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Light is incident from air onto the surface of a liquid. The angle of incidence is \(53.0^{\circ},\) and the angle of refraction is \(34.0^{\circ} .\) At what angle of incidence would the reflected light be 100\(\%\) polarized?

Short Answer

Expert verified
At the Brewster's angle \( \theta_B = \tan^{-1}(n_2) \).

Step by step solution

01

Understand Brewster's Angle

Brewster's angle, also known as the polarization angle, is the angle at which light is reflected completely polarized. It occurs when the reflected and refracted rays are perpendicular to each other. At this angle, the refracted and incident angles add up to 90 degrees.
02

Apply Brewster's Law

Brewster’s law states that the tangent of the Brewster's angle (\( \theta_B \)) is equal to the refractive index of the material (\( n \)) relative to the refractive index of the incident medium. The formula is given as: \( \tan(\theta_B) = n \).
03

Calculate the Refractive Index of the Liquid

From Snell's Law, \( n_1 \sin(\theta_i) = n_2 \sin(\theta_r) \). Here, \( n_1 = 1 \) (since the light is in air first) and \( \theta_i = 53^{\circ} \), \( \theta_r = 34^{\circ} \). Calculate \( n_2 \) as follows:\[ n_2 = \frac{\sin(53^{\circ})}{\sin(34^{\circ})} \]
04

Calculate Brewster's Angle

Once you have the value of \( n_2 \), you use Brewster's Law to find \( \theta_B \):\[ \theta_B = \tan^{-1}(n_2) \] This will give you the angle of incidence at which the light is 100% polarized.

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Key Concepts

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

Snell's Law
Snell’s Law is a fundamental principle guiding how light refracts when passing from one medium to another. This law emphasizes the relationship between the angles of incidence and refraction, as well as the refractive indices of the two media involved.

When light enters a new medium, like from air into water, it bends due to a change in its speed. Snell’s Law mathematically expresses this bending:
  • The formula is: \( n_1 \sin(\theta_i) = n_2 \sin(\theta_r) \).
  • Here, \( n_1 \) and \( n_2 \) are the refractive indices of the first and second media, respectively.
  • \( \theta_i \) and \( \theta_r \) are the angles of incidence and refraction.
This law helps in calculating unknown parameters regarding the light’s path as it traverses different substances.

In the problem we handled, the light was transitioning from air (\( n_1 = 1 \)) to a liquid with a known angle of incidence (\(53^{\circ}\)). By utilizing Snell’s Law, we computed the refractive index for the liquid. Understanding this relationship is crucial for explaining many natural phenomena and practical applications.
Refractive Index
The refractive index is a measure that describes how light propagates through a medium. The change in speed when light enters a new material impacts the angle at which it travels, which the refractive index quantifies.

When light moves from one substance into another, for instance, from air into water, the refractive index will determine how much the light will bend or refract.
  • The formula \( n = \frac{c}{v} \) shows this, where \( c \) is the speed of light in a vacuum and \( v \) is the speed of light in the material.
  • A high refractive index indicates that light travels slower in that material compared to a vacuum.
During calculations in our exercise, we determined the refractive index of the liquid using Snell's Law. The angles given (incidence and refraction) allowed us to find the refractive index of the liquid by rearranging Snell's equation.

Understanding the refractive index helps to predict how light behaves across various substances, leading to applications in lenses, glasses, and even detailed light analysis techniques.
Polarization
Polarization refers to the orientation of light's oscillations. In unpolarized light, oscillations occur in multiple planes. However, when light becomes polarized, its vibrations are limited to a specific direction.

One particular way to achieve polarization is through reflection. Brewster's angle is key in understanding this phenomenon. It states that light reflecting off a surface at a certain angle becomes perfectly polarized. This idea connects to the Brewster’s Angle and law, enabling us to determine when light will be 100% polarized.
  • The reflected light is polarized when the reflected and refracted light are at a right angle to each other.
  • At Brewster’s angle, the angle of reflection and the angle of incidence sum up to \(90^{\circ}\).
  • Utilizing the refractive index, Brewster's angle can be calculated with the formula \( \tan(\theta_B) = n \).
This principle is widely used in designing sunglasses and camera lenses to reduce glare. Moreover, understanding polarization helps in advanced optics like microscopy and laser physics.

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Most popular questions from this chapter

A slide projector has a converging lens whose focal length is 105.00 mm. (a) How far (in meters) from the lens must the screen be located if a slide is placed 108.00 mm from the lens? (b) If the slide measures 24.0 \(\mathrm{mm} \times 36.0 \mathrm{mm}\) what are the dimensions (in mm) of its image?

A microscope for viewing blood cells has an objective with a focal length of 0.50 cm and an eyepiece with a focal length of 2.5 cm. The distance between the objective and eyepiece is 14.0 cm. If a blood cell subtends an angle of 2.1 \(\times 10^{-5}\) rad when viewed with the naked eye at a near point of \(25.0 \mathrm{cm},\) what angle (magnitude only) does it subtend when viewed through the microscope?

A ray of sunlight is passing from diamond into crown glass; the angle of incidence is \(35.00^{\circ} .\) The indices of refraction for the blue and red components of the ray are: blue \(\left(n_{\text { diamood }}=2.444, n_{\text { cromgless }}=1.531\right)\) and red \(\left(n_{\text { dianood }}=2.410, n_{\text { crown glass }}=1.520\right) .\) Determine the angle between the refracted blue and red rays in the crown glass.

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A woman can read the large print in a newspaper only when it is at a distance of 65 cm or more from her eyes. (a) Is she nearsighted (myopic) or farsighted (hyperopic), and what kind of lens is used in her glasses to correct her eyesight? (b) What should be the refractive power (in diopters) of her glasses (worn 2.0 cm from the eyes), so she can read the newspaper at a distance of 25 cm from the eyes?

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