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If an equiconvex lens is divided equally, we will obtain a planoconvex lens. Let assume focal length of our equiconvex lens is 'f'. I would like to know what will be the focal length of planoconvex lens that we obtained after dividing equiconvex lense. Will it be twice of 'f'? Or will it be 0.5 of 'f'? How to determine it?

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  • $\begingroup$ Just think about the refraction and the resulting deviation of light at each face of the two types of lenses. $\endgroup$ – Farcher Aug 25 '16 at 14:22
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Using the Lens-Maker's equation: $$ \frac{1}{f} = \frac{n_{lens} - n_o}{n_o}\left(\frac{1}{R_{left}}-\frac{1}{R_{right}}\right)$$ And because we have an equiconvex lens, $R_{right}=-R_{left}$ (the negative sign is from the sign convention of the equation), so we have: $$ \frac{1}{f_{equi}} = \frac{n_{lens} - n_o}{n_o}\left(\frac{1}{R_{left}}+\frac{1}{R_{left}}\right) = 2\frac{n_{lens} - n_o}{n_o}\frac{1}{R_{left}}.$$ So $f_{equi}=\frac{R_{left}n_o}{2(n_{lens} - n_o)}$ and for the planoconvex lens, $R_{right} \approx \infty$ so the equation becomes: $$ \frac{1}{f_{plano}} = \frac{n_{lens} - n_o}{n_o}\left(\frac{1}{R_{left}}+\frac{1}{\infty}\right) = \frac{n_{lens} - n_o}{n_o}\frac{1}{R_{left}}.$$ So when you cut the lens in half, the focal length is actually doubled when you cut the lens, since $f_{plano}=\frac{R_{left}n_o}{n_{lens} - n_o}=2 f_{equi}$.

For more info on the Lens-Maker's equation, you can look here for an explanation.

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Focal lengths are first order properties of lenses. An equi-convex lens is ideal for imaging an object 2*focal length away and forming an image 2*f behind. If you cut the equi-convex lens right down the middle such that you get a convex-plano lens, the focal length will double and it will basically collimate the object that was originally 2*f away. Shape factor can be changed while keeping the focal length a constant and you can watch the spherical aberration change as a function of shape factor for a given f. But that's another story. Bottom line, I agree with first answer. Focal length will double.

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