How Many Photons Reach the Retina?

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Homework Help Overview

The discussion revolves around calculating the number of photons that reach the retina from a distant incandescent light bulb. The original poster expresses confusion regarding the relevance of distance and pupil diameter in this context, seeking clarity on how to approach the problem.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants explore the concept of photon distribution from a light source and the fraction of photons that enter the eye. Questions arise regarding the geometry of the eye and how to accurately represent the relevant areas for this calculation.

Discussion Status

Some participants have provided insights into the geometry of the problem, suggesting the use of a sphere to represent the distribution of photons. There is an ongoing exploration of how the surface area of the pupil relates to the total area from which photons can be detected. The discussion is active, with participants questioning assumptions and clarifying concepts.

Contextual Notes

Participants note the importance of considering the pupil's diameter and the experimental conditions under which the human eye can detect photons. There is uncertainty about whether the detection capability is measured per unit area or across the entire surface of the eye.

kingwinner
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How many photons?

1) A typical incandescent light bulb emits 3x10^18 visible-light photons per second. Your eye, when it's fully dark adapted, can barely see the light from an incandescent light bulb 10 km away. How many photons per second are incident at the image point on your retina? The diameter of a dark-adapted pupil is 6mm.

Some relevant formulas I found from the textbook are E=hf, c=(lambda)(frequency), lambda=h/(mv)
But none of the above formulas seem to apply by any means to this weird question. Why does the distance 10km matter? Why does the diameter of an eye matter? How can I even use them?

I am really lost and confused. I hope that someone would be nice enough to help me out. Thank you so much!
 
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Imagine that the light bulb is sending out all those photons in all directions equally. What fraction would make it into your eye at a distance of 10 km? One way to figure it out is to think of a huge 10 km radius sphere surrounding the light bulb--compared to the surface area of that sphere, how big is the surface area of your pupil?
 
Doc Al said:
Imagine that the light bulb is sending out all those photons in all directions equally. What fraction would make it into your eye at a distance of 10 km? One way to figure it out is to think of a huge 10 km radius sphere surrounding the light bulb--compared to the surface area of that sphere, how big is the surface area of your pupil?
Thanks for your reply!


But what is the surface area of the eye? Should it be considered full sphere, a semi-sphere (half sphere), or something else?
 
The photons have to go through the pupil. That's the relevant area, a disk 6mm in diameter.
 
Thanks for your help! I got it!
 
I was curious as to the answer. I used to do research in biophysics/vision. Under really good experimantal conditions, the human eye can detect between 1 and 5 photons, an amazing sensitivity when you consider the dynamic range of the system.
 


Hi all,
discussion is goin on nicely...
I had que. whether this number of photons that our eye can detect is per unit area or on the whole surface of eye...?
 


agtee said:
Hi all,
discussion is goin on nicely...
I had que. whether this number of photons that our eye can detect is per unit area or on the whole surface of eye...?

Most of these experiments I recall were done in dark adapted volunteers using all of the eye. But in theory, all you need is capture of one photon by one rhodopsin photopigment molecule which sets into motion a biochemical cascade with gain measured in the millions.
 

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