How Does Distance from Infrared Source Affect Photocell Output?

i25959341
Messages
17
Reaction score
0

Homework Statement



investigate how output from photocell depends on distance from infrared point source
am required to design a labratory experiment to investigate how the output from a photocell depends on its distance from a point source of infrared radiation.

i am proposing my idea and i want ur comments on it
ermm i wud connect a photocell in series with an ammeter with a voltage meter. The photocell/ LDR/ Photodiode watever u call it, wud be directly pointed toward my light source. Therefore my circuit wud have a change in resistance wen my light source hit the LDR or a photocell. Using that information i cud vary the distance and obsevre the difference in resistance and therefore the strenght of my signal. i wonder if it s possible to calculate number of photons hitting my LDR?
The experiment wud be performed in a dark room and with safety google due to the dange of infrared light.
my thought on this experiment wud be since light diffract sometimes therefore the amount of photons hitting my LDR wud varied with the distances. Yet the question said how the output from a photo cell depends on its distance from a POINT SOUCE of infre-red radiation, so i have doubt whether it wud diffract a substantial amount or not due the confusion caused by the wording of the question. Nevertheless, that s my simple plan at the moment and i wud be really gratefull if some1 cud spot any wrong doing in my ignorant planning
 
Last edited:
Physics news on Phys.org
'A photoresistor is made of a high resistance semiconductor. If light falling on the device is of high enough frequency, photons absorbed by the semiconductor give bound electrons enough energy to jump into the conduction band. The resulting free electron (and its hole partner) conduct electricity, thereby lowering resistance.'
therefore does that mean wen more photon hitting the LDR then the Amp increase since it allowed more charged particle to go around the circuit? it sound like photo electric effect??
THEREFORE, since amp is number of charged particle per second or something. the current of an infrared hitting my resistor and the current of my resistor without any light, the differences is the number of electrons released by the infrared. and since according to photoelectric effect each electron cud only be released by one photon then the number of electrons= the number photon hitting my LDR?assuming each photon has archeieved the threshold frequency or something? but i think that wud depend on the sensitivity of my LDR right? so really i don't have to go through all those confusing intensity equation and stuff
 
Last edited:
Thread 'Need help understanding this figure on energy levels'
This figure is from "Introduction to Quantum Mechanics" by Griffiths (3rd edition). It is available to download. It is from page 142. I am hoping the usual people on this site will give me a hand understanding what is going on in the figure. After the equation (4.50) it says "It is customary to introduce the principal quantum number, ##n##, which simply orders the allowed energies, starting with 1 for the ground state. (see the figure)" I still don't understand the figure :( Here is...
Thread 'Understanding how to "tack on" the time wiggle factor'
The last problem I posted on QM made it into advanced homework help, that is why I am putting it here. I am sorry for any hassle imposed on the moderators by myself. Part (a) is quite easy. We get $$\sigma_1 = 2\lambda, \mathbf{v}_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \sigma_2 = \lambda, \mathbf{v}_2 = \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \\ 0 \end{pmatrix} \sigma_3 = -\lambda, \mathbf{v}_3 = \begin{pmatrix} 1/\sqrt{2} \\ -1/\sqrt{2} \\ 0 \end{pmatrix} $$ There are two ways...
Back
Top