A biconvex lens of focal length f and radii o | vedclass

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(A) According to the Lens Maker's Formula, the focal length $f'$ of a lens in a medium is given by $\frac{1}{f'} = (\frac{n_{1}}{n_{2}} - 1)(\frac{1}{

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Either as a convex or as a concave lens depending solely on $R$

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(A) According to the Lens Maker's Formula, the focal length $f'$ of a lens in a medium is given by $\frac{1}{f'} = (\frac{n_{1}}{n_{2}} - 1)(\frac{1}{R_{1}} - \frac{1}{R_{2}})$.For a biconvex lens, $R_{1} = R$ and $R_{2} = -R$, so $(\frac{1}{R_{1}} - \frac{1}{R_{2}}) = \frac{2}{R}$.Thus, $\frac{1}{f'} = (\frac{n_{1}}{n_{2}} - 1)(\frac{2}{R})$.If $n_{1} > n_{2}$, then $(\frac{n_{1}}{n_{2}} - 1) > 0$, so $f' > 0$, and the lens behaves as a convex (converging) lens.If $n_{1} Therefore, the behavior of the lens depends on the relative values of $n_{1}$ and $n_{2}$.

$A$ biconvex lens of focal length $f$ and radii of curvature of both surfaces $R$ is made of a material of refractive index $n_{1}$. This lens is placed in a liquid of refractive index $n_{2}$. How will this lens behave?

Similar Questions

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