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Chapter 24: Problem 57

A plate of crown glass \((n=1.52)\) is covered with a layer of water. A beam of light traveling in air is incident on the water and partially transmitted. Is there any angle of incidence for which the light reflected from the water- glass interface will have maximum linear polarization? Justify your answer mathematically.

Short Answer

Expert verified
Yes, the angle of incidence is approximately 48.37 degrees for maximum polarization.

Step by step solution

01

Understand the Problem

We need to determine if there is an angle of incidence for light in air that results in maximum linear polarization when reflecting from a water-crown glass interface. To solve this, we'll use Brewster's Law, which indicates that maximum polarization occurs when the reflected and refracted light are perpendicular.
02

Apply Brewster's Law

Brewster's Law states that the tangent of the Brewster angle (\(\theta_B\)) equals the refractive index of the second medium divided by the first medium's refractive index. In this case, we look for maximum polarization at the water-glass interface, so we need the refractive indices of water (approximately 1.33) and crown glass (1.52).
03

Calculate Polarization Condition

For maximum polarization, the reflected light must be perpendicular to the refracted light at the water-glass interface. At this interface, Brewster's angle is:\[\theta_B = \tan^{-1}\left(\frac{n_2}{n_1}\right)\]where \(n_2 = 1.52\) (glass) and \(n_1 = 1.33\) (water).
04

Solve for Brewster's Angle

Calculate \(\theta_B\) using:\[\theta_B = \tan^{-1}\left(\frac{1.52}{1.33}\right)\]First, compute the ratio:\(\frac{1.52}{1.33} \approx 1.142857\). Thus,\[\theta_B = \tan^{-1}(1.142857) \approx 48.37^\circ.\]
05

Assess for Achievability

Since \(\theta_B = 48.37^\circ\), an angle of incidence exists for which the light is maximally polarized. This condition is achievable because these angles occur within the practical range (0 to 90 degrees) of incidence angles in typical optical setups.

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Key Concepts

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

Angle of Incidence
The angle of incidence is the angle between the incident ray of light and the normal (a line perpendicular) to the surface it strikes.
This angle is crucial when discussing the behavior of light as it encounters different media, such as air, water, or glass.
  • In optics, this angle is essential in determining how light refracts or reflects when passing through or bouncing off a surface.
  • For the case of maximum polarization, the angle of incidence must be equal to Brewster's angle. This is the specific angle where reflected light is perfectly polarized.
  • Understanding the angle of incidence helps predict how much light will reflect off a surface versus refract into it, like what happens at the water and glass interface in our example.
Refractive Index
The refractive index of a material is a measure of how much it slows down light, compared to light's speed in a vacuum.
It is crucial in optics for understanding how light bends as it passes from one medium into another.
  • Brewster's Law uses the refractive indices of two media to determine the angle of incidence for maximum polarization.
  • In our example, we have refractive indices of 1.33 for water and 1.52 for crown glass.
  • The ratio of these indices determines the Brewster angle, signaling where light transitions through different mediums with maximum efficiency in polarization.
Maximum Polarization
Maximum polarization occurs when light reflects off a surface at Brewster's angle, an angle where the reflected and refracted light are at 90 degrees to each other.
  • Brewster's Law is key to finding this angle. It tells us that the tangent of the Brewster angle is equal to the refractive index of the second medium divided by the first.
  • For the water-glass interface, this calculation shows us that the angle for maximum polarization is approximately 48.37 degrees.
  • This phenomenon is utilized in polarizing lenses and certain photographic filters to reduce glare by blocking polarized light.
Crown Glass
Crown glass is a type of optical glass with a lower refractive index, typically around 1.52, making it ideal for precise optical lenses and prisms.
  • It is known for its relatively low dispersion compared to other types of glass, which reduces chromatic aberration in optical systems.
  • Crown glass's refractive index plays a pivotal role in calculations involving polarization phenomena like the one in our exercise.
  • This glass type is commonly used in the manufacturing of lenses, ensuring high-quality visual clarity and light manipulation.

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Most popular questions from this chapter

Two parallel slits are illuminated with monochromatic light, and an interference pattern is observed on a screen. (a) If the distance between the slits were decreased, would the distance between the maxima (1) increase, (2) remain the same, or (3) decrease? Explain. (b) If the slit separation is \(1.0 \mathrm{~mm}\), the wavelength is \(640 \mathrm{nm}\), and the distance from the slits to the screen is \(3.00 \mathrm{~m}\), what is the separation between adjacent interference maxima? (c) What if the slit separation is \(0.80 \mathrm{~mm} ?\)

At what angle will the second-order maximum be seen from a diffraction grating of spacing \(1.25 \mu \mathrm{m}\) when illuminated by light of wavelength \(550 \mathrm{nm} ?\)

The critical angle for total internal reflection in a certain media boundary is \(45^{\circ} .\) What is the polarizing (Brewster) angle for light externally incident on the same boundary?

Light of wavelength \(550 \mathrm{nm}\) in air is normally incident on a glass plate \((n=1.5)\) whose thickness is \(1.1 \times 10^{-5} \mathrm{~m} .\) (a) What is the thickness of the glass expressed in terms of the wavelength of light in glass? (b) How many reflected waves will experience the \(180^{\circ}\) phase shift? (c) Will the reflected waves interfere constructively or destructively?

A thin air wedge between two flat glass plates forms bright and dark interference bands when illuminated with normally incident monochromatic light. (See Fig. 24.9.) (a) Show that the thickness of the air wedge changes by \(\lambda / 2\) from one bright band to the next, where \(\lambda\) is the wavelength of the light. (b) What would be the change in the thickness of the wedge between bright bands if the space were filled with a liquid with an index of refraction \(n ?\)
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