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Laser Safety. If the eye receives an average intensity greater than \(1.0 \times 10 ^ { 2 } \mathrm { W } / \mathrm { m } ^ { 2 }\) , damage to the retina can occur. This quantity is called the damage threshold of the retina. (a) What is the largest average power (in mW) that a laser beam 1.5 \(\mathrm { mm }\) in diameter can have and still be considered safe to view head-on? (b) What are the maximum values of the electric and magnetic fields for the beam in part (a)? (c) How much energy would the beam in part (a) deliver per second to the retina? (d) Express the damage threshold in W/cm?

Short Answer

Expert verified
(a) 0.177 mW; (b) \( E_{max} \approx 274.56 \text{ V/m}, B_{max} \approx 9.15 \times 10^{-7} \text{ T} \); (c) 0.177 J/s; (d) 0.01 \( \text{W/cm}^2 \).

Step by step solution

01

Determine Safe Power for Part (a)

To find the largest safe average power, we first calculate the area of the laser beam using the formula for the area of a circle: \( A = \pi \left( \frac{d}{2} \right)^2 \). For a diameter \( d = 1.5 \, \text{mm} = 1.5 \times 10^{-3} \, \text{m} \), this becomes:\[ A = \pi \left( \frac{1.5 \times 10^{-3}}{2} \right)^2 = \pi \times (0.75 \times 10^{-3})^2 \approx 1.77 \times 10^{-6} \, \text{m}^2. \]The largest average power, \( P \), that is safe, is given by the intensity times the area:\[ P = I \times A = 100 \, \text{W/m}^2 \times 1.77 \times 10^{-6} \, \text{m}^2 \approx 1.77 \times 10^{-4} \, \text{W} = 0.177 \, \text{mW}. \]
02

Calculate Maximum Electric and Magnetic Fields for Part (b)

For the maximum electric field, \( E_{max} \), the relationship between intensity \( I \) and the electric field is given by:\[ I = \frac{1}{2} c \varepsilon_0 E^2 \] where \( c = 3 \times 10^8 \, \text{m/s} \) is the speed of light and \( \varepsilon_0 = 8.85 \times 10^{-12} \, \text{F/m} \) is the permittivity of free space.Solving for \( E \), we have:\[ E_{max} = \sqrt{\frac{2I}{c \varepsilon_0}} = \sqrt{\frac{2 \times 100}{3 \times 10^8 \times 8.85 \times 10^{-12}}} \approx 274.56 \, \text{V/m}. \]The magnetic field \( B_{max} \) is found using \( B = \frac{E}{c} \):\[ B_{max} = \frac{274.56}{3 \times 10^8} \approx 9.15 \times 10^{-7} \, \text{T}. \]
03

Calculate Energy Delivered to Retina for Part (c)

Energy delivered by the beam per second is simply the power calculated in Part (a). Therefore:\[ \,E = P = 0.177 \, \text{mW} = 0.177 \, \text{J/s}. \]
04

Express Damage Threshold in W/cm² for Part (d)

To convert the intensity from \( \text{W/m}^2 \) to \( \text{W/cm}^2 \), note that 1 \( \text{m}^2 = 10^4 \text{cm}^2 \). Thus:\[ \text{Intensity in W/cm}^2 = \frac{100 \, \text{W/m}^2}{10^4} = 0.01 \, \text{W/cm}^2. \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Damage Threshold
The damage threshold in the context of laser safety refers to the level of intensity that a laser beam can possess before causing harm to biological tissues, such as the retina. It's a vital measure for regulating laser use, especially in environments requiring direct eye exposure. In this exercise, the danger threshold is given as \(1.0 \times 10 ^ { 2 } \text{ W/m}^2\).
This threshold determines the maximum safe limit of laser exposure for the eye, preventing retinal damage. Understanding the damage threshold is crucial for designing and using lasers safely, ensuring they operate within limits protective to human health.
Electric Field
The electric field component of a laser beam plays a significant role in its behavior and interaction with materials. It is a part of the electromagnetic wave, which in this context, is crucial to determining the beam's intensity and potential impact.
The relationship between intensity \( I \) and electric field \( E \) is given by:
\[ I = \frac{1}{2} c \varepsilon_0 E^2 \]
where \( c \) is the speed of light and \( \varepsilon_0 \) is the permittivity of free space.
By rearranging and solving for \( E \), we can determine the maximum electric field for a safe laser beam, given the intensity does not exceed the damage threshold. In this example, it is calculated as approximately 274.56 V/m.
Magnetic Field
In a laser beam, the magnetic field is another component of the electromagnetic wave, acting perpendicular to the electric field. Though generally not as influential in the effects we directly notice, like heating or movement induced by light, it's crucial for the full understanding of electromagnetic waves.
We can express the magnetic field \( B \) as a function of the electric field \( E \) using:
\[ B = \frac{E}{c} \]
This shows the interdependence of these fields within the laser beam. In the given problem, for an electric field of approximately 274.56 V/m, the magnetic field would be around 9.15 \times 10^{-7} \text{ T}. Such calculations ensure we recognize the laser beam's full characteristics while considering safety.
Energy Delivery
Energy delivery by a laser beam is the total energy that it transfers to a surface, like the retina, over time. In laser safety, understanding energy delivery is critical to prevent exceeding the damage threshold.
The energy delivery rate, or power, of a laser can be calculated using its average intensity and the area it impacts. Given a safe power level calculated in the exercise as 0.177 mW or 0.177 J/s, this tells how much energy is imparted per second to the retina when exposed to such intensity.
  • This insight is fundamental for designing laser devices that remain within safety compliance.
It ensures protective measures can be accurately assessed and deployed.
Intensity Conversion
Intensity conversion involves changing the units used to express the laser's intensity, which is necessary for comparisons and applications in different contexts.
In this exercise, the provided intensity is in \( \text{W/m}^2 \), but it is useful to convert it to \( \text{W/cm}^2 \) for practical applications or standards compliance.
Since 1 \( \text{m}^2 = 10^4 \text{cm}^2 \), converting the initial intensity, \( 100 \, \text{W/m}^2 \), to \( \text{W/cm}^2 \), we get:
\[ \frac{100 \, \text{W/m}^2}{10^4} = 0.01 \, \text{W/cm}^2 \]
This ease in conversion facilitates global use of data and regulatory standards, ensuring a straightforward comparison of safety thresholds across various measurement units.

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Most popular questions from this chapter

An electromagnetic wave has an electric field given by \(\vec { \boldsymbol { E } } ( y , t ) = \left( 3.10 \times 10 ^ { 5 } \mathrm { V } / \mathrm { m } \right) \hat { \boldsymbol { k } } \cos \left[ k y - \left( 12.65 \times 10 ^ { 12 } \mathrm { rad } / \mathrm { s } \right) t \right]\) (a) In which direction is the wave traveling? (b) What is the wave-length of the wave? (c) Write the vector equation for \(\vec { \boldsymbol { B } } ( y , t )\)

A small helium-neon laser emits red visible light with a power of 4.60\(\mathrm { mW }\) in a beam that has a diameter of 2.50\(\mathrm { mm }\) . (a) What are the amplitudes of the electric and magnetic fields of the light? (b) What are the average energy densities associated with the electric field are the magnetic field? (c) What is the total energy contained in a \(1.00 - \mathrm { m }\) length of the beam?

The electric-field component of a sinusoidal electromagnetic wave traveling through a plastic cylinder is given by the equation \(E = ( 5.35 \mathrm { V } / \mathrm { m } ) \cos \left[ \left( 1.39 \times 10 ^ { 7 } \mathrm { rad } / \mathrm { m } \right) x - ( 3.02 \times \right.\) \(10 ^ { 15 } \mathrm { rad } / \mathrm { s } ) t ]\) . (a) Find the frequency, wavelength, and speed of this wave in the plastic. (b) What is the index of refraction of the plastic? (c) Assuming that the amplitude of the electric field does not change, write a comparable equation for the electric field if the light is traveling in air instead of in plastic.

Electromagnetic waves propagate much differently in conductors than they do in dielectrics or in vacuum. If the resistivity of the conductor is sufficiently low (that is, it is a sufficiently good conductor), the oscillating electric field of the wave gives rise to an oscillating conduction current that is much larger than the displacement current. In this case, the wave equation for an electric field \(\vec { E } ( x , t ) = E _ { y } ( x , t ) \hat { J }\) propagating in the \(+ x\) -direction within a conductor is $$\frac { \partial ^ { 2 } E _ { y } ( x , t ) } { \partial x ^ { 2 } } = \frac { \mu } { \rho } \frac { \partial E _ { y } ( x , t ) } { \partial t }\( where \)\mu$$ is the permeability of the conductor and \(\rho\) is its resistivity. (a) A solution to this wave equation is $$E _ { y } ( x , t ) = E _ { \max } e ^ { - k _ { c } x } \cos \left( k _ { C } x - \omega t \right)$$ where \(k _ { \mathrm { C } } = \sqrt { \omega \mu / 2 \rho } .\) Verify this by substituting \(E _ { y } ( x , t )\) into the above wave equation. (b) The exponential term shows that the electric field decreases in amplitude as it propagates. Explain why this happens. (Hint: The field does work to move charges within the conductor. The current of these moving charges causes \(i ^ { 2 } R\) heating within the conductor, raising its temperature. Where does the energy to do this come from? \(( \mathrm { c } )\) Show that the electric-field amplitude decreases by a factor of 1\(/ e\) in a distance \(1 / k _ { \mathrm { C } } = \sqrt { 2 p / \omega \mu } ,\) and calculate this distance for a radio wave with frequency \(f = 1.0 \mathrm { MHz }\) in copper (resistivity \(1.72 \times 10 ^ { - 8 } \Omega \cdot\) m; permeability \(\mu = \mu _ { 0 } )\) . since this distance is so short, electromagnetic waves of this frequency can hardly propagate at all into copper. Instead, they are reflected at the surface of the metal. This is why radio waves can- not penetrate through copper or other metals, and why radio reception is poor inside a metal structure.

A source of sinusoidal electromagnetic waves radiates uniformly in all directions. At 10.0\(\mathrm { m }\) from this source, the amplitude of the electric field is measured to be 1.50\(\mathrm { N } / \mathrm { C }\) . What is the electric-field amplitude at a distance of 20.0\(\mathrm { cm }\) from the source?

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