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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 \(\mathrm{mW}\) ) 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 \(\mathrm{W} / \mathrm{cm}^{2} .\)

Short Answer

Expert verified
(a) 0.1767 mW (b) \(E = 274.3\, \text{V/m}, B = 9.143 \times 10^{-7}\, \text{T}\) (c) 0.1767 mJ/s (d) \(1.0 \times 10^{-2}\, \text{W/cm}^2\)

Step by step solution

01

Understanding the Safety Limit

The damage threshold for the retina is given as an intensity of \(1.0 \times 10^2 \text{ W/m}^2\). This means that the laser beam can have an average intensity up to \(1.0 \times 10^2 \text{ W/m}^2\) to be safely viewed head-on.
02

Calculating Safe Power for the Laser Beam

First, we calculate the area of the laser beam using the formula for the area of a circle: \(A = \pi r^2\). Since the diameter is \(1.5 \text{ mm}\), the radius \(r\) is \(0.75 \text{ mm} = 0.00075 \text{ m}\). Now, \[ A = \pi (0.00075)^2 \approx 1.767 \times 10^{-6} \text{ m}^2. \] Using the intensity formula \(I = \frac{P}{A}\), where \(P\) is power, we rearrange to find \(P = IA\). Substituting the values: \[ P = (1.0 \times 10^2 \text{ W/m}^2)(1.767 \times 10^{-6} \text{ m}^2) \approx 1.767 \times 10^{-4} \text{ W}. \] Converting to milliwatts (1 W = 1000 mW): \(P \approx 0.1767 \text{ mW}\).
03

Electric and Magnetic Fields for the Safe Power

The intensity of an electromagnetic wave is given by \(I = \frac{1}{2}\epsilon_0 c E^2\) for the electric field, where \(\epsilon_0\) is the permittivity of free space \((8.854 \times 10^{-12} \text{ C}^2/ ext{N}\cdot\text{m}^2)\), and \(c\) is the speed of light \((3 \times 10^8 \text{ m/s})\). Solving for \(E\) gives: \[ E = \sqrt{\frac{2I}{c\epsilon_0}} \approx \sqrt{\frac{2 \times 1.0 \times 10^2}{3 \times 10^8 \times 8.854 \times 10^{-12}}} \approx 274.3 \text{ V/m}. \] The magnetic field \(B\) is related to \(E\) by \(B = \frac{E}{c}\), so: \[ B \approx \frac{274.3}{3 \times 10^8} \approx 9.143 \times 10^{-7} \text{ T}. \]
04

Energy Delivered per Second

The energy delivered per second is simply the power, which we found to be approximately \(0.1767 \text{ mW}\) or \(1.767 \times 10^{-4} \text{ J/s}\), as energy per second in watts is the same as power in joules per second.
05

Expressing Damage Threshold in Different Units

To convert \(1.0 \times 10^2 \text{ W/m}^2\) to \(\text{W/cm}^2\), note that \(1 \text{ m}^2 = 10^4 \text{ cm}^2\), thus \[ \frac{1.0 \times 10^2 \text{ W/m}^2}{10^4 \text{ cm}^2} = 1.0 \times 10^{-2} \text{ W/cm}^2. \]

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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 refers to the maximum intensity of a laser beam that the eye can safely withstand without suffering injury. For the human retina, this value is set at \(1.0 \times 10^2 \; \mathrm{W/m}^2\). This number indicates the level of light energy per unit area that is considered safe. When this threshold is exceeded, the intense laser light can cause permanent damage to the delicate tissues of the retina, potentially leading to vision impairment. It is crucial to understand and respect this boundary to prevent eye injuries. The specification of the damage threshold is a protective guideline for users of laser devices, ensuring they avoid exposure beyond safe limits.
Intensity Calculation
To figure out how much power a laser beam can safely have, one must calculate the beam's intensity. Intensity is a measure of power per unit area and is calculated with the formula \(I = \frac{P}{A}\), where \(I\) is intensity, \(P\) is power, and \(A\) is the area. First, determine the area of the beam's cross-section if the diameter is given. For a beam with a diameter of \(1.5\; \mathrm{mm}\), the radius would be half of that, \(0.75\; \mathrm{mm}\), which converts to \(0.00075\; \mathrm{m}\). The area \(A\) of the beam can be calculated using the formula for the area of a circle: \(A = \pi r^2\). Once the area is known, you can find the maximum safe power \(P\) that the beam can have while staying within the damage threshold: \[ P = I \cdot A. \] In our case, keeping the power under \(1.767 \times 10^{-4} \; \mathrm{W}\) (or \(0.1767\; \mathrm{mW}\)) ensures safety. Ensuring these calculations are correct protects the user from exceeding the safe intensity level.
Electric and Magnetic Fields
In laser beams, like all electromagnetic waves, electric and magnetic fields are present. Their magnitudes can be calculated using the beam's intensity. The electric field's maximum value, \(E\), is determined by the relation \(I = \frac{1}{2}\epsilon_0 c E^2\), where \(\epsilon_0\) is the permittivity of free space, and \(c\) is the speed of light. Solving for \(E\) gives:
  • \(E = \sqrt{\frac{2I}{c\epsilon_0}}\)
This allows us to calculate \(E\) once \(I\) is known. The magnetic field, \(B\), is related to \(E\) by the relation \(B = \frac{E}{c}\). These calculations are essential for understanding the full effects of laser radiation, as both fields can interact with matter in different ways. By determining these fields, you can ensure that the laser operates within safe limits for both electric and magnetic effects, preventing possible hazards.
Energy Delivery to Retina
The energy that a laser beam delivers to the retina per second is equivalent to the power of the beam. For our laser, this is approximately \(0.1767\; \mathrm{mW}\), which is the same as \(1.767 \times 10^{-4} \; \mathrm{J/s}\). This calculation is important to assess how much energy the eye can receive without damage. In practical terms, the energy delivery per second should not surpass the power determined to be safe. By calculating the energy delivery correctly, users can verify that no excess radiation is reaching the eye, maintaining safety according to established threshold limits. This is especially crucial for scenarios requiring direct observation of laser beams, such as in scientific experiments or medical treatments, where precision and safety are paramount. Understanding energy delivery helps prevent accidental overexposure and protects eye health.

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Most popular questions from this chapter

A sinusoidal electromagnetic wave emitted by a cellular phone has a wavelength of \(35.4 \mathrm{~cm}\) and an electric-field amplitude of \(5.40 \times 10^{-2} \mathrm{~V} / \mathrm{m}\) at a distance of \(250 \mathrm{~m}\) from the phone. \(\mathrm{Cal}-\) culate (a) the frequency of the wave; (b) the magnetic-field amplitude; (c) the intensity of the wave.

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