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If you have completely polarized light of intensity \(150 \mathrm{W} / \mathrm{m}^{2},\) what will its intensity be after passing through a polarizing filter with its axis at an \(89.0^{\circ}\) angle to the light's polarization direction?

Short Answer

Expert verified
The intensity of the light after passing through the polarizing filter with its axis at an \(89.0^\circ\) angle to the light's polarization direction is approximately \(0.045 \mathrm{W/m^{2}}\).

Step by step solution

01

Write down the given information

We have the following information: - Intensity of the incident light, \(I_0 = 150 \mathrm{W/m^{2}}\) - Angle between the light's polarization direction and the polarizing filter's axis, \(\theta = 89.0^\circ\)
02

Use Malus's Law to calculate the transmitted intensity

According to Malus's Law, the intensity of the transmitted light is: \(I = I_0 \cdot \cos^2 \theta\)
03

Convert the angle to radians

In order to use the cosine function, we need to convert the angle to radians: \(\theta_{rad} = \frac{\pi \cdot\theta}{180} = \frac{\pi (89.0)}{180} \mathrm{rad}\)
04

Calculate the transmitted intensity

Now, plug in the values for \(I_0\) and \(\theta\) to get the transmitted intensity: \(I = 150 \cdot \cos^2 \left(\frac{\pi (89.0)}{180}\right) \mathrm{W/m^{2}}\) Calculate the cosine value and square it: \(\cos^2 \left(\frac{\pi (89.0)}{180}\right) \approx 0.000300249\) Finally, multiply the initial intensity by the calculated value: \(I \approx 150 \cdot 0.000300249 = 0.04503735 \mathrm{W/m^{2}}\)
05

Write the final answer

The intensity of the light after passing through the polarizing filter with its axis at an \(89.0^\circ\) angle to the light's polarization direction is approximately \(0.045 \mathrm{W/m^{2}}\).

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Key Concepts

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

Malus's Law
Understanding the behavior of light as it interacts with materials like polarizing filters is fascinating and vital in various optics applications. One fundamental concept in this realm is Malus's Law. This law mathematically describes how the intensity of completely polarized light changes as it passes through a polarizing filter.

According to Malus's Law, the intensity of the transmitted light, denoted as \( I \), is directly related to the intensity of the incident light, represented as \( I_0 \), and the angle \( \theta \) between the light's original polarization direction and the axis of the filter. The relationship is expressed as:\[ I = I_0 \cdot \cos^2(\theta) \]
This equation means that the transmitted light's intensity depends on the cosine of the angle, squared. When the angle \( \theta \) is zero, which means the polarizing filter is aligned with the light's polarization direction, the transmitted intensity will be maximal, equal to the incident intensity \( I_0 \). As \( \theta \) increases, the transmitted intensity decreases, following the cosine squared pattern until it becomes zero when \( \theta \) is 90 degrees, at which point the filter completely blocks the polarized light.

Utilizing this law allows us to predict and calculate the behavior of polarized light in various scientific and engineering applications, such as photography, display technology, and even in studies of the Earth's atmosphere.
Intensity of Transmitted Light
The intensity of transmitted light is a crucial concept in optics as it quantifies how much light passes through a medium or filter. In the case of our exercise, it's the amount of light that makes it through a polarizing filter.

The intensity after passing through such a filter isn't arbitrary; it follows specific physical laws, such as Malus's Law, which we've just explored. Intensity is typically measured in watts per square meter (\(\mathrm{W/m^2}\)), as it signifies the power of the light in relation to the area it covers.

When a polarizing filter is involved, the intensity of the transmitted light is determined by the orientation of the filter compared to the light's polarization direction. If the polarizing axis is parallel to the polarization direction of the light, there is no reduction in intensity, aside from minor losses due to reflection and absorption in the filter material itself. However, if the filter is perpendicular, all of the light is blocked and the transmitted intensity drops to zero. For angles in between, Malus's Law gives us the exact intensity reduction to expect, which we calculate using the angle converted to radians, a natural unit for angular measurements in these calculations.

Understanding how to calculate this reduction in intensity can be incredibly beneficial for designing systems where controlling the amount of light is essential, such as in sensors, cameras, and various scientific instruments.
Angle of Polarization
The angle of polarization is a central concept when discussing the fate of polarized light as it encounters various optical elements. This angle is the orientation, often denoted as \( \theta \), at which polarized light meets a polarizing filter or another interface at which polarization can occur.

The incident angle greatly influences light behavior. If the polarized light is aligned with the filtering axis of a polarizer (angle of 0 degrees), it passes through with minimal loss. Conversely, if it is perpendicular (angle of 90 degrees), the light is fully blocked. Malus's Law helps us to determine the outcome for any other angle, indicating that the resultant intensity is a cosinusoidal function of this angle, specifically the cosine squared.

Importance in Applications

For practical purposes, the angle of polarization matters greatly in many technologies. For example, polarized sunglasses take advantage of this by having a filter oriented to block horizontally polarized light, which is commonly reflected off horizontal surfaces, thereby reducing glare and improving visual clarity.

In our exercise, we've been dealing with a nearly critical angle—89 degrees. When such a high angle is used, the transmitted intensity significantly drops, resulting in almost complete blockage of the light. This tells us about the sensitivity of polarized light transmission to the angle of polarization, emphasizing the importance of precise alignment in polarizing filters for their intended use cases.

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Most popular questions from this chapter

From his measurements, Roemer estimated that it took 22 min for light to travel a distance equal to the diameter of Earth's orbit around the Sun. (a) Use this estimate along with the known diameter of Earth's orbit to obtain a rough value of the speed of light. (b) Light actually takes \(16.5 \mathrm{min}\) to travel this distance. Use this time to calculate the speed of light.

At the end of Example \(1.7,\) it was stated that the intensity of polarized light is reduced to \(90.0 \%\) of its original value by passing through a polarizing filter with its axis at an angle of \(18.4^{\circ}\) to the direction of polarization. Verify this statement.

A ray of 610 -nm light goes from air into fused quartz at an incident angle of \(55.0^{\circ} .\) At what incident angle must \(470 \mathrm{nm}\) light enter flint glass to have the same angle of refraction?

(a) On a day when the intensity of sunlight is \(1.00 \mathrm{kW} / \mathrm{m}^{2},\) a circular lens \(0.200 \mathrm{m}\) in diameter focuses light onto water in a black beaker. Two polarizing sheets of plastic are placed in front of the lens with their axes at an angle of \(20.0^{\circ} .\) Assuming the sunlight is unpolarized and the polarizers are \(100 \%\) efficient, what is the initial rate of heating of the water in \(^{\circ} \mathrm{C} / \mathrm{s},\) assuming it is \(80.0 \%\) absorbed? The aluminum beaker has a mass of 30.0 grams and contains 250 grams of water. (b) Do the polarizing filters get hot? Explain.

Is it possible that total internal reflection plays a role in rainbows? Explain in tems of indices of refraction and angles, perhaps referring to that shown below. Some of us have seen the formation of a double rainbow; is it physically possible to observe a triple rainbow?

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