# Lens diameter calculation

## Homework Statement

How we can determine The diameter of the lens of the camera approximitly that this picture took with that?

( we know that lens of the camera made by some parts but we count them as a thin one in Practical calculations )

## Homework Equations

m = q / p 1/f = 1/q + 1/p f = f/p-f f : Focal length q = Distance between of the object and the lens
p = Distance between of the image and the lens m = Magnification

## The Attempt at a Solution

i tried to find lens Magnification by calculating the distance between every centimeter of the ruler and then find its Magnification ( m ) . after that i used this formula 1/f = 1/p + 1/q
so i got f Terms p now i dont know how i must find its diameter .

#### Attachments

• Capture.PNG
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Bystander
Homework Helper
Gold Member
There's almost a zero depth of field in the photo which suggests a very small f-stop.

yeh we can realize that the object is near 2F so can it help us to soving the problem ? cos objent can not be in a very far distance

I Think The diameter of the lens depends to

Lens thickness.
The question is not so much information at our fingertips but
We can answer this question with different assumptions. good luck guys

Bystander
Homework Helper
Gold Member
diameter of the lens of the camera approximitly
"f-stop" --- ratio of focal length to aperture.

i have a Question : diameter means thickness of lens or its diagnal ?!

Bystander
Homework Helper
Gold Member
Diameter is orthogonal to thickness.

hmmm a good question , i think finding the thickness of the lens its umpossible or its hard and if we want to solve it we must know high level formula ( more than high school ) so probebly it means diagnal

I think we should estimate the distance of the camera to the object. Otherwise, we can calculate the diameter of the lens.

its obviously so question is how ? You think this formula can help ? f = m ( q - p ) / (m+1)^2 ?

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m=1/2=q/p so p=q/2 and 1/f=1/p+1/q so f=3/2p and And what else do we have? nothing ....

What a hard question ...

The image magnification image is easily calculated according to this pic (= 0.5 ). is it another way to reach him
To obtain the diameter of the lens?

berkeman
Mentor
@BrokenSoul -- Do you know where this picture was taken? Are you familiar with the scenery outside of the window in the background?

no i do not know where is that place but i guess it has a INTOLERABLE distance

If you assume that the lens the picture was taken with is a so-called thin lens and make the approximation of parallel rays coming from the background (because of the far distance) there is a rough solution to your problem:
(1) (Q/P)=(q/p) (Q=real object size, P=virtual object size, q=distance between object and lens, p=distance between image and lens)
(2) (1/f)=(1/p)+(1/q) (f=focal length)
Combination of (1) and (2): (1/f)=(1/p)+(P/Qp) => (p/f)=1+(P/Q) => (3) p=f+(Pf/Q)
Only the diameter D of the lens limits the size of the unsharp bokeh spots (because f-number is f/1.4 in my calculation). The approximately parallel rays forming these spots spread over a far distance (compared to the size of the lens) before they get refracted by the lens, converge in the image point and finally depict the spots on the photograph which have the diameter d.
The ratio of D and f equals the ratio of d and s (=p-f [distance between image point and image]):
(4) (D/f)=(d/s)=(d/p-f) => D=(df/p-f)
Application of (3) leads to (5) D=(df/f+(Pf/Q)-f)=(Qd/P)
Accordingly the formula is D=Qd/P
Q is the real and P the virtual ruler size. Insert the diameter of a random bokeh spot in the middle of your photograph for d.