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

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Homework Help Overview

The discussion revolves around calculating the distance from a pinhole camera to a 20 ft tree in order to produce a 1-inch image. Participants express concerns about missing information necessary for solving the problem.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants question the necessity of the distance from the pinhole to the film and discuss the relevance of the size of the pinhole. Some suggest using similar triangles and trigonometric functions to explore the relationships involved.

Discussion Status

There is an ongoing exploration of the problem, with participants sharing their thoughts on the importance of certain distances and the implications of similar triangles. While some guidance has been offered regarding the use of geometric principles, there is no explicit consensus on the necessary information or approach.

Contextual Notes

Participants note that the length of the camera and the distance from the pinhole to the film are critical pieces of information that are not provided in the original question.

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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, coming 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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