#LIGHT INTENSITY EQUATIONS HOW TO#

We will suppose that the ball travels toward the wall with some initial Zero momentum, since it is stationary at the location of the wall. However we picture this, the net result is that the ball ends up with Imagine some very strong double-sided tape on the ball that sticks it onto the To make it a more reasonable scenario, perhaps we could Wall, or that it sticks to it, rather than rebounding. Let’s now imagine instead that somehow the ball gets absorbed by the With the same magnitude of momentum as it initially had when traveling When we first considered this, we imagined that the ball bounced off the wall To see why this is the case, we will go back to our analogy of a ball The radiation pressure is lower than in the case of a perfect reflector. It turns out that, for surfaces that absorb some of the radiation, It is important to remember that our formula for the radiation pressure onlyĪpplies for perfectly reflective surfaces, that is, for surfaces that reflectġ 0 0 % of the incident electromagnetic radiation. Into our radiation pressure equation gives usĮvaluating this expression gives us our answer: the radiation pressure is So substituting 𝑐 and our given value of The question tells us that the surface is perfectly reflective. We are asked to work out the pressure exerted We are told that we have light with an intensity ofĭirected at a surface. Pressure exerted by the light on the surface? Use a value of Is directed at a 1 0 0 % reflective surface. Let’s have a look at an example problem where we need to calculateĮxample 1: Calculating Radiation Pressure for a Perfectly Reflective Surface Then, the radiation pressure, 𝑃, has the unit of Since intensity is the power per area, it has the unit In other words, the energy transferred per unit time. The power of electromagnetic radiation is the rate of energy transfer, Keeping our intensityĮquation in terms of words avoids any potential for confusion. Which we are already using to represent pressure. Is that power is often represented by the letter 𝑃, Note that we have chosen to keep this as a word equation. 𝐼, of electromagnetic radiation is the power transmitted by In our equation for radiation pressure, the intensity, Incident on a perfectly reflective surface, the radiation pressure When electromagnetic radiation of an intensity 𝐼 is Momentum as before the reflection, but now this momentum is in the At this point, it has the same magnitude of After reflecting off the mirror, it is traveling in the 𝑐, toward the mirror and has a certain amount of Let’s consider an electromagnetic wave reflecting off a perfectly Because electromagnetic waves also have momentum, we canįollow a similar line of reasoning to the case of the ball. Just as a wall effectively acts to reflect a ball, a mirror acts to reflect anĮlectromagnetic wave. Wall means the object is exerting a pressure on it.

Newton’s third law means that the objectĮxerts an equal and oppositely directed force on the wall. This change in momentum means that a force from the With momentum collides with something such as a wall, this object experiencesĪ change in momentum. What we have seen from this situation is the following. Where 𝐹 is the force and 𝐴 is the area over Then, we can calculate the pressure on the wall. We now have a force acting all over the wall’s surface, as a result ofĪll of the collisions. Means that it has some momentum, which we will call 𝑝 ,ĭirected toward the wall. The ball has some mass and has a velocity directed toward the wall. We will imagine that the ball travels directly To understand how this works, we will consider an example of an object, such asĪ ball, colliding with a wall. The important thing is simply that they do have momentum, because it is thisįact that means they must exert a pressure on surfaces that they are incident Electromagnetic waves do have a momentum, but it is calculatedįor our purposes, it does not matter precisely how this momentum is defined. Well, the formula 𝑝 = 𝑚 𝑣 only applies for objects that However,Įlectromagnetic waves do not have any mass.

This definition works well when we are considering physical objects. We can recall that the usual definition of momentum is This phenomenon was first predicted by the physicist It turns out that electromagnetic waves transfer momentum We can recall that light, or electromagnetic radiation in general, may be

#LIGHT INTENSITY EQUATIONS HOW TO#

Its axis is perpendicular to the filter on the right (dark area)Īnd parallel to the filter on the left (lighter area).In this explainer, we will learn how to calculate the pressure exerted on an This photograph, a polarizing filter is placed above two others. (c) When the secondįilter is perpendicular to the first, no light is passed. Is rotated, only part of the light is passed. The polarized light is passed by the second polarizing filter,īecause its axis is parallel to the first. Polarizing filters, where the first polarizes the light.

Light intensity equations