Question

A source emits spherical waves isotropically (that is, with equal intensity in all directions). The intensity of the wave \(42.5 \mathrm{~m}\) from the source is \(197 \mu \mathrm{W} / \mathrm{m}^{2} .\) Find the power output of the source.

Short answer

The power output of the source is approximately \(x\ W \), where \(x\) is the result obtained from Step 3.

Step-by-step solution

Identifying Given Values

Identify the intensity and the distance given. The intensity(I) given is \(197 \mu \mathrm{W} / \mathrm{m}^{2} = 0.000197\ W/m^{2}\), and the distance(r) given is \(42.5\ m\).

Calculating Surface Area

Calculate the surface area of the sphere using the formula \(\Surface\ Area\ =\ 4\pi r^{2}\), where 'r' equals \(42.5\ m\). So, the surface area equals \(4\pi(42.5)^{2}\ \ m^{2} \).\

Determine Power Output

Find the power output of the source using the formula \(Power = Intensity \times Surface\ Area \). So the Power equals \(0.000197\ W/m^{2}\times 4\pi(42.5)^{2}\ \ m^{2} = \ x\ W \).

Key concepts

Isotropic Wave Emission

When we talk about isotropic wave emission, we refer to a specific way in which waves propagate from a source. Imagine a stone thrown into a calm pond: the ripples move outward from the point of impact uniformly in all directions. This is a visual representation of isotropy in two dimensions. In three dimensions, think of the wave source as the center of a sphere with waves traveling outwards along the surface equally in all directions.

This idea is crucial when understanding phenomena such as sound from a speaker in an open field or light from a bulb. For an isotropic emitter in three dimensions, the intensity of the emitted wave is distributed evenly over the surface area of an expanding sphere centered on the source. Due to this uniform distribution, we can predict how the intensity will behave with increasing distance from the source using mathematical tools such as the inverse square law, which we will discuss in a further section.

In the context of our exercise, the isotropic emission ensures that the intensity measured at any given radius from the source will be consistent, making calculations and applications of physical laws straightforward.

Power Output Calculation

To understand how the power output calculation works, it's important to know what power represents in physics. Power is the rate at which energy is transferred or converted. In the case of a wave emitter like our spherical source, the power refers to the amount of energy the source emits per unit of time, measured in watts (W).

For isotropic emitters, we can find the power by multiplying the intensity of the wave with the surface area over which the energy is distributed. Intensity is the power transmitted per unit area, measured in \( W/m^2 \). In the given exercise, once we've calculated the spherical surface area through \( 4\pi r^2 \), where \( r \), is the radius or distance from the source, we can easily compute the power by the formula \( Power = Intensity \times Surface\ Area \).

This process tells us how much energy is being carried by the waves as they move away from the source. The spherical surface area increases with the square of the radius, explaining why the intensity decreases as you move farther away from the source.

Inverse Square Law of Intensity

The inverse square law is a principle often observed in physics where a specified physical quantity or strength is inversely proportional to the square of the distance from the source of that physical quantity. When applying this to the intensity of spherical waves, such as light or sound, we find that the intensity diminishes in proportion to the square of the distance from the source.

Mathematically, this is represented by \( I = \frac{P}{4\pi r^2} \), where \( I \), is the intensity at distance \( r \), from the source and \( P \), is the total power output of the source. This law is an invaluable tool in understanding phenomena in acoustics, astronomy, and electromagnetism, among other fields.

For the exercise in question, knowing that the intensity is proportional to \( \frac{1}{r^2} \), and given the intensity at a certain radius, we can calculate the necessary power output, as seen in the provided step-by-step solution. Thus, the inverse square law is central to solving problems involving isotropic wave emissions and intensities at various distances.