Estimate the radiation pressure of your finger tip

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Homework Help Overview

The discussion revolves around estimating the radiation pressure exerted by a 100-W light bulb at a distance of 5.0 cm, assuming complete absorption of light. Participants are tasked with calculating the force on a fingertip with a specified area based on this pressure.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the calculation of intensity using the formula for power over the surface area of a sphere. There are questions about the relationship between pressure, intensity, and the speed of light. Some participants express uncertainty about how to incorporate photon momentum into their calculations.

Discussion Status

The discussion is active, with participants offering insights into the calculations needed to determine intensity and pressure. Some have suggested breaking the problem into stages for clarity, while others are exploring the implications of photon momentum and energy. There is no explicit consensus yet, but various approaches are being considered.

Contextual Notes

Participants are working under the assumption that all power from the bulb reaches the fingertip and are considering the implications of using average wavelength and temperature of the bulb's filament in their calculations. There is also mention of unit consistency and conversion challenges.

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Homework Statement



Estimate the radiation pressure due to a 100-W bulb at a distance of 5.0 cm from the center of the bulb. Assume that light is complitely absorbed.

Estimate the force exerted on your fingertip if you place it at this point. Assume area of the fingertip to be 1.5 cm2.

Homework Equations



P = I/c
I = Power/(4*pi*d2)??

The Attempt at a Solution


I don't know how to solve for intensity. I think it should have units of W/m^2 so I would guess it would be 100/(4*pi*.05*.05) = Power/(4*pi*d2), since 4*pi*d*d would be the area of a sphere of radius d but I doubt it is correct.

For the second part it would just be the pressure I get in the first part times 0.00015 (1.5 cm2 in m2) for force.

Thanks in advance!
 
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Yes, you need the surface area of a sphere radius=5cm.
Then you have the power/area at that distance ( since all the power from the bulb reaches the surface of this sphere)
Now you need to know how many photons/sec from the bulb and how much momentum/photon.

It's probably easier to do this in stages rather than just write a single equation - be carefull of the units.
 
mgb_phys said:
Yes, you need the surface area of a sphere radius=5cm.
Then you have the power/area at that distance ( since all the power from the bulb reaches the surface of this sphere)
Now you need to know how many photons/sec from the bulb and how much momentum/photon.

It's probably easier to do this in stages rather than just write a single equation - be carefull of the units.

So am I correct in saying that pressure is intensity over the speed of light? If so, am I also correct in my formula for intensity?

I'm not quite sure how I work in momentum/second (momentum/photon and photons/second).
 
The momentum of a photon is h / wavelength - where h is Planks constant.
The energy of a photon is h * frequency or h * speed of light / wavelength.

You will have to estimate an average wavelength for the light bulb (hint the filament is around 2-3000K + Wein's law)
 
mgb_phys said:
The momentum of a photon is h / wavelength - where h is Planks constant.
The energy of a photon is h * frequency or h * speed of light / wavelength.

You will have to estimate an average wavelength for the light bulb (hint the filament is around 2-3000K + Wein's law)

How is this not giving me more than I need? The title's radiation is just about light not radiation per se.

My units work out if I have (100 W/(4*pi*.05*.05 m2)/c = 1.something*10-5 pascals which are the units I want for the question (thank you masteringphysics)
 
100/(4*pi*.05*.05) is the intensity, ie the power/area at that distance.
To go from power to momentum you need to work out the number of photons and the momentum of each.
 
mgb_phys said:
100/(4*pi*.05*.05) is the intensity, ie the power/area at that distance.
To go from power to momentum you need to work out the number of photons and the momentum of each.

Why can't I simply divide intensity by the speed of light which would make my units work out?

I submitted my solutions; Pressure = /frac{/frac{100}{4/pi .05^2}}{c} worked.
 
Last edited:
Queue said:
Why can't I simply divide intensity by the speed of light which would make my units work out?

I submitted my solutions; Pressure = /frac{/frac{100}{4/pi .05^2}}{c} worked.

That should be \frac{\frac{100 W}{4pi .05^2 m^2}}{c}

And for some reason this is coming out as mu_0*8/(2*pi*(.0447213595)) for some reason...
 

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