Light

Problem : Compute the energy density of a 2.0 mW laser beam with a diameter of 2mm.

Solution for Problem 1 >>

The energy outputted by the laser per unit time (second) is 0.002 J/s. The area of the beam is Πr ² = Π×(0.001)² = 3.14×10^-6 m ² . We want the energy per volume, so just by dimensional analysis we can say J/m ² = J/s×1/m ²×s/m , so the energy density is given by u = power×1/area×1/c = 0.002× = 2.12×10^-6 J/m ³ .

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Problem : The power per unit area incident on the earth due to sunlight is about 1.4 kW/m ² . What is the approximate magnitude of the electric field vector on sunlight near the earth ( μ ₀âÉá4Π×10^-7 N.s ² /C ² )?

Solution for Problem 2 >>

Since the electric and magnetic field vectors are perpendicular, we can write the expression for the incident power per unit area as S = . We can assume the fields are of equal magnitude, so E ² E = 4.19×10^-2 V/m.

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Problem : Write an expression (in SI units) for the magnetic field of a 10 MHz light wave traveling in the negative z direction and an electric field amplitude of .

Solution for Problem 3 >>

First let us consider the direction of the wave. We know that must yield the direction and is given as in the positive x direction. Thus must point in the negative y direction by the right hand rule. Thus we want an equation of the form . The amplitude must be given by B ₀ = E ₀/c = 0.1/3.0×10⁸ = 3.33×10^-10 . σ = 2Πν = 2Π×10⁷ and k = 2Π/λ = 2Πν/c = 2Π×0.033 . So the expression is:

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Problem : Consider the following electric and magnetic waves:

		=
		=

Solution for Problem 4 >>

First we have â??. which is clearly satisfied since the magnetic field is traveling in the y -direction. Similarly â??. . Now â??× , since the magnetic field only has a component in the x -direction, and this is the only non-zero partial derivative. This should be equal to ?ü ₀ ε ₀ = . These components are equal if and only if = E ₀ ?ü ₀ ε ₀ . Similarly â??× . This needs to be equal to - = . This is true if E ₀/v = B ₀ . Now ?ü ₀ ε ₀ = 1/c ² , so the first condition gives us that B ₀/v = E ₀/c ² . Combining with the second condition we have 1/v = E ₀/(c ² B ₀) = v/c ² . Thus the condition is satisfied if v = ±c . The two conditions required, then, are v = ±c and E ₀ = cB ₀ .

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Problem : Consider a 60W light bulb. Assume that all radiation given off is in the form of light of wavelength 650 nm. Calculate the number of photons given off per second.

Solution for Problem 5 >>

Each second the total energy given off is 60 Joules. Each photon has an energy given by E = hν where ν = c/λ = 3.0×10⁸/(650×10^-9) = 4.62×10¹⁴ Hz. So E = 6.626×10^-34×4.62×10¹⁴ = 3.06×10^-19 Joules per photon. Thus the total number of photons must be 60/(3.06×10^-19) 1.96×10²⁰ photons emitted every second!

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