### Theory:

Magnification

The magnification produced by a lens, like that produced by spherical mirrors, is defined as the ratio of the image height to the object height. The letter $$m$$ is used to indicate magnification. If $$h$$ is the height of the object and $$h′$$ is the height of the image produced by a lens, then

$m=\frac{\mathit{Height}\phantom{\rule{0.147em}{0ex}}\mathit{of}\phantom{\rule{0.147em}{0ex}}\mathit{the}\phantom{\rule{0.147em}{0ex}}\mathit{image}}{\mathit{Height}\phantom{\rule{0.147em}{0ex}}\mathit{of}\phantom{\rule{0.147em}{0ex}}\mathit{the}\phantom{\rule{0.147em}{0ex}}\mathit{object}}=\frac{{h}^{\prime }}{h}$

The object-distance u and the image-distance v are also related to the magnification produced by a lens. This relationship is given by

$\mathit{Magnification}\left(m\right)=\frac{{h}^{\prime }}{h}=\frac{v}{u}$

Lens maker's formula:

Transparent materials are used to make lenses. Any optically transparent material will have a refractive index. The focal length of a lens is proportional to the distance between the object and the image, according to the lens formula. A lens maker must be familiar with the radii of curvature of the lens. This clearly demonstrates the need for an equation that connects the lens's radii of curvature, the refractive index of the lens's given material and the lens's required focal length. One such equation is the lens maker's formula. It is written as

$\frac{1}{f}=\left(\mathrm{\mu }-1\right)\left(\frac{1}{{R}_{1}}-\frac{1}{{R}_{2}}\right)$
where $$µ$$ is the refractive index of the material of the lens; $\frac{1}{{R}_{1}}$ and ${R}_{2}$  are the radii of curvature of the two faces of the lens; $$f$$ is the focal length of the lens.