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

  • Thread starter Thread starter rdiprimio
  • Start date Start date
  • Tags Tags
    Camera
AI Thread Summary
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.
rdiprimio
Messages
1
Reaction score
0
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.
 
Physics news on Phys.org
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!
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
Thread 'Correct statement about a reservoir with an outlet pipe'
The answer to this question is statements (ii) and (iv) are correct. (i) This is FALSE because the speed of water in the tap is greater than speed at the water surface (ii) I don't even understand this statement. What does the "seal" part have to do with water flowing out? Won't the water still flow out through the tap until the tank is empty whether the reservoir is sealed or not? (iii) In my opinion, this statement would be correct. Increasing the gravitational potential energy of the...
Thread 'A bead-mass oscillatory system problem'
I can't figure out how to find the velocity of the particle at 37 degrees. Basically the bead moves with velocity towards right let's call it v1. The particle moves with some velocity v2. In frame of the bead, the particle is performing circular motion. So v of particle wrt bead would be perpendicular to the string. But how would I find the velocity of particle in ground frame? I tried using vectors to figure it out and the angle is coming out to be extremely long. One equation is by work...
Back
Top