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Unpolarized light of intensity \(20.0 \mathrm{~W} / \mathrm{cm}^{2}\) is incident on two polarizing filters. The axis of the first filter is at an angle of \(25.0^{\circ}\) counterclockwise from the vertical (viewed in the direction the light is traveling), and the axis of the second filter is at \(62.0^{\circ}\) counterclockwise from the vertical. What is the intensity of the light after it has passed through the second polarizer?

Short Answer

Expert verified
The final intensity of the light after it passed through both polarizers is \(6.4 \mathrm{~W}/\mathrm{cm}^{2}\).

Step by step solution

01

Apply the first polarizer

As the light is initially unpolarized, the first polarizer will reduce its intensity by a factor of 1/2, not depending on the angle of its axis. So the intensity \(I_1\) after the first polarizer is \(I_1 = (1/2) \times 20.0 \mathrm{~W} / \mathrm{cm}^{2} = 10.0 \mathrm{~W} / \mathrm{cm}^{2}\). Now the light is polarized along the axis of the first polarizer, at an angle of \(25.0^{\circ}\) to the vertical.
02

Find the angle for the second polarizer

The angle for the second polarizer is given relative to the vertical, but we need it relative to the direction of polarization after the first polarizer. Subtracting, we find this angle is \(62.0^{\circ}- 25.0^{\circ} = 37.0^{\circ}\).
03

Apply the second polarizer

We can now apply Malus's law for the second polarizer. The outgoing intensity \(I_2\) is \(I_2 = I_1 \cdot \cos^{2}(37.0^{\circ}) = 10.0 \mathrm{~W} / \mathrm{cm}^{2} \times \cos^{2}(37.0^{\circ}).\)
04

Calculate Final Intensity

Evaluate the cosine, square it, and multiply to find \(I_2 \). This gives \(I_2= 6.4 \mathrm{~W}/\mathrm{cm}^{2}\). This is the intensity of the light after passing through both polarizers.

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Key Concepts

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

Malus's Law
Imagine light as a wave, dancing through space in all directions. This dance, while beautiful, isn't always as orderly as one would like, especially when trying to study how light interacts with materials. This is where 'Malus's Law' steps in, a simple yet profound principle that allows us to predict the fate of this light dance when it encounters polarizing filters.

In technical terms, Malus's Law states that the intensity (\( I \)) of polarized light passing through a polarizing filter is directly proportional to the square of the cosine of the angle between the light's initial polarization direction and the axis of the filter. Mathematically, it is expressed as \[ I = I_0 \times \text{cos}^2(\theta) \] where \( I_0 \) is the initial intensity of the light and \( \theta \) is the angle in question. This equation is pivotal when dealing with multiple filters, as it helps us understand how light dims when crossing each successive layer, as seen in the textbook problem we explored.

It's worth noting that Malus's Law applies only to polarized light. If the incoming light is unpolarized, we first consider the effect of the initial polarizing filter which generally cuts the light intensity to half regardless of the angle.
Polarizing Filters
Take a pair of polarized sunglasses, for instance. They aren't just tinted glass; they contain 'polarizing filters' that are like selective gatekeepers, deciding which light waves can pass through. This filtering is due to the unique structure of polarizing material that aligns in one direction, blocking waves that don't match its orientation.

Essentially, these filters convert unpolarized light, which vibrates in multiple planes, into polarized light, wiggling along a single plane. By doing so, they reduce glare and unwanted reflections – a principle exploited in photography and broadcasting to enhance picture quality.

When multiple filters are placed in a series, as in the given exercise, the first polarizer will let through light that aligns with its axis. Subsequent filters will then only allow light through that aligns with their axes as well, which further reduces the light's intensity. The angle between the axes of these filters is crucial to how much light will ultimately make it through the series.
Light Intensity
Imagine basking in the sunlight on a bright day compared to a cloudy one; the difference you feel is all about 'light intensity.' It tells us how much energy a light wave carries over an area in a given time. In science lingo, it's the power carried by light per unit area, often measured in watts per square centimeter (\( W/cm^2 \
In the context of our exercise, understanding light intensity is important because it helps us quantify the effects of polarizing filters on light. The initial intensity is the starting point; after passing through the polarizing filters, the intensity changes, reflecting the filtering effect. When we talk about reducing glare on a camera lens or on your sunglasses, what we’re really doing is manipulating the light's intensity.

This concept also explains why the light becomes dimmer after it passes through each filter - because we measure the rate at which light energy is falling on a surface, and as the polarizers do their job, they 'turn down the tap,' so to speak, leading to a lower intensity reading. Using Malus's Law in conjunction with this concept provides a powerful tool for predicting how much light will survive after traversal through polarizing media.

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Most popular questions from this chapter

Light Pipe. Light enters a solid pipe made of plastic having an index of refraction of 1.60 . The light travels parallel to the upper part of the pipe (Fig. E33.15). You want to cut the face \(A B\) so that all the light will reflect back into the pipe after it first strikes that face. (a) What is the largest that \(\beta\) can be if the pipe is in air? (b) If the pipe is immersed in water of refractive index 1.33 , what is the largest that \(\beta\) can be?

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