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A parallel beam of light in air makes an angle of 47.5\(^\circ\) with the surface of a glass plate having a refractive index of 1.66. (a) What is the angle between the reflected part of the beam and the surface of the glass? (b) What is the angle between the refracted beam and the surface of the glass?

Short Answer

Expert verified
(a) The angle between the reflected beam and the glass surface is 47.5°. (b) The angle between the refracted beam and the glass surface is 65.4°.

Step by step solution

01

Understanding the Problem

We have a parallel beam of light incident on a glass surface at an angle of 47.5° to the surface normal. The refractive index of the glass is 1.66. We are to find the angles of the reflected and refracted beams in relation to the glass surface.
02

Calculating the Angle of Reflection

According to the law of reflection, the angle of incidence is equal to the angle of reflection. The angle of incidence is the angle relative to the normal, which is 90° minus the angle with the surface. Therefore, the angle of incidence is 90° - 47.5° = 42.5°. The angle of reflection is also 42.5° from the normal, so the angle between the reflected beam and the surface is 47.5°, same as the angle of incidence with the surface.
03

Applying Snell's Law for Refraction

Snell's Law describes the refraction of light as it passes through different media. The formula is \( n_1 \sin(\theta_1) = n_2 \sin(\theta_2) \), where \( n_1 \) and \( n_2 \) are the refractive indices of air and glass respectively, and \( \theta_1 \) and \( \theta_2 \) are the angles of incidence and refraction from the normal. Here, \( n_1 = 1 \), \( \theta_1 = 42.5° \), and \( n_2 = 1.66 \).
04

Calculating the Angle of Refraction

Using Snell's law, substitute the known values: \( 1 \cdot \sin(42.5°) = 1.66 \cdot \sin(\theta_2) \). Solve for \( \sin(\theta_2) \):\[ \sin(\theta_2) = \frac{\sin(42.5°)}{1.66} \approx 0.4163.\]This gives \( \theta_2 \approx \arcsin(0.4163) \approx 24.6° \). This angle is with respect to the normal line.
05

Finding the Angle with the Surface for Refracted Beam

The refracted beam's angle from the normal is \( 24.6° \), so the angle with the surface is \( 90° - 24.6° = 65.4° \).

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Key Concepts

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

Law of Reflection
The law of reflection is a fundamental principle of optics that applies when light reflects off a surface. It states that the angle of incidence is equal to the angle of reflection.
This means when a light beam hits a surface, the angle at which it arrives (the incidence angle) is the same as the angle at which it leaves the surface (the reflection angle).
Imagine a mirror: if a beam of light hits it at 30°, it reflects away at 30° on the opposite side of the normal line, which is an imaginary line perpendicular to the surface.
  • **Critical in design:** This principle is crucial in optical technologies like telescopes and cameras.
  • **Simple understanding:** Think of playing pool; the angle you hit the ball against the side rail is how it will bounce off.
For this exercise, we know the incident angle with the normal is 42.5°, making the reflected angle 47.5° with the glass surface, due to the angle with the perpendicular line being equal on both sides.
Refractive Index
The refractive index is a measure of how much a ray of light bends, or refracts, as it passes between different media.
Every material has a unique refractive index, which tells you how slower light travels in that medium compared to a vacuum.
The refractive index is a ratio given by the speed of light in a vacuum divided by the speed of light in the material.
For example, a refractive index of 1.66 means light travels 1.66 times slower in the glass than in a vacuum.
  • **Material property:** This number changes based on the substance, like water (1.33), glass (around 1.5–1.9), or diamond (2.42).
  • **Key in lens crafting:** Knowing these values helps engineers design lenses that correct vision or focus light properly.
In our exercise, the glass's refractive index helps us apply Snell's Law to find the new angle when light enters the glass.
Angle of Incidence
The angle of incidence is a crucial aspect whenever light interacts with surfaces. It is defined as the angle between the incoming light ray and the normal line of the surface.
To find this angle, use the surface angle and subtract it from 90°, the angle of the normal.
In our scenario, the light beam hits the glass at 47.5° to the surface, meaning its incidence angle is 42.5° from the normal line.
  • **Design importance:** Understanding and calculating this angle helps in creating devices like reflective coatings and solar panels.
  • **Everyday application:** If sunlight streams through a window and forms a rectangle on the floor, the angle of incidence dictates the size and position of that light patch.
The angle of incidence not only determines reflection, but also is a key factor in calculating the angle of refraction using Snell's Law, which helps predict how the light beam will behave as it travels into a different medium like glass.

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Most popular questions from this chapter

A beam of polarized light passes through a polarizing filter. When the angle between the polarizing axis of the filter and the direction of polarization of the light is \(\theta\), the intensity of the emerging beam is \(I\). If you now want the intensity to be \(I/2\), what should be the angle (in terms of \(\theta\)) between the polarizing angle of the filter and the original direction of polarization of the light?

The vitreous humor, a transparent, gelatinous fluid that fills most of the eyeball, has an index of refraction of 1.34. Visible light ranges in wavelength from 380 nm (violet) to 750 nm (red), as measured in air. This light travels through the vitreous humor and strikes the rods and cones at the surface of the retina. What are the ranges of (a) the wavelength, (b) the frequency, and (c) the speed of the light just as it approaches the retina within the vitreous humor?

A parallel beam of unpolarized light in air is incident at an angle of 54.5\(^\circ\) (with respect to the normal) on a plane glass surface. The reflected beam is completely linearly polarized. (a) What is the refractive index of the glass? (b) What is the angle of refraction of the transmitted beam?

A thin layer of ice (\(n\) = 1.309) floats on the surface of water (\(n\) = 1.333) in a bucket. A ray of light from the bottom of the bucket travels upward through the water. (a) What is the largest angle with respect to the normal that the ray can make at the ice-water interface and still pass out into the air above the ice? (b) What is this angle after the ice melts?

A flat piece of glass covers the top of a vertical cylinder that is completely filled with water. If a ray of light traveling in the glass is incident on the interface with the water at an angle of \(\theta_a = 36.2{^\circ}\), the ray refracted into the water makes an angle of 49.8\(^\circ\) with the normal to the interface. What is the smallest value of the incident angle \(\theta_a\) for which none of the ray refracts into the water?

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