Respuesta :
Answer:
f / 8
Explanation:
The f-number is the ratio of the lens's focal length to its diameter and is a measure of the intensity of the light reaching the CCD. The smaller the greater the intensity of the light
I = D / f
f-number = f / D
I = 1 / f-number
the energy deposited in the CCD is proportional to the intensity of the light and the exposure time (Δt)
E = I Δt
E = 1 /f-number Δt
in the exercise they indicate that the system is well exposed (the image is sharp and clear) for fo-number = f / 16
Δt = exposure time = 1/120 s
E = [tex]\frac{1}{120} \ \ \frac{1}{fo-number}[/tex]
indicate that the exposure time has been changed to Δt = 1/60 s, which should be the f₁-number
E = [tex]\frac{1}{60} \ \ \frac{1}{f_1-number}[/tex]
the energy deposited must be the same per location we can equal the expressions
[tex]\frac{1}{120} \ \frac{1}{f_o-number} = \frac{1}{60} \ \frac{1}{f_1-number}[/tex]
f₁-number = [tex]\frac{60}{120} \ \ \frac{1}{fo-number}[/tex]
f₁-number = ½ 16
f₁-number = 8
so the system should be set to f / 8
