A camera lens has a focal length of 180.0 mm and an aperture diameter of 16.36 mm. (a) What is the ƒ-number of the lens? (b) If the correct exposure of a certain scene is 1/30 s at f/11, what is the correct exposure at f/2.8?

An optical lens is a transparent transmissive optical component that is used to either converge or diverge the light emitting from a object. These transmitted light forms an image of that object.

A camera uses light-sensitive film to form an image on it where the intensity of the fallen light is proportional to f-number. f-number shows the capability of lens to collect light.

The f-number is defined as the ratio of focal length of a lens to its aperture diameter, and it is given by

$\mathit{f}\mathbf{-}\mathit{n}\mathit{u}\mathit{m}\mathit{b}\mathit{e}\mathit{r}\mathbf{=}\frac{\mathbf{f}}{\mathbf{d}}$

Focal length of the lens, f = 180 mm

Aperture diameter, D = 16.36 mm

F – number for the lens is given by

$$\begin{gathered} f-number=\frac{f}{d} \\ =\frac{180mm}{16.36mm} \\ =11 \end{gathered}$$

Therefore, the f-number for the given lens is 11 and it can be represented by f-stop formula as f/11.

Correct exposure of a certain scene at f/11, t_f/11 = (1/30) sec.

The ratio of intensity of two f-numbers (f/11) and (f/2.8) is equals to square of the aperture diameter of each intensity:

Thus, the correct exposure at f/2.8 will be given by

$$\begin{gathered} t_{f/2.8}=\frac{f/11}{f/2.8}\left(t_{f/11}\right) \\ =\left(6.47\times {10}^{-2}\right)\left(\frac{1}{30}s\right) \\ =2.16\times {10}^{-3}s \\ =2.16ms \end{gathered}$$

Therefore, the correct exposure at f/2.8 is 2.16 milliseconds.