Q:

A​ 35-mm camera allows you to control the diameter d of the aperture by changing the​ f-stop, which varies inversely with the diameter according to the formula equals F=44 over d . a. If f =4​, find the diameter. b. If the diameter is quadrupled​, by what factor will the f dash stop ​change? c. The radius r is​ one-half of the diameter. Describe the proportional relationship between the radius and the diameter. d. Area A is proportional to the radius squared according to the equation A equals pi r^2. What is the equation that relates area and​ diameter? e. What is the equation that relates area and​ f-stop?

Accepted Solution

A:
Answer:a) 11b) Factor of 1/4c) r = d/2d) [tex]A=\frac{\pi d^{2}}{4}[/tex]e) [tex]F=\frac{22\sqrt{\pi}}{\sqrt{A}}[/tex]Step-by-step explanation:Part a)The relation between F dash stop and the diameter of the aperture is given by:[tex]F=\frac{44}{d}[/tex]We have to find the value of diameter, if F = 4. Substituting F = 4 in above equation, we get:[tex]4=\frac{44}{d}\\\\ 4d=44\\\\ d=\frac{44}{4}\\\\  d=11[/tex]This means, if f-stop is at F=4, the diameter of the aperture would be 11 mm.Part b)Let the new diameter be represented by d' and new F-stop by F'. Since, the diameter is quadrupled, the new diameter is 4 times the previous one. i.e.d' = 4dNow, the formula of F' with new diameter will be:[tex]F'=\frac{44}{d'}[/tex]Using the value of d', we get:[tex]F'=\frac{44}{4d}[/tex][tex]F'=\frac{1}{4} \times \frac{44}{d}\\\\  F'=\frac{1}{4} \times F[/tex]This means, the dash stop will change by a factor fo 1/4 when the diameter is quadrupled.Part c)Radius is defined as one half of the diameter. If r represents the radius, the relation between radius r and diameter d can be expressed as:[tex]r=\frac{d}{2}[/tex]Part d)Area of lens (Circle) is given as:[tex]A=\pi r^{2}[/tex]Using the value of r from previous part, we get:[tex]A= \pi (\frac{d}{2} )^{2}\\\\ A=\frac{\pi d^{2}}{4}[/tex]Part e)In order to find the relation between Area and F-stop, we replace the value of d with its equivalent expression in terms of Area.From the previous part:[tex]A=\frac{\pi d^{2}}{4}\\\\ 4A=\pi d^{2}\\\\d^{2}=\frac{4A}{\pi}\\\\ d=\sqrt{\frac{4A}{\pi}}\\\\ d=2\sqrt{\frac{A}{\pi} }[/tex]Using this value of d, in formula of F, we get:[tex]F=\frac{44}{2\sqrt{\frac{A}{\pi}}}\\\\ F=\frac{22}{\sqrt{\frac{A}{\pi}}}\\\\ F=\frac{22\sqrt{\pi}}{\sqrt{A}}[/tex]