Laser Pointer Wave Energy and eyes seeing

[TFM]

Homework Statement

A typical “laser pen” pointer has a power output of 3mW at 670nm. If the angular divergence of the beam is 2mrad, estimate the maximum distance at which it could be seen by a day-adapted human eye. You can assume that in bright light, a normal eye will give a signal from 660nm light if 4 * 10^5 photons arrive in 0.1s.

Homework Equations

E = hf = \frac{hc}{\lambda}

Power = \frac{E}{t}

The Attempt at a Solution

Okay, I have used the above equations for the 660nm light

The energy from each photon at that wavelength is 3*10^{-19} Joules

Since 4 * 10^5 Photons are arriving in 0.1s, that means a total of 1.2*10^{-12} Joules.

Now the power of this beam is 1.2*10^{-13} Watts.

I am slightly unsure what to do from here, the power I have calculated for the 660nm Wave seems to be a LOT smaller to the 670nm Wave, which is 3mW, or 0.003 Watts.

Any suggestions/ideas about what to do from here...?

Thanks in advance,

TFM

 

[Carid]

If you are getting 1.2*10^{-12} Joules in 0.1 sec, how many joules arrive in 1 second.
This will give you the power in watts. Check your figure.

Next, of course this value is less than the power of the laser; our eyes are pretty sensitive, but the question asks you how far away do you have to be before the laser light is only just visible. The beam spreads out so the number of photons reaching your eye declines with distance. Think about how we can calculate the rate of reduction of light intensity with distance.

 

[TFM]

Okay yeah, it should have been

1.2 * 10^{-11} Watts

Still a lot smaller then the 0.003 Watts from the other laser.

TFM

 

[Carid]

Check out my modified reply above.

 

[TFM]

Okay, well light obeys the inverse square law, so the intesity will fall like so:

I \propto \frac{1}{d^2}

It is also getting wider with a divergence of 2 mrad.

 

[mgb_phys]

Those two statements are the same thing,
If the beam is spreading with 2mRad divergence then it's making a cone with that angle (note divergence is normally the full angle ie diameter/distance)
Work out how much lower the intensity must be, and so the area the beam footprint must cover and so the distance away it must be - you can consier the beam to be a point at the laser

 

[TFM]

Do we use the initial Power of the 670nm of 0.003 Watts for this?

 

[mgb_phys]

Correct, either work out how many photons/s this is an compare with the minimum rate - or work out what power the minimum rate represents.

 

[TFM]

Okay so with the 660nm Wave,

The photons per second is calculated from:

0.003 watts = 0.003 Joules/S

in 1 S = 0.003 Joules.

Using

E = \frac{hc}{\lambda}

we get that the energy of a 660 nm Wave photon is 3.01 * 10^{-19}

This means that the Laser is emitting 9.97 * 10^{15} photons per second.

this number is a lot bigger then the minimum required for the 670 (4 * 10^5)

So what is the best way now to use this finromation to get the distance?

Is it to do with seing how far the beam has to travel for the density of these photons to gbecome equal to the 670 min value?

TFM

 

[mgb_phys]

This means that the Laser is emitting 9.97 * 10^{15} photons per second.
this number is a lot bigger then the minimum required for the 670 (4 * 10^5)

Looks reasonable, so you know the beam has to be reduced by around 10¹⁰
(careful your number detected is in 0.1s not per second)

Now for the distance,
The light is forming a cone from the laser with apex angle 2mRad.
So the number of photons you calculated is passing through the flat base of this cone every second.
What proportion pass though an eye at this distance?
It's just the ratio of the area of the pupil to the area of the cone (assume 5mm diameter for the pupil - the number doesn't really matter as long as you show your working)

Then just work out what size the cone must be for roughly 10^-10 of the photons to go through the eye.

 

[TFM]

Okay so let me see. we have the Laser is emitting 9.97 * 10^{15} photons per second. The eye only needs 4 * 10^6 photons per second.

Now this means that the area of the pupil can be 4.01 * 10^{-10} the size of the totral area.

So assuming a pupil diameter of 5mm (0.005m), this means that the area of the pupil is 0.0019m. so the total area of the whole circle is 4.89*10^6] m (does this look right, it seem very large?)

This means that the radius of the cirle is 1248m.

So using triginometery, the distance is 624054m (6.2 * 10^5m)

Does this look right. will the fact that I have seemingly mixed the 660nm and 670 nm make any difference?

TFM

 

[mgb_phys]

Not unreasonable - you can check you answer by workign out how many photons/mm^2 the 3mW gives at this distance, and so how many go through a 5mm dia opening/s
(if anything the given eye senstivity is a bit pessimistic, a 6th mag star is <1000/photons/s in your eye!)

The 660/670 isn't important, your eye's sensitivity is a bit less at 670 than 660 but you aren't given any data for the curve.

 

[TFM]

Okay, so that looks to be the right answer?

Excellent, Thanks for all your assistance,

TFM