Photons - Trying to find distance

  • Thread starter Thread starter Nitric
  • Start date Start date
  • Tags Tags
    Photons
AI Thread Summary
The discussion focuses on calculating the distance from which the naked eye can detect a 60-W lightbulb, given that the eye can detect as few as five photons per second. Participants emphasize the need to determine the number of photons emitted by the bulb and how they disperse in space, forming an imaginary sphere around the light source. The conversation highlights the importance of using the energy of the bulb and the energy of individual photons to find the relevant formulas. By applying the known detection threshold of five photons per second, one can derive the distance by manipulating the equations related to photon emission and detection. This approach illustrates a common physics problem-solving technique involving assumptions and calculations to find unknown variables.
Nitric
Messages
21
Reaction score
0
1. The night-adapted eye can detect as few as several photons per second -- say five to be definite. Assuming that the pupil of the eye is 0.6 cm in diameter, from what distance would it be possible for the naked eye to detect a 60-W lightbulb?



2. I'm not sure which equation to use. In the section where the problem is from, I get teh follow equations (which don't help me)
E=pc, E=hf, h=6.63x10^-34, p=h/lamda




3. Not sure which formula to use
 
Physics news on Phys.org
Well, you will need to know how many photons the light emits per second. You can do this by picking a frequency near the low end of the visible range (to end up with the minimum), and using the energy of the bulb and the energy of the photon. Guess or look up the energy efficiency of a light bulb.

The next thing is how the photons from the bulb spread out in space. They go out in all directions equally, so light up an imaginary sphere at the distance where the eye is. The question is what portion of that lit up area is caught by the eye.
 
It says 5 photons/sec in the question.
 
Yes, you know the 5 per sec answer so you can use the expression for the # of photons hitting the eye - which has a d in it - to solve for the distance.

Physics often works that way. You pretend you know something like the distance to the source, work out other quantities based on that unknown, and when you find one you know the answer for, you can solve for the unknown. Very useful trick!
 
Thread 'Voltmeter readings for this circuit with switches'
TL;DR Summary: I would like to know the voltmeter readings on the two resistors separately in the picture in the following cases , When one of the keys is closed When both of them are opened (Knowing that the battery has negligible internal resistance) My thoughts for the first case , one of them must be 12 volt while the other is 0 The second case we'll I think both voltmeter readings should be 12 volt since they are both parallel to the battery and they involve the key within what the...
Thread 'Trying to understand the logic behind adding vectors with an angle between them'
My initial calculation was to subtract V1 from V2 to show that from the perspective of the second aircraft the first one is -300km/h. So i checked with ChatGPT and it said I cant just subtract them because I have an angle between them. So I dont understand the reasoning of it. Like why should a velocity be dependent on an angle? I was thinking about how it would look like if the planes where parallel to each other, and then how it look like if one is turning away and I dont see it. Since...
Back
Top