WEBVTT
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Light is directed at a 100 percent reflective surface.
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The light exerts a pressure of 3.50 times 10 to the negative six newtons per meter squared on the surface.
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What is the intensity of the light?
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Use a value of 3.00 times 10 to the 8 meters per second for the speed of light in vacuum.
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This question is about light reflecting from a surface.
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Weβre told that the light exerts a pressure on the surface of 3.50 times 10 to the negative six newtons per meter squared.
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Letβs label this pressure as capital π.
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In order to understand why the light exerts a pressure like this, letβs recall that even though light waves donβt have mass, they can still transfer momentum.
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Imagine that we have a load of light waves like this one colliding with and reflecting off a surface.
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Then those light waves experience a change in momentum.
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A momentum change β³π that occurs during a time interval β³π‘ means that there is a force πΉ equal to β³π divided by β³π‘.
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So when the light waves reflect, they must be exerting a force on the surface.
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Whenever a force is exerted across a surface with a particular area, this results in a pressure on the surface equal to the force divided by the area.
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So in this way the light waves exerted pressure on the surface.
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This is known as a radiation pressure.
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For a perfectly reflective surface, that is, a surface that reflects 100 percent of the light incident on it, the radiation pressure π exerted on the surface by light with an intensity of πΌ is equal to two times πΌ divided by π, the speed of light.
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In this question, weβre told that the light is directed at a 100 percent reflective surface.
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This means that this equation for the radiation pressure does apply in this case.
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Weβre asked to work out the intensity of the light.
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We know the value of the radiation pressure, and weβre told to take the speed of light as 3.00 times 10 to the 8 meters per second.
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So if we rearrange this equation to make πΌ the subject, we can then sub in our values for the pressure π and the speed of light π to calculate the lightβs intensity.
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In order to make πΌ the subject of the equation, we first need to multiply both sides of the equation by the speed of light π.
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Then on the right-hand side of the equation, the π in the numerator cancels with the π in the denominator.
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This gives us an equation that says two times πΌ is equal to π times π.
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Then we divide both sides of the equation by two.
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On the left-hand side, the two in the numerator cancels with the two in the denominator.
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This gives us an equation that says intensity πΌ is equal to the speed of light π multiplied by the radiation pressure π divided by two.
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We can now take our values for π and π and sub them into this equation.
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Doing that gives us this expression for the intensity πΌ.
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We can notice that the speed of light is given in units of meters per second, which is the SI base unit for speed.
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And the radiation pressure has units of newtons per meter squared, which is the SI base unit for pressure.
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This means that the light intensity weβre going to calculate will be in the SI base unit for intensity.
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This is the watt per meter squared.
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When we evaluate this expression for πΌ, it gives us our answer to the question that the intensity of the light is equal to 525 watts per meter squared.