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Chapter 34: Problem 46

At a beach the light is generally partially polarized due to reflections off sand and water. At a particular beach on a particular day near sundown, the horizontal component of the electric field vector is \(2.3\) times the vertical component. A standing sunbather puts on polarizing sunglasses; the glasses eliminate the horizontal field component. (a) What fraction of the light intensity received before the glasses were put on now reaches the sunbather's eyes? (b) The sunbather, still wearing the glasses, lies on his side. What fraction of the light intensity received before the glasses were put on now reaches his eyes?

Short Answer

Expert verified
a) 15.9% b) 84.1%

Step by step solution

01

- Define the initial electric field components

Let the vertical component of the electric field be represented as \(E_v\) and the horizontal component as \(E_h\). According to the problem, \(E_h = 2.3 E_v\).
02

- Calculate the initial light intensity

The intensity of light is proportional to the square of the magnitude of the electric field components. Thus, the total initial intensity \(I_i\) is given by:\[I_i = E_h^2 + E_v^2\]Substituting \(E_h = 2.3 E_v\), we get:\[I_i = (2.3 E_v)^2 + E_v^2 = 5.29 E_v^2 + E_v^2 = 6.29 E_v^2\]
03

- Determine the light intensity with sunglasses (horizontal component eliminated)

When the horizontal component \(E_h\) is eliminated by the polarizing sunglasses, the remaining light intensity is only due to the vertical component \(E_v\):\[I_r = E_v^2\]
04

- Calculate the fraction of light intensity now reaching the eyes (sunglasses on)

The fraction of the original intensity that reaches the eyes is given by:\[\text{Fraction} = \frac{I_r}{I_i} = \frac{E_v^2}{6.29 E_v^2} = \frac{1}{6.29} \approx 0.159\]So, approximately 15.9% of the initial light intensity reaches the sunbather's eyes.
05

- Light intensity when lying on the side

When the sunbather lies on his side, the orientation of the polarizing sunglasses changes. Now, the horizontal component originally becomes vertical. Thus, the intensity that reaches the eyes is again considering the vertical component which is now \(2.3 E_v\):\[I_r = E_h^2 = (2.3 E_v)^2 = 5.29 E_v^2\]
06

- Calculate the fraction of light intensity now reaching the eyes (lying on the side)

The new fraction of the original intensity now reaching the eyes while lying on the side is:\[\text{Fraction} = \frac{I_r}{I_i} = \frac{5.29 E_v^2}{6.29 E_v^2} = \frac{5.29}{6.29} \approx 0.841\]So, approximately 84.1% of the initial light intensity reaches the sunbather's eyes when he lies on his side.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Electric Field Components
Light is an electromagnetic wave that has both electric and magnetic field components. For simplicity, we often focus on the electric field, as it plays a crucial role in many optical phenomena. The electric field in light waves can be broken down into vertical (E_v) and horizontal (E_h) components. Each component interacts differently with materials and influences how light behaves when it encounters reflections or polarizing filters. In this scenario, understanding the relationship between these components is key to solving the problem. At the beach, the horizontal component (E_h) of the electric field is 2.3 times stronger than the vertical component (E_v).
Light Intensity
Light intensity is the measure of the power or energy of light received per unit area. It's directly proportional to the square of the magnitude of the electric field components. Mathematically, if you have a total electric field magnitude, the intensity (I) is given by:I = E^2=(E_h^2 + E_v^2dot E_v^2). For this situation, the initial intensity (I_i) is a combination of both the vertical and horizontal components of the electric field. Before polarizing sunglasses are applied, the sunbather receives the combined intensity from both components, which we have calculated as:I_i = 6.29 E_v^2.
Polarizing Sunglasses
Polarizing sunglasses are designed to block certain orientations of light waves. This characteristic is especially useful in reducing glare from reflective surfaces like water or sand. In our case, the sunglasses eliminate the horizontal component (E_h) of the electric field. Thus, once the sunglasses are on, only the vertical component (E_v) contributes to the light intensity received by the eyes. It's important to note that this elimination leads to a significant drop in perceived light intensity, calculated to be about 15.9% of the original intensity.
Vertical and Horizontal Components
The vertical and horizontal components of the electric field (E_v and E_h) describe how the light wave oscillates in different directions. When light reflects, particularly from horizontal surfaces like the ground or water, it increases its horizontal component. In this exercise, the horizontal component (E_h) is 2.3 times the vertical component (E_v). The sunglasses that only allow vertical components to pass through exploit this property by reducing glare and improving visual comfort.
Light Reflection
When light reflects off surfaces like sand or water, it becomes partially polarized, mainly strengthening its horizontal component due to the angle of reflection. This principle is why polarizing sunglasses are so effective at beaches: they block this horizontally polarized light, reducing glare. In the given scenario, reflections cause the horizontal component of the reflected light to be stronger, making E_h = 2.3 E_v. By accounting for this, we understand how much light intensity is reduced when specific components are filtered out.

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Most popular questions from this chapter

High-power lasers are used to compress a plasma (a gas of charged particles) by radiation pressure. A laser generating pulses of radiation of peak power \(1.5 \mathrm{GW}\) is focused onto \(1.0 \mathrm{~mm}^{2}\) of high-electron- density plasma. Find the pressure exerted on the plasma if the plasma reflects all the light pulses directly back along their paths.

Galileo tried to measure the speed of light by having two people stand on hills about \(5 \mathrm{~km}\) apart. Each would hold a shuttered lantern. The first would open his lantern and when the second saw the light, he would open his lantern. The first person would then measure how much time it took between the time he first opened his lantern and the time he saw the light returning. (a) How much time would it take the light to travel between the two hills? (b) Is this a good way to measure the speed of light? Support your argument with a brief explanation that includes some quantitative discussion of the uncertainty in the measurement.

An electromagnetic wave with frequency 400 terahertz travels through vacuum in the positive direction of an \(x\) axis. The wave is polarized, with its electric field directed parallel to the \(y\) axis, with amplitude \(E^{\max }\). At time \(t=0\), the electric field at point \(P\) on the \(x\) axis has a value of \(+E^{\max } / 4\) and is decreasing with time. What is the distance along the \(x\) axis from point \(P\) to the first point with \(E=0\) if we search in (a) the negative direction and (b) the positive direction of the \(x\) axis?

The intensity of direct solar radiation th?t is not absorbed by the atmosphere on a particular summer day is \(100 \mathrm{~W} / \mathrm{m}^{2}\). How close would you have to stand to a \(1.0 \mathrm{~kW}\) electric heater to feel the same intensity? Assume that the heater radiates uniformly in all directions.

A plane electromagnetic wave, with wavelength \(3.0 \mathrm{~m}\), travels in vacuum in the positive \(x\) direction with its electric field \(\vec{E}\), of amplitude \(300 \mathrm{~V} / \mathrm{m}\), directed along the \(y\) axis. (a) What is the frequency \(f\) of the wave? (b) What are the direction and amplitude of the magnetic field associated with the wave? (c) What are the values of \(k\) and \(\omega\) if \(\vec{E}=\vec{E}^{\max } \sin (k x-\omega t) ?\) (d) What is the time-averaged rate of energy flow in watts per square meter associated with this wave? (e) If the wave falls on a perfectly absorbing sheet of area \(2.0 \mathrm{~m}^{2}\), at what rate is momentum delivered to the sheet and what is the radiation pressure exerted on the sheet?
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