According to the Lensmaker's formula,

$$\frac{1}{f} = (μ-1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$$

Apparently, focal length $f$ is inversely proportional to the refractive index of the medium $μ$. Since, Optical Power $P = \frac{1}{f}$. This should imply that optical power is directly proportional to the refractive index. However, today I got told by my teacher that these two are independent. Where have I gone wrong in my reasoning? Is the fact provided to me correct and if yes, how?

  • $\begingroup$ Please talk to your teacher and present your formula. $\endgroup$
    – Semoi
    Commented Mar 6, 2021 at 7:40

1 Answer 1


I don't know what your teacher thought about while saying this, but this is not true. The power of the lens does depend on the refractive index $\mu_{lm}=\mu_l/\mu_m$ of the lens which is clear from the formula, you have given. $$P=\frac{1}{f}=(\mu_{lm}-1)\left( \frac{1}{R_1}-\frac{1}{R_2}\right)$$

  • $\begingroup$ I understand that, another side question though. We know that, for a thin lens, $f=\frac{R}{2}$. The lens is cut off from a bigger sphere with radius $R$. I gather that it is because of the light rays bending earlier in a denser region leading to the point of focus being a bit nearer. Am I right? Do the dimensions of that bigger sphere from which the lens was cut off appear to be apparently differently as well, so the apparent radius changes as well? $\endgroup$
    – Doodoo28
    Commented Mar 5, 2021 at 15:42
  • 1
    $\begingroup$ "for a thin lens, 𝑓=𝑅/2" Not so. This relationship is for a concave mirror of large radius of curvature. $\endgroup$ Commented Mar 5, 2021 at 18:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.