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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?

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

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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. $$P=\frac{1}{f}=(\mu_{lm}-1)\left( \frac{1}{R_1}-\frac{1}{R_2}\right)$$

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  • $\begingroup$ 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
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    $\begingroup$ "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

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