7.1 Thin lens

Now we can turn to the case of lenses. A lens can be considered the combination of two spherical interfaces. To solve an optical problem usingmultiple interfaces or lenses, one considers each one by one. For example one finds the image created by the first surface and then uses it as the object ofthe second surface.

$$$$ $$$$ Now we take the thin lens case,that is $d\to 0$ $$\frac{n_1}{s_o}+\frac{n_3}{s_i}=\frac{n_2-n_1}{R_1}+\frac{n_3-n_2}{R_2}$$ That equation would work for making prescription swim goggles for example, howevermost of the time $n_1=n_3$ (namely air for eyeglasses). So making that the case we get $$\frac{n_1}{s_o}+\frac{n_1}{s_i}=\left(n_2-n_1\right)\left[\frac{1}{R_1}-\frac{1}{R_2}\right]$$ which in the case of air (n=1) is $$\frac{1}{s_o}+\frac{1}{s_i}=\left(n_l-1\right)\left[\frac{1}{R_1}-\frac{1}{R_2}\right]\text{.}$$ That is The lensmaker's formula(where $n_l$ is the index of refraction of the lens material)Now we can find the foci $$f_i=\underset{s_i\to \infty }{lim}$$ $$f_o=\underset{s_o\to \infty }{lim}$$ Which we see from the lensmaker's formula must be the same so lets drop thesubscripts: $$\frac{1}{f}=\left(n_l-1\right)\left[\frac{1}{R_1}-\frac{1}{R_2}\right]$$ and $$\frac{1}{s_o}+\frac{1}{s_i}=\frac{1}{f}$$ which is the Gaussian Lens Formula

A convex lens will have a positive focal length $f$ .

We can use the same equation as before for the magnification, $$m=\frac{h_i}{h_o}=-\frac{n_1{s}_i}{n_2{s}_o}\text{.}$$ but now note that $n_1=n_2$ so the equation becomes

$$m=\frac{h_i}{h_o}=-\frac{s_i}{s_o}\text{.}$$ You can examine the figure above to verify that this is true.

We can also consider a concave lens which has a negative focal length. Notice that in this case the image is upright and virtual. Notice that in this case $s_i$ is negative.