Optical Instruments — Question 2

Question 2

Changing the f-number trades aperture area against exposure duration

An ideal camera lens has focal length f=56 mm and f-number N=f/D, where D is aperture diameter. For a fixed scene and image scale, assume image irradiance is proportional to aperture area. Compare N=2.8 and N=5.6.

1. Find the two aperture diameters and irradiance ratio.

2. If the wider aperture requires 1/250 s, find the exposure duration at the smaller aperture for equal received energy.

Question 2 - Solution

Changing the f-number trades aperture area against exposure duration

(a) Method and calculation

$$\boxed{D_{2.8}=56/2.8=20\,\mathrm{mm},\quad D_{5.6}=56/5.6=10\,\mathrm{mm}}.$$$$\boxed{E_{5.6}/E_{2.8}=(10/20)^2=1/4}.$$

(b) Interpretation and further calculation

$$Q\propto Et\Rightarrow t_{5.6}=4t_{2.8}.$$$$\boxed{t_{5.6}=4/250\,\mathrm s=0.0160\,\mathrm s}.$$

Physical check

The aperture diameter halves but its area falls by four; this is a geometrical exposure model with constant transmission.