Answer:
a) 11
b) Factor of 1/4
c) r = d/2
d) [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' = 4d
Now, 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]
