal foci are the same distance either side of the optical centre, even if the lens surfaces have different curvatures. It would be a co-incidence if either centre of curvature were twice the focal length from the optical centre, and if the curvatures were equal, the centres of curvature would certainly not be at this distance, as it would need the lens material to have too high a refractive index. As far as I know, with the exception of diamond, transparent materials have refractive indices less than 2. $\endgroup$ Commented Dec 27, 2021 at 13:17

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