Pinhole Camera Question: Calculating Distance for 1-Inch Image of a 20 Ft Tree

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To calculate the distance needed for a 1-inch image of a 20 ft tree in a pinhole camera, the distance from the pinhole to the film is essential. The size of the pinhole affects image sharpness and brightness, but the distance from the pinhole to the film influences image size. Using the concept of similar triangles can help solve the problem, as the relationship between the object's height and the image size is proportional. While some participants noted that the size of the opening might be irrelevant, others clarified that it does have an impact on image quality. Ultimately, understanding these principles is key to resolving the question effectively.
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How far would you have to stand from 20 ft tree so the image in a pinhole camera is 1 inch long? I feel like the teacher left out a piece of information...please help I am stumped.
 
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rdiprimio said:
How far would you have to stand from 20 ft tree so the image in a pinhole camera is 1 inch long? I feel like the teacher left out a piece of information...please help I am stumped.

I am stumped as well.The length of the camera is needed.
 
Yes, you're missing information. You need the distance from the pinhole to the film. However, you can make a simple formula and just call this distance "x". Draw two rays. One going through the pinhole and hitting the middle of your film, and another one, comming off of the top of your object and hitting the bottom of your film.
 
I remember asking my physics teacher this question (it wasn't a physics problem for class, just a personal question) and i remember him saying by way of similar triangles that both a) the size of the opening is irrelevant (expected) and b) the length from the hole to the film is irrelevant (blew my mind, still don't know how that works). Try messing with trig functions and creating triangles is my advise!
 
mg0stisha said:
I remember asking my physics teacher this question (it wasn't a physics problem for class, just a personal question) and i remember him saying by way of similar triangles that both a) the size of the opening is irrelevant (expected) and b) the length from the hole to the film is irrelevant (blew my mind, still don't know how that works). Try messing with trig functions and creating triangles is my advise!

You do need the distance from the hole to the film.The greater the distance the bigger the image.The size of the hole has some relevance as well...ignoring diffraction a small hole gives a sharp dim image and a large hole gives a blurred bright image.It is true that you can use similar triangles to solve the problem.
 
I probably just remembered wrong then, it makes complete sense that the distance from hole-to-film would matter. I did figure out how to solve this by way of similar triangles, not that this helps the thread starter in any way!
 
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