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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

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(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

Three polarizing filters are stacked, with the polarizing axis of the second and third filters at 23.0\(^\circ\) and 62.0\(^\circ\), respectively, to that of the first. If unpolarized light is incident on the stack, the light has intensity 55.0 \(\mathrm {W/cm}^2\) after it passes through the stack. If the incident intensity is kept constant but the second polarizer is removed, what is the intensity of the light after it has passed through the stack?

(a) A tank containing methanol has walls 2.50 cm thick made of glass of refractive index 1.550. Light from the outside air strikes the glass at a 41.3\(^\circ\) angle with the normal to the glass. Find the angle the light makes with the normal in the methanol. (b) The tank is emptied and refilled with an unknown liquid. If light incident at the same angle as in part (a) enters the liquid in the tank at an angle of 20.2\(^\circ\) from the normal, what is the refractive index of the unknown liquid?

Optical fibers are constructed with a cylindrical core surrounded by a sheath of cladding material. Common materials used are pure silica \(n_2 = 1.4502\) for the cladding and silica doped with germanium \(n_1 = 1.4652\) for the core. (a) What is the critical angle \(\theta_{crit}\) for light traveling in the core and reflecting at the interface with the cladding material? (b) The numerical aperture (NA) is defined as the angle of incidence \(theta_i\) at the flat end of the cable for which light is incident on the core-cladding interface at angle \(\theta_{crit}\) (\(\textbf{Fig. P33.46}\)). Show that sin \(\theta_i\) =\( \sqrt {n^2_1 - n^2_2}\) . (c) What is the value of \(\theta_i\) for \(n_1\) = 1.465 and \(n_2\) = 1.450?

When the sun is either rising or setting and appears to be just on the horizon, it is in fact below the horizon. The explanation for this seeming paradox is that light from the sun bends slightly when entering the earth’s atmosphere, as shown in Fig. \(\textbf{P33.51}\). Since our perception is based on the idea that light travels in straight lines, we perceive the light to be coming from an apparent position that is an angle \(\delta\) above the sun's true position. (a) Make the simplifying assumptions that the atmosphere has uniform density, and hence uniform index of refraction \(n\), and extends to a height \(h\) above the earth's surface, at which point it abruptly stops. Show that the angle \(\delta\) is given by $${ \delta = \mathrm{arcsin} ({{nR}\over{R + h}}}) - \mathrm{arcsin}({{{R}\over R + h}}) $$ where \(R\) = 6378 km is the radius of the earth. (b) Calculate \(\delta\) using \(n\) = 1.0003 and \(h\) = 20 km. How does this compare to the angular radius of the sun, which is about one quarter of a degree? (In actuality a light ray from the sun bends gradually, not abruptly, since the density and refractive index of the atmosphere change gradually with altitude.)

Light with a frequency of \(5.80 \times 10^{14}\) Hz travels in a block of glass that has an index of refraction of 1.52. What is the wavelength of the light (a) in vacuum and (b) in the glass?

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