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Pinhole Size for Eclipse

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#1 Alex McConahay

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Posted 15 March 2017 - 09:04 AM

I want to make a pinhole solar eclipse viewer....You have seen them, I am sure, or the pictures from them. You punch some holes in a piece of stiff material, probably in some cute pattern, and the sun goes through and makes pinhole images.

 

So here is my question-----What is the ideal size of the pinhole?

 

What is the effect of different size pinholes?

 

Alex

Here is an example from 2006 in Egypt....(Sorry, don't know/remember who did the work on the pinholes, but I took the picture of it, it was so cute.)

 

 

Attached Thumbnails

  • EgyptSphinxPinhole.jpg


#2 CP Kuiper

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Posted 15 March 2017 - 10:47 AM

The ideal hole size is determined by how far from the hole you want to see the projected image.

That distance will also determine how large the projected image is.

Don't forget the Inverse Square Law.



#3 Don W

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Posted 15 March 2017 - 02:09 PM

You can experiment on the full sun. Try some different size holes and project them at different lengths.



#4 Alex McConahay

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Posted 15 March 2017 - 04:42 PM

>>>>>You can experiment on the full sun.

 

Maybe this is a silly question, but how do I tell whether I am getting a focused image of the sun, or just a spot of light the hole is letting through? 

 

Alex



#5 Don W

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Posted 17 March 2017 - 11:18 PM

I believe the edges of the sun would be in focus, or sharp.



#6 Ernest_SPB

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Posted 18 March 2017 - 12:16 AM

Diameter of pinhole d strongly defines o - minimal diameter of details on projected image. Taking into account diffraction it could be calculated like this:

 

o = d + L/(750*d),

 

where L is distance from pinhole to screen.

 

It is easy to calculate optimal dopt (that makes minimal o):

 

dopt = SQRT(L/750) = SQRT(L)/27

 

and the minimal diameter of details on screen:

 

omin = 2*SQRT(L/750) = SQRT(L)/14

 

E.g. for distance L = 1 meter (1000 mm): omin = SQRT(1000)/14 = 2.3 mm

 

Let's compare the omin with diameter of solar disc D = L/115.

 

For the same distance 1000 mm D = 1000/115 = 8.7 mm - just 3 times large 

 

Let's introduce merit value "quality" of pinhole projection like this:

 

Q = D/omin = 14* L/(150*SQRT(L)) ~ SQRT(L)/10 = 1.4*omin

 

We can see that Q slowly grows with larger L 

 

L(mm)   D     dopt  omin   Q 

1000     8.7   1.2    2.3   3.3

2000    18     1.7    3.3   4.6

3000     26    2.0    4.1   5.7

4000     35    2.3    4.7   6.6

5000     44    2.6    5.3   7.3


Edited by Ernest_SPB, 18 March 2017 - 12:31 AM.



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