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?
[Physics] Is focal length of a planoconvex lens twice to the focal length of equiconvex lens from which the planoconvex lens was made
lenses
Related Solutions
We have to consider the individual refractive surfaces only and not the planoconvex lens as a whole, but it so happens, that the final answer can be simply expressed in terms of the focal length the planoconvex lens would have without the silvering.
For a general derivation, consider a lens with radius ($R_1,R_2$), with $R_2$ silvered. Cartesian sign convention are used here.
For an object distance (from the first surface, but for a thin lens, the distance is same for both the surfaces) $u$, we get for the first surface:-
$$\frac{\mu}{v_1}=\frac{\mu-1}{R_1}+\frac{1}{u}$$
(where $\mu$ is the refractive index of lens and that of air is $1$)
Now for surface two, a mirror, $v_1$ is the object:-
$$\frac1{v_2}=\frac2{R_2}-\frac{1}{v_1}$$
Now the second image $v_2$ acts as the object for first surface again, but now with inverted sign conventions.
$$-\frac 1v=\frac{1-\mu}{-R_1}+\frac{\mu}{v_2}$$
using the value of $v_1$ and $v_2$ from preceding equations you come up with
$$\frac 1v+\frac 1u=\frac{2\mu}{R_2}-\frac{2(\mu-1)}{R_1}$$
This is similar to a mirror with focal length given by the left hand side of the equation. It can be rearranged to
$$\frac 1f=-\bigg(\frac{2(\mu-1)}{R_1}-\frac{2(\mu-1)}{R_2}\bigg)+\frac 2{R_2}=\frac{-2}{f_{\text{lens}}}+\frac{1}{f_{\text{mirror}}}$$
which is the required equation.
The focal length tell us how much the light rays will be bent. Think of the light rays as a paper cone, just like the one you get when you buy a snow cone. The mouth of the cone is the size of the lens, and the point of the cone is the focal point. The length of the cone is the focal length. Now picture a cone where the point is very close to the mouth. It would be a very steep short cone. The light rays would come in at very steep angles, and after they crossed the focal point they would spread out quickly. In fact, they would make another cone leaving the focal point that matches the cone that entered the focal point. The rays would continue on spreading out wider and wider. Now picture a cone that is very long, say three feet, a novelty snow cone, notice how these rays come into focus at a much smaller angle, and after they pass the focal point they will form another long cone on there way out. Well, there you have it, the focal length tells us how long the snow cone is. I hope your not offended by my simple examples and language in this answer. It's how I think of focal length. With a little imagination you can see how different cone sizes would be applicable to different applications.
Best Answer
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.