If we label the pressure as capital 𝑃, then for a force 𝐹 acting over a surface with an area 𝐴, then we can write this equation in terms of symbols as capital 𝑃 is equal to 𝐹 divided by 𝐴. Now, whenever there’s a force acting over a surface with a particular area, this causes a pressure on the surface equal to the force divided by the area.

In this question, we’ve got light waves exerting a force 𝐹 over a surface with an area 𝐴.

What this means is that when the light waves reflect, they must be exerting a force on the surface. This momentum change means that there is a force, since whenever something’s momentum changes by an amount Δ𝑝 over a time Δ𝑡, there’s a force 𝐹 equal to Δ𝑝 divided by Δ𝑡. This means that if we have a load of light waves like this one colliding with and reflecting off a surface, then those light waves experience a change in momentum. We can recall that even though light waves don’t have mass, they can still transfer momentum. But first, let’s begin by thinking about why it is that the light exerts this force on the surface. We’re being asked to work out the intensity of the light. We’re also told that the force exerted by the light on the surface is 2.00 times 10 to the negative six newtons. We’re told that the area of this surface is 4.00 meters squared. This question is asking us about light reflecting from a surface. What is the intensity of the light? Use a value of 3.00 times 10 to the eight meters per second for the speed of light in vacuum. The light exerts a force of 2.00 times 10 to the negative six newtons on the surface. The surface has an area of 4.00 meters squared. The number of photons will change when the t Δ time is exceeded.Light is directed at a 100 percent reflective surface. The number of photons not changing is directly related to t Δ time. In the figure, pay attention to the fact thatĭ 1 and d 2 distances are determined by t Δ time. Then, an object that receivesĮ 0 amount of energy will receive E 2= E 0.c/(c+v) energy if it is in motion and if it is moving away, and it will receiveĮ 1= E 0.c/(c-v) if it is approaching. Therefore, we can define the equation: E X. The change in the energy will be directly proportional to wavelength change. The fact that the number of photons does not change doesn’t mean that the object will get the same light intensity in all these cases because energies of the photons will change although their numbers stay the same. In case the object comes to the light source, photons will reach the target, which travels at (c-v) speed, atĭ 2 distance and the number of photons will, again, not change. However, this time, the speed of the photons is (c+v) and these photons will arrive at the object atĭ 1 distance instead of d 0 distance and the number of photons will not change. When the object is atĭ 0 distance, n number of photons again choose the object as their arrival target and set out towards it. Now, let’s think of the situation the object moves away. As a result, n number of photons coverĭ 0 distance in t Δ =d 0/c time and reach their target objects. T 0 choose this object as their arrival targets. Assume that n of the photons that are emitted at a time such as The fact that the object is drawn as a round shape is not important. Think of an object that stands still atĭ 0 distance to the light source. We can see how (c+v) (c-v) mathematics interferes in the figure below. When there is movement, (c+v) (c-v) mathematics steps in the situation. However, both equations above are valid for targets that are not in motion relative to light source. Since Light Intensity is a result primarily based on the number of photons that make up the light, we can write the equation above by presenting the real reason as below: There is the following equation in line with this rule. We can think that in practice when we double the distance, the light intensity will decrease fourfold. The general rule is as follows: Intensity of a light that a point source emits around it decreases inversely proportional to the square of the distance. As you move a circle from the light source, amount of light that passes through it in a unit of time decreases.

It is a well-known topic that light intensity changes depending on distance. LIGHT INTENSITY, DISTANCE AND (C+V) (C-V) MATHEMATICS

Light intensity equation