Question

The intensity from sunlight on a bright sunny day is around \(0.1 \frac{\mathrm{W}}{\mathrm{cm}^{2}}\). Laser power can be confined to a very small spot size. Assume a laser produces a beam with spot size \(1 \mathrm{~mm}^{2}\). For what laser power in watts will the intensity of the beam be equivalent to the intensity from sunlight on sunny day? Staring at the sun can damage an eye, so staring at a laser beam of this intensity is dangerous for the same reason.

Short answer

The laser power needed is 1 mW to have the same intensity as sunlight on a bright day.

Step-by-step solution

Understand the Given Data

We are given that the intensity of sunlight on a bright sunny day is \(0.1 \frac{\text{W}}{\text{cm}^2}\), and the laser's spot size is \(1\text{ mm}^2\). We need to calculate the power of the laser that will produce the same intensity as sunlight.

Convert Units of Spot Size

The laser's spot size is given in square millimeters and must be converted into square centimeters because the intensity is given in \(\text{cm}^2\). We use the conversion \(1\text{ cm}^2 = 100\text{ mm}^2\), so \(1 \text{ mm}^2 = 0.01 \text{ cm}^2\).

Set Up the Equation for Intensity

The formula for intensity is \(\text{Intensity} = \frac{\text{Power}}{\text{Area}}\). We want the laser's intensity, \(\frac{P}{0.01 \text{ cm}^2}\), to equal the intensity of sunlight, \(0.1 \frac{\text{W}}{\text{cm}^2}\).

Solve for Laser Power

By setting the laser intensity equal to the sunlight intensity, we have:\[ \frac{P}{0.01} = 0.1 \]Solving for \(P\), we multiply both sides by \(0.01\):\[ P = 0.1 \times 0.01 = 0.001 \text{ W} \]

Interpret the Result

The laser power required to match the intensity of sunlight is \(0.001 \text{ W}\), or \(1\text{ mW}\). This intensity can be harmful to the eye, similar to staring at the sun.

Key concepts

Sunlight Intensity

Sunlight intensity is a measure of how much solar energy is received over a specific area. On a bright sunny day, this intensity can reach approximately \(0.1 \frac{\mathrm{W}}{\mathrm{cm}^{2}}\).
Since this intensity specifies energy over a specific area, it's a useful measure to understand the energy density of light coming from the sun.
This value represents natural sunlight when the sun is bright and directly overhead, such as around midday.

• This intensity is crucial for understanding the solar energy available for applications such as solar panels and light exposure on plants.
• It also serves as a reference point when dealing with artificial light sources such as lasers, which can be more intense but concentrated over a smaller area.

Power and Area Conversion

To compare laser power to sunlight intensity, we need to understand power and area conversions. In this exercise, we convert the area of the laser's spot. The given spot size is \(1 \mathrm{~mm}^{2}\), which we convert to square centimeters because the intensity is given in \(\text{cm}^2\).
Remember, converting units ensures consistency in calculations. Here, the conversion rate is \(1 \text{ cm}^2 = 100 \text{ mm}^2\), thus \(1 \text{ mm}^2 = 0.01 \text{ cm}^2\).

• This step allows us to apply the intensity formula \(\text{Intensity} = \frac{\text{Power}}{\text{Area}}\) accurately.
• Such conversions are standard practices in physics to maintain uniform units across calculations.

Using correct conversions enhances calculation accuracy and the understanding of how area size affects intensity calculations.

Safety in Optics

Safety in optics is paramount when dealing with intense light sources, including lasers. The exercise indicates that the intensity of a laser matching sunlight can be dangerous. Staring into a laser beam with such intensity, even for a short period, poses significant risks to eye health, similar to staring directly at the sun.

• Both sunlight and laser beams at this intensity can cause permanent eye damage.
• Lasers produce coherent light, meaning the light waves travel in unison and are more focused, making them particularly dangerous over small areas.

Preventive measures include:

• Never looking directly into laser beams.
• Using appropriate eye protection, especially in laboratory settings.
• Being aware of reflections, as reflected beams can also pose risks.

In optics, understanding the power and intensity of light sources is crucial not just theoretically, but also for ensuring safety.