Competition - post a photo you've taken of nature >>

Physcis help needed

The minimum energy of visible light of wavelength 500 to 600 nm that can just be detected by the eye is 10^-17 assume only 1% of the power input to a 60W light bulb is emitted between these wavelengths estimate the maximum distance that it might be possible to see the lit bulb. the pupil of the eye will be at its maximum diameter say 8mm. at its max sensitivity the eye only detects 8% of the light incident on it

answer is 140km but i don't see how they got to it

Reply 1

lerjj

Original post by jamesbird18

What have you done so far? You need to do it in steps: what is the total output of the bulb?

Reply 2

jamesbird18

0.01 x 60? I was going to divide that by the surface area and set it equal to the min energy leaving the radius as 'r' but that wouldn't make any sense i wouldn't have any unknowns so i'm kinda stuck at what to do next

Reply 3

jamesbird18

Original post by lerjj

What have you done so far? You need to do it in steps: what is the total output of the bulb?

•

1 minute ago

•

Report post

[INDENT]0.01 x 60? I was going to divide that by the surface area and set it equal to the min energy leaving the radius as 'r' but that wouldn't make any sense i wouldn't have any unknowns so i'm kinda stuck at what to do next[/INDENT]

Reply 4

lerjj

Original post by jamesbird18

•

1 minute ago

•

Report post

0.01 x 60? I was going to divide that by the surface area and set it equal to the min energy leaving the radius as 'r' but that wouldn't make any sense i wouldn't have any unknowns so i'm kinda stuck at what to do next

I'm not sure I know this stuff TBH, and I'm too busy to go online and learn it as well at the moment. If it's 60W at 1% when that's a start. I'd probably try working out everything except the stuff about the eye for now, then get around to that later. If noone else has, I'll try and do this tomorrow.

Reply 5

lerjj

Ok, my maths is probably not good enough to do this actually...

What you want to do is basically draw out a triangle with one vertex representing the bulb and the other two the diameter of the eye. (It's really a cone, but I think the triangle is easier to draw...)

The total output of the bulb is going to be 1% of 60W=0.6W, but the amount landing on the eye will now be \frac{\theta}{360}

Again, the issue is partially my maths as I don't know whether this still holds given that its actually a cone of light, not a 2D cross section of it.

As you go further out, the angle will decrease. I would do this in a few steps:
Find a function for the angle depending upon distance from the eye
Find the required angle for the energy to be below the minimum sensitivity you posted.
Substitute that into your function to find the maximum distance.

If you still need help then you'll have to wait while I teach myself the necessary geometry.

EDIT: actually, a better way of looking at it (that I can do) is to say the pupil will be a circle of diameter 8mm on the surface area of a sphere of radius d centred on the bulb. Work out the fraction of the sphere taken up by the eye and proceed from there. The triangle thing is nice for getting a handle on the situation but misleading when it comes to working out the numbers.