Problem with electromagnetic waves

In summary, to suspend a spherical glass bead with a mass of 1 microgram and a density of 0.200 g/cm^3 in the Earth's gravitational field, a radiation intensity would be needed. This can be determined by calculating the volume of the bead from its mass and density, and then using the energy density and speed of light to find the intensity. The magnitude of the force needed to suspend the bead is equal to its weight, which can be calculated using pressure and area, taking into account the curved surface. This can also be understood in terms of the relationship between energy and momentum for light.
  • #1
andrew410
59
0
Lasers have been used to suspend spherical glass beads in the Earth's gravitational field. a) A black bead has a mass of 1 micrograms and a density of 0.200 g/cm^3. Determine the radiation intensity needed to support the bead. b) If the beam has a radius of 0.200 cm, what is the power required for this laser.

I know that density = mass/volume, so I can get the volume from the density and mass. I know that the energy density = energy/volume. Also, I know that intensity = speed of light*energy density. How do I get the energy or am I doing this wrong? Please help...any help would be great! thx! :)
 
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  • #2
Well, the magnitude force needed to suspend the micro-ball must = mg.

The magnitude of the force is related to Pressure x area, but remember this is a curved surface, and one needs the force operating anti-parallel to gravity.

In another of your posts, there is a discussion of radiation (light) pressure. The pressure is simply due to momentum transfer. Consider the relationship between energy and momentum for light.
 
  • #3


a) To determine the radiation intensity needed to support the bead, we can use the equation for gravitational force:

F = mg = (1 microgram)(9.8 m/s^2) = 9.8 x 10^-6 N

This force must be balanced by the radiation pressure from the laser beam. The equation for radiation pressure is:

P = I/c, where P is pressure, I is intensity, and c is the speed of light.

Since we want to find the intensity, we can rearrange the equation to solve for I:

I = Pc

Now, we need to find the power of the laser in order to calculate the intensity. This will be done in part b.

b) To find the power required for the laser, we can use the equation:

P = Fv, where P is power, F is force, and v is velocity.

In this case, the velocity is the speed of light, so we can rewrite the equation as:

P = Fc

Substituting in the value for force from part a, we get:

P = (9.8 x 10^-6 N)(3.00 x 10^8 m/s) = 2.94 x 10^-3 W

Therefore, the power required for the laser to support the black bead is 2.94 x 10^-3 W.

Now, to find the intensity, we can plug this value into the equation from part a:

I = Pc = (2.94 x 10^-3 W)(3.00 x 10^8 m/s) = 882 W/m^2

This is the radiation intensity needed to support the black bead.
 

1. What are electromagnetic waves?

Electromagnetic waves are a form of energy that can travel through space in the form of oscillating electric and magnetic fields. They are created by the movement of charged particles and can range in wavelength from very long radio waves to very short gamma rays.

2. What are the dangers associated with electromagnetic waves?

Electromagnetic waves can be harmful in high doses, such as in the case of ionizing radiation like X-rays and gamma rays. Prolonged exposure to non-ionizing radiation, such as from cell phones and Wi-Fi, has also been a topic of debate and research, but there is currently no conclusive evidence of negative health effects.

3. How are electromagnetic waves used in everyday life?

Electromagnetic waves have a wide range of uses in everyday life. They are used in communication systems, such as radio and television broadcasting, cell phones, and Wi-Fi. They are also used in medical imaging technologies like MRI and in cooking with microwaves. Additionally, they play a crucial role in the functioning of our planet's ecosystem, such as through photosynthesis and the Earth's magnetic field.

4. What is the difference between electromagnetic waves and light?

Light is just one type of electromagnetic wave, specifically in the visible range of the electromagnetic spectrum. Electromagnetic waves encompass a wider range of wavelengths, including radio waves, microwaves, infrared, ultraviolet, X-rays, and gamma rays. The main difference between light and other electromagnetic waves is the wavelength and frequency.

5. Can electromagnetic waves be blocked or shielded?

Yes, electromagnetic waves can be blocked or shielded with certain materials. Metals are good at blocking electromagnetic waves because they have free electrons that can absorb and dissipate the energy of the waves. Additionally, materials like lead and concrete are commonly used to shield against harmful radiation. However, some electromagnetic waves, like radio waves, can pass through certain materials and are used in communication technologies.

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