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Prove that if \(I\) is the intensity of light transmitted by two polarizing filters with axes at an angle \(\theta\) and \(I^{\prime}\) is the intensity when the axes are at an angle \(90.0^{\circ}-\theta\) then \(I+I^{\prime}=I_{0}, \quad\) the original intensity. (Hint: Use the trigonometric identities \(\cos 90.0^{\circ}-\theta=\sin \theta \quad\) and \(\left.\cos ^{2} \theta+\sin ^{2} \theta=1 .\right)\)

Short Answer

Expert verified
We are given that the intensity of light transmitted by two polarizing filters with axes at an angle \( \theta \) is \(I\) and at an angle \(90.0^{\circ}-\theta\) is \(I'\). Using Malus' Law and the trigonometric identity \( \cos(90^\circ - \theta) = \sin \theta \), we get \( I = I_0 \cos^2{\theta} \) and \( I' = I_0 \sin^2{\theta} \). Adding the intensities, we obtain \( I + I' = I_0 \cos^2{\theta} + I_0 \sin^2{\theta} \). Finally, using the trigonometric identity \( \cos^2 \theta + \sin^2 \theta = 1 \), we conclude that the intensities of light transmitted by the two polarizing filters add up to the original intensity: \( I + I' = I_0 \).

Step by step solution

01

Write down the intensities

Given that the intensity of light transmitted by two polarizing filters with axes at an angle \( \theta \) is \(I\) and at an angle \(90.0^{\circ}-\theta\) is \(I'\). Using Malus' Law, we have: \( I = I_0 \cos^2{\theta} \) \( I' = I_0 \cos^2{(90^\circ - \theta)} \)
02

Use the trigonometric identity

Now, we'll use the given trigonometric identity \( \cos(90^\circ - \theta) = \sin \theta \), so we have: \( I' = I_0 \cos^2{(90^\circ - \theta)} = I_0 \sin^2{\theta} \)
03

Add the intensities

Next, we'll add the intensities of light transmitted by the two polarizing filters: \( I + I' = I_0 \cos^2{\theta} + I_0 \sin^2{\theta} \)
04

Use the trigonometric identity

We'll now use the trigonometric identity \( \cos^2 \theta + \sin^2 \theta = 1 \) , so we have: \( I + I' = I_0 (\cos^2{\theta} + \sin^2{\theta}) \) \( I + I' = I_0 \cdot 1 \)
05

Conclusion

Therefore, the intensities of light transmitted by two polarizing filters do indeed add up to the original intensity: \( I + I' = I_0 \) This completes the proof.

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Key Concepts

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

Malus' Law
When light passes through a polarizing filter, its intensity gets reduced based on the angle between the light's polarization direction and the axis of the filter. Malus' Law provides a mathematical way to describe this effect. It states that the intensity (\( I \)) of polarized light after passing through a filter is given by:
  • \( I = I_0 \cos^2{\theta} \),
where \( I_0 \) is the initial intensity of light, and \( \theta \) is the angle between the light's polarization direction and the filter's axis. This equation shows that as \( \theta \) increases, the intensity decreases. This law is crucial for understanding how polarized sunglasses work, where lenses reduce glare by blocking horizontally polarized light. It also plays a significant role in scientific instruments that measure light properties.
Trigonometric Identities
Trigonometric identities are essential tools in mathematics that help us relate the angles and sides of triangles. In the context of polarization and Malus' Law, two identities are particularly useful:
  • \( \cos(90^\circ - \theta) = \sin \theta \)
  • \( \cos^2 \theta + \sin^2 \theta = 1 \)
These identities simplify equations and calculations in trigonometry. For instance, in our exercise, we rearranged the angle \( 90^\circ - \theta \) using the identity \( \cos(90^\circ - \theta) = \sin \theta \) to find the intensity in terms of a sine function. This conversion is handy because it shows the complementary relationship between sine and cosine for any angle. Furthermore, the Pythagorean identity \( \cos^2 \theta + \sin^2 \theta = 1 \) is used to demonstrate that the sum of the intensities in the exercise equals the original light intensity \( I_0 \), validating the total energy conservation in light transmission.
Polarizing Filters
Polarizing filters are devices designed to control the polarization of light. They allow light waves of a specific polarization to pass through while blocking others. These filters are commonly made from materials arranged in a particular alignment to achieve this effect.
  • The use of polarizers is widespread in optical devices for reducing glare, such as photography lenses and sunglasses.
  • In scientific settings, they help in experiments involving light properties and behaviors.
When stacking two polarizing filters, the angle between their axes significantly affects how much light is transmitted. Through the application of Malus' Law, we learned that the intensity of light transmitted changes with this angle. For example, when two filters are aligned (angle is 0 degrees), the light intensity remains maximum. However, at 90 degrees, almost no light gets through. This understanding of polarizers is essential for their application in various technologies and scientific studies.

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Most popular questions from this chapter

(a) A narrow beam of light containing yellow (580 nm) and green (550 nm) wavelengths goes from polystyrene to air, striking the surface at a \(30.0^{\circ}\) incident angle. What is the angle between the colors when they emerge? (b) How far would they have to travel to be separated by \(1.00 \mathrm{mm}\) ?

(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.

What is Brewster's angle for light traveling in water that is reflected from crown glass?

A light ray falls on the left face of a prism (see below) at the angle of incidence \(\theta\) for which the emerging beam has an angle of refraction \(\theta\) at the right face. Show that the index of refraction \(n\) of the glass prism is given by. $$n=\frac{\sin \frac{1}{2}(\alpha+\phi)}{\sin \frac{1}{2} \phi}$$ where \(\phi\) is the vertex angle of the prism and \(\alpha\) is the angle through which the beam has been deviated. If \(\alpha=37.0^{\circ}\) and the base angles of the prism are each \(50.0^{\circ},\) what is \(n ?\)

Shown below is a ray of light going from air through crown glass into water, such as going into a fish tank. Calculate the amount the ray is displaced by the glass \((\Delta x), \quad\) given that the incident angle is \(40.0^{\circ}\) and the glass is \(1.00 \mathrm{cm}\) thick.

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