Problem 56 - Physics for Scientists and Engineers with Modern Physics

Question

(II) A low-power laser used in a physics lab might have a power of 0.50 \(\mathrm{mW}\) and a beam diameter of 3.0 \(\mathrm{mm}\) . Calculate \((a)\) the average light intensity of the laser beam, and \((b)\) compare it to the intensity of a lightbulb emitting15 \(\mathrm{W}\) of light viewed from a distance of 2.0 \(\mathrm{m} .\)

Step-by-Step Solution

Answer

Laser intensity is 70.7 W/m²; bulb intensity at 2m is 0.298 W/m².

Step 1: Calculate the Area of the Laser Beam

The diameter of the laser beam is given as 3.0 mm. To find the area, we first convert this to meters: Diameter = 3.0 mm = 0.003 m.Using the formula for the area of a circle, \( A = \pi \left(\frac{d}{2}\right)^2 \), where \( d \) is the diameter, we find:\[ A = \pi \left(\frac{0.003}{2}\right)^2 = \pi \times (0.0015)^2 = 7.07 \times 10^{-6} \; \text{m}^2.\]

Step 2: Calculate the Average Light Intensity of the Laser

The power of the laser is given as 0.50 mW, which is equivalent to \( 0.50 \times 10^{-3} \text{ W} \). The intensity \( I \) is calculated using the formula \( I = \frac{P}{A} \), where \( P \) is power and \( A \) is area:\[ I = \frac{0.50 \times 10^{-3} \text{ W}}{7.07 \times 10^{-6} \; \text{m}^2} \approx 70.7 \; \text{W/m}^2.\]

Step 3: Calculate the Intensity of the Lightbulb at 2m Distance

The power of the lightbulb is 15 W, and the intensity at a distance \( r \) from a point source is given by \( I = \frac{P}{4\pi r^2} \). Here, the distance is 2 m:\[ I = \frac{15}{4\pi (2)^2} = \frac{15}{16\pi} \approx 0.298 \; \text{W/m}^2.\]

Step 4: Compare the Intensities

The average light intensity of the laser beam is approximately 70.7 W/m², whereas the intensity of the lightbulb at a distance of 2 meters is approximately 0.298 W/m². Thus, the laser beam's intensity is much higher than that of the lightbulb when viewed from 2 meters away.

Key Concepts

Laser Beam Intensity

Laser beam intensity refers to how much power a laser delivers over a specific area. It's essentially how concentrated the laser's light is at a given point. To find the intensity of a laser beam, we use the formula \( I = \frac{P}{A} \), where \( P \) is the power of the laser in watts, and \( A \) is the area over which this power is spread, typically measured in square meters (\( \text{m}^2 \)).

• The power, given in the example, is 0.50 mW, or \( 0.50 \times 10^{-3} \text{ W} \).
• The area is calculated using the area of a circle formula \( A = \pi \left( \frac{d}{2} \right)^2 \).
• In this case, with a diameter of 3.0 mm (or 0.003 m), the area becomes \( 7.07 \times 10^{-6} \; \text{m}^2 \).

Thus, the laser beam intensity amounts to \( 70.7 \; \text{W/m}^2 \). This result shows us that even small lasers can achieve high intensities due to their focused energy on a small area.

Lightbulb Intensity Comparison

When comparing a laser's intensity to a lightbulb, one must consider how their energies are distributed. A lightbulb generally disperses its light uniformly over a large area, unlike the concentrated laser beam.

• The lightbulb in the problem emits 15 W of power, and we examine its intensity at a distance of 2 m.
• Using the formula \( I = \frac{P}{4\pi r^2} \), we substitute \( r = 2 \) meters to find the area over which the light spreads.
• Solving, we find the lightbulb's intensity is approximately \( 0.298 \; \text{W/m}^2 \).

This illustrates that while a laser, such as the one in our example, has a much higher intensity over its small beam area, a lightbulb dissipates its light more, lowering its intensity over larger distances.

Area of a Circle

The area of a circle is an extremely useful formula, especially when dealing with problems involving light and optics. The formula is \( A = \pi \left(\frac{d}{2}\right)^2 \), where \( d \) represents the diameter of the circle.

• Start by finding the radius, which is half the diameter (\( \frac{d}{2} \)).
• Square the radius to find the area part of the formula.
• Multiply by \( \pi \), approximately 3.14159, to complete the calculation.

For example, with a diameter of 3.0 mm, converting to meters gives 0.003 m, and the area calculates to \( 7.07 \times 10^{-6} \; \text{m}^2 \).
Understanding this provides a foundation not only for calculating light intensities but also for other applications like material distribution and geometry.