What is the power in dioptres of a camera lens that has a $\mathbf{50}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{mm}$focal length?

Geometrical optics, often known as ray optics, is an optics model that describes light propagation using rays. In geometric optics, a ray is an abstraction that can be used to approximate the routes along which light propagates under particular conditions.

The power in diopters of lenses is inversely proportional to the focal length of the lens, where the power is evaluated by:

$\mathbf{P}\mathbf{=}\frac{\mathbf{1}}{\mathbf{f}}$

We know the focal length of the camera's lens, and then we directly substitute it in the above equation to find the power of the lens. We should also keep in mind that in order to find the power of the lens in the proper unit of diopter, we are supposed to convert the given focal length from unit mm to unit o m.

So, we convert it by multiplying by a conversion factor of ${\mathbf{10}}^{\mathbf{‐}\mathbf{3}}$.

Putting the numerical values and we obtain:

$$\begin{gathered} \mathit{P}\mathbf{=}\frac{\mathbf{1}}{\mathbf{f}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{1}}{\mathbf{50}\mathbf{\times }{\mathbf{10}}^{\mathbf{‐}\mathbf{3}}\mathbf{}\mathbf{m}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{20}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{D} \end{gathered}$$

Therefore, the power is:$\mathbf{20}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{D}$