# Physics3, question on intensity, power, wavelength

• Yroyathon
In summary, a scientist is studying the behavior of individual photons and needs to decrease the intensity of her laser beam to a level where there is no more than one photon in her apparatus at any given time. Using the equations for power and energy of a photon, and the fact that the beam must contain 1 photon per L1 m of beam length, the intensity can be calculated by dividing the power by the area, which is equal to the product of the speed of light, Planck's constant, and the speed of light again, divided by the product of the beam length, wavelength, and area.
Yroyathon
hi folks, here's another problem I've made a few incorrect guesses on. On my other post I may have not mentioned the potential for errors in using various SI units, but I think in general I'm careful enough so I'll continue to not worry about that in this post. So assume I know enough to convert grams to kilograms, and nanometers to meters, etc.

## Homework Statement

A scientist wishes to study the behavior of individual photons. To do that, she must decrease the intensity of her i_1 mm2 laser beam -- the laser emits radiation with wavelength w_1 nm -- to a level at which there is no more than one photon in her apparatus at any given time. The path length of the light beam from source to detector is L_1 m. What should be the intensity?

## Homework Equations

for intensity I've used I=P/A, power over area.
for Power of the laser in terms of the photons, I used P = N * E, where N is the number of photons and E is the energy of a photon. for calculating the energy of a photon I used E = h*c/w_1, where h is Planck's constant, c is the speed of light, and w_1 is the wavelength. so in this problem, P = N*6.6*10^(-34)*(3*10^8))/(w_1*10^(-9)).

## The Attempt at a Solution

Hmm. I guess I just realized I wasn't sure what to use for A in the equation for intensity. I think in my first guess I just set A=1, without really thinking.

In my guess I realized I didn't use the distance L_1, which I feel is bad. I wasn't really sure how to incorporate it into the problem.

So I'm considering the quantity of L_1/c, which might be how long it takes a photon to reach the detector? but I'm not certain of this.

I'm not getting this right because using my conceptual understanding of the scenario and the equations i have at hand, I don't have a complete cohesive picture.

Any tips would be appreciated. Thanks.

,Yroyathon

I'm guessing that "in her apparatus" means in the L1 m between the source and the apparatus.

So basically, the beam must contain 1 photon per L1 m of beam length. Can you use that information to say how many photons per second the beam would have?

exactly. I'm thinking the quantity L_1/c is the number of photons per second the beam would have. the units seem to match up too, and intuitively i think this number should be pretty small, as it is.

well that did not work out. hmmm. i'll think on it more, trying to understand it conceptually, but as always any suggestions would be great.

It should have worked out. From L1/c for the number of photons per sec, and also the energy per photon, you can get
power=energy/sec​
and from there use the area to get the intensity. Looking at the units will be a good way to check the final answer.

Aaaack, hold on a minute!

1/L1 is the number of photons per m.

c/L1 is the number of photons per second.

(also, I'm going to rename i_1 as A, which was a misnomer in my head since that figure is not an intensity but an area)

I got it! it turns out that I had that one figure upside down. the number of photons per second the beam should have is very large actually (my intuition was that it should be small).

so N=c/L_1 was what worked out.

So with I=P/A, and P=NE, and E=hc/w_1, and N=c/L_1, i get

I=P/A=NE/A=(c/L_1)*(h*c/w_1)/A=h*c*c/(L_1*w_1*A)

exactly!

## What is intensity in physics?

Intensity in physics refers to the amount of energy that passes through a certain area over a given period of time. It is often measured in watts per square meter (W/m²) and can be used to describe the strength of a wave or the brightness of a light source.

## How is intensity related to power?

Intensity and power are closely related in physics. Power is the rate at which energy is transferred, while intensity is the amount of energy per unit area. Therefore, the higher the power of a wave or light source, the higher the intensity will be.

## What is the formula for calculating intensity?

The formula for intensity is I = P/A, where I is intensity, P is power, and A is the area through which the energy is passing. This means that intensity is directly proportional to power and inversely proportional to area.

## What is the relationship between intensity and wavelength?

The relationship between intensity and wavelength depends on the type of wave. For example, in electromagnetic waves, the intensity is inversely proportional to the square of the wavelength. In sound waves, the intensity is directly proportional to the square of the wavelength.

## How does intensity affect the human body?

High intensities of certain types of waves, such as ultraviolet light or sound waves, can have harmful effects on the human body. For example, high intensity ultraviolet light can cause sunburns and skin cancer, while high intensity sound waves can damage the ear and cause hearing loss. It is important to be aware of the intensity levels of different types of waves and take precautions to protect oneself from potential harm.

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