/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 54 At what angle above the horizon ... [FREE SOLUTION] | 91Ó°ÊÓ

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At what angle above the horizon is the Sun if light from it is completely polarized upon reflection from water?

Short Answer

Expert verified
The angle above the horizon at which light from the sun becomes completely polarized upon reflection from water is equivalent to the Brewster's Angle, which is approximately \(53.1^\circ\).

Step by step solution

01

Identify the Values

Firstly, identify the values given in the problem. To calculate the Brewster's angle, we need the refractive indices of water and air. The refractive index of light in air is approximately \(n_1 = 1.00\) and in water is \(n2 = 1.33\).
02

Apply Brewster's Angle Formula

The formula for Brewster's Angle is \(\theta_B = arctan(n_2/n_1)\). Substituting provided values into the formula, \(\theta_B = arctan(1.33 / 1.00)\)
03

Calculate Brewster's Angle

By performing the calculation for the tangent inverse of the ratio between the refractive indices, the Brewster's Angle, and accordingly the angle above the horizon for the sun, can be found. \( \theta_B = arctan(1.33) \)

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Key Concepts

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

Polarization of Light
When we talk about polarization of light, we're referring to the process by which the oscillations of a light wave are confined to a single plane. Light waves are transverse waves, meaning their oscillations are perpendicular to the direction of the wave's travel.

Light normally oscillates in all directions perpendicular to its path. However, when light is polarized, it only oscillates in one direction. This can occur naturally, such as when sunlight reflects off a surface like water at a certain angle—this is known as Brewster's angle.

Polarized light has several practical applications, including in sunglasses which block horizontally polarized light to reduce glare from surfaces like roads and water. Cameras also use polarizing filters to enhance contrast and saturation in photographs by filtering out specific polarized light.
Refractive Index
The refractive index of a material, denoted as n, is a dimensionless number that describes how light propagates through that medium. It is defined as the ratio of the speed of light in a vacuum to the speed of light in the material. The refractive index determines how much the path of light is bent, or refracted, when entering a material.

Materials with a higher refractive index will cause light to bend more as it enters them from air, which has a refractive index close to 1. This bending can be observed when a straw seems to be bent at the surface of a glass of water; the water has a higher refractive index than the air, causing the light to bend.
Snell's Law
Now let's explore Snell's law, a fundamental principle in optics that describes the relationship between the angles of incidence and refraction when light crosses the boundary between two media with different refractive indices. Snell's Law is given by the equation
\[ n_1 \sin(\theta_1) = n_2 \sin(\theta_2) \]
where n_1 and n_2 are the refractive indices of the first and second medium, and \theta_1 and \theta_2 are the angles of incidence and refraction, respectively.

Understanding Snell's law is critical for calculating Brewster's angle, as it directly relates to the angles at which light is refracted when moving from air (or any medium) into water. When applied to Brewster's angle, Snell's law helps in defining the specific angle at which reflected light will be perfectly polarized.

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Most popular questions from this chapter

A thin film of oil \((n=1.25)\) is located on smooth, wet pavement. When viewed from a direction perpendicular to the pavement, the film reflects most strongly red light at \(640 \mathrm{~nm}\) and reflects no green light at \(512 \mathrm{~nm}\). (a) What is the minimum thickness of the oil film? (b) Let \(m_{1}\) correspond to the order of the constructive interference and \(m_{2}\) to the order of the destructive interference. Obtain a relationship between \(m_{1}\) and \(m_{2}\) that is consistent with the given data.

Unpolarized light passes through two Polaroid sheets. The transmission axis of the analyzer makes an angle of \(35.0^{\circ}\) with the axis of the polarizer. (a) What fraction of the original unpolarized light is transmitted through the analyzer? (b) What fraction of the original light is absorbed by the analyzer?

Nonreflective coatings on camera lenses reduce the loss of light at the surfaces of multilens systems and prevent internal reflections that might mar the image. Find the minimum thickness of a layer of magnesium fluoride \((n=1.38)\) on flint glass \((n=1.66)\) that will cause destructive interference of reflected light of wavelength \(550 \mathrm{~nm}\) near the middle of the visible spectrum.

A soap bubble \((n=1.93)\) having a wall thickness of \(120 \mathrm{~nm}\) is floating in air. (a) What is the wavelength of the visible light that is most strongly reflected? (b) Explain how a bubble of different thickness could also strongly reflect light of this same wavelength. (c) Find the two smallest film thicknesses larger than the one given that can produce strongly reflected light of this same wavelength.

Light of wavelength \(5.40 \times 10^{2} \mathrm{~nm}\) passes through a slit of width \(0.200 \mathrm{~mm}\). (a) Find the width of the central maximum on a screen located \(1.50 \mathrm{~m}\) from the slit. (b) Determine the width of the first-order bright fringe.

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