Question

(a) You direct a 16.0 mW laser beam toward a surface at normal incidence. The surface is made of a material that completely absorbs the radiation. What is the radiation pressure on the surface if the cross section of the beam has a diameter of 2.85 mm when it arrives at the surface? How is the pressure related to the intensity of the incident beam? How is the intensity related to the power and cross sectional area? N/m² (b) You now direct the same laser beam, with the same power and beam cross section, at normal incidence on a mirror where 100% of the beam is reflected from the surface. What is the total pressure on it due to the incident and reflected energies? By what factor will the pressure change when 100% of the beam is reflected? N/m²

          (a) You direct a 16.0 mW laser beam toward a surface at normal incidence. The surface is made of a material that completely absorbs the radiation. What is the radiation pressure on the
surface if the cross section of the beam has a diameter of 2.85 mm when it arrives at the surface?

How is the pressure related to the intensity of the incident beam? How is the intensity related to the power and cross sectional area? N/m²
(b) You now direct the same laser beam, with the same power and beam cross section, at normal incidence on a mirror where 100% of the beam is reflected from the surface. What is
the total pressure on it due to the incident and reflected energies?

By what factor will the pressure change when 100% of the beam is reflected? N/m²
        
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(a) You direct a 16.0 mW laser beam toward a surface at normal incidence. The surface is made of a material that completely absorbs the radiation. What is the radiation pressure on the
surface if the cross section of the beam has a diameter of 2.85 mm when it arrives at the surface?

How is the pressure related to the intensity of the incident beam? How is the intensity related to the power and cross sectional area? N/m²
(b) You now direct the same laser beam, with the same power and beam cross section, at normal incidence on a mirror where 100% of the beam is reflected from the surface. What is
the total pressure on it due to the incident and reflected energies?

By what factor will the pressure change when 100% of the beam is reflected? N/m²

Added by Albert W.

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University Physics with Modern Physics
University Physics with Modern Physics
Hugh D. Young 14th Edition
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please help me with (a) and (b) You direct a 6.0 mW aoe beam toward a surface at normal (ncidence, The surface is made of a matenat that conpletely absorbs the radiation what is the radt suface tr tha cross section af the beam has a diametec of .s mm when it arrives at the surtace sureonthe SYou now drect the same r bmwith the same power and beam ro ctionat normal incidence on a mrorwhe 100 f the bam i rcted rom the utace wnat the total presure on it due to the incdent and reflected ener
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Let the total force exerted by the beam on the target surface be $F$ and the power liberated there be $P$. Then, using the result of the previous problem we see $$ F=N p=\frac{N}{c} \sqrt{T\left(T+2 m_{0} c^{2}\right)}=\frac{I}{e c} \sqrt{T\left(T+2 m_{0} c^{2}\right)} $$ since $I=N e, N$ being the number of particles striking the target per second. Also, $$ P=N\left(\frac{m_{0} c^{2}}{\sqrt{1-v^{2} / c^{2}}}-m_{0} c^{2}\right)=\frac{I}{e} T $$ These will be, respectively, equal to the pressure and power developed per unit area of the target if $I$ is current density.

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Transcript

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00:02 Okay, so the laser beam of problem 64 has power of 2 .5 millawatts and a diameter of 3 .4 millimeters.
00:23 So in this problem, we want to calculate how much energy does a surface there was absorbing the light.
00:37 Receive in 12 seconds.
00:39 So we have a time of 12 seconds and we want to calculate how much energy is absorbed.
00:50 So okay let's remember that the energy for definition is the it's just the power multiply by the time.
01:04 Therefore we can say that the power of this laser which is 2 .5 times 10 to the minus 3 because we have a milli watts in here remember mealy watts in here is going to be multiply by 12 seconds therefore we have that the energy absorber by the surface is just 0 .03 jails that's the answer to the first item the second item we want to calculate the radiation pressure exerted by the beam.
01:46 So, okay, this is a little difficult, but we just need to go in the chapter and get back the equation.
01:54 So we know that the pressure, pressure is going to be equal the average intensity divided by c...
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