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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) \(47.5^\circ\); (b) \(65.97^\circ\).

Step by step solution

01

Understand Reflection and Refraction Angle

For the reflection of light, the angle of incidence is equal to the angle of reflection. Given that the angle between the incoming light beam and the surface of the glass is \(47.5^\circ\), the angle of incidence is actually the angle with respect to the normal (perpendicular to the surface), which is \(90^\circ - 47.5^\circ = 42.5^\circ\). Since the reflection angle is the same as the incidence angle with respect to the normal, calculate the angle between the reflected light and the surface: \(90^\circ - 42.5^\circ = 47.5^\circ\).
02

Apply Snell's Law for Refraction

Snell's Law relates the angle of incidence \( \theta_1 \) and the angle of refraction \( \theta_2 \) using the refractive indices \( n_1 \) and \( n_2 \): \( n_1 \sin(\theta_1) = n_2 \sin(\theta_2) \). For air, \( n_1 = 1 \), \( \theta_1 = 42.5^\circ \), and for glass, \( n_2 = 1.66 \). Plug these values into Snell's Law: \( \sin(\theta_2) = \frac{1}{1.66} \sin(42.5^\circ) \). Calculate \( \theta_2 \) using the inverse sine function.
03

Solve for Refracted Angle from the Surface

Let us first calculate \( \theta_2 \) using the equation from Snell's Law, which results in \( \theta_2 \approx 24.03^\circ \). This is the angle with respect to the normal. Thus, the angle between the refracted beam and the glass surface is \(90^\circ - \theta_2 = 65.97^\circ\).

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Key Concepts

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

Reflection Angle
When a beam of light hits a surface and bounces back, the angle at which it reflects is known as the **reflection angle**. One fundamental principle is that the angle of incidence (the angle the incoming ray makes with a line normal to the surface) is always equal to the angle of reflection. This is true no matter the surface or medium involved.

In the given problem, the light hits the glass surface at an angle of incidence of \(42.5^{\circ}\) with respect to the normal. Since light follows the law of reflection, the reflection angle is also \(42.5^{\circ}\) with respect to the normal. Now, if we need to determine the angle the reflected beam makes with the surface of the glass, simply subtract the incidence angle from \(90^{\circ}\) like so: \(90^{\circ} - 42.5^{\circ} = 47.5^{\circ}\).
  • **Key Point**: The reflection angle relative to the surface is the same as the original incidence angle due to the law of reflection.
  • **Application**: Helps in designing optical devices and understanding natural phenomena like the brightness of lakes and the glare from windows.
Refractive Index
The **refractive index** is a dimensionless number that indicates how much light slows down and bends as it passes from one medium into another. This key optical property is represented by the symbol \( n \).

In our context, air has a refractive index close to \(1\), and typical glass has a refractive index of about \(1.66\). This increase in refractive index suggests that light travels more slowly and bends more sharply in glass compared to air.
  • **Role**: Helps to calculate the change in direction using Snell's Law, which is vital in lenses, eyeglasses, and cameras.
  • **Insight**: A higher refractive index means light will bend more sharply, causing effects like magnification or focusing.
Angle of Refraction
The **angle of refraction** is the angle a light ray makes with the normal after it enters a new medium. Snell's Law comes in handy to determine this angle, mathematically expressed as \( n_1 \sin(\theta_1) = n_2 \sin(\theta_2) \), where \( \theta_1 \) and \( \theta_2 \) are the angles with the normal in the original and the new medium respectively.

In solving exercises like ours, calculate the angle of refraction \(\theta_2\) by plugging in the values for air and glass into Snell's Law and using the inverse sine function for the new angle with respect to the normal. For our particular problem, \(\theta_2 \approx 24.03^{\circ}\).

To find the angle of refraction with respect to the glass surface, you then perform \(90^{\circ} - \theta_2\), resulting in \(65.97^{\circ}\).
  • **Observation**: The beam bends towards the normal when entering a medium with higher refractive index.
  • **Application**: Understanding refraction is crucial in fields like optics, physics, and even in the creation of corrective lenses for vision abnormalities.

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Most popular questions from this chapter

In a material having an index of refraction \(n,\) a light ray has frequency \(f,\) wavelength \(\lambda,\) and speed \(v .\) What are the frequency, wavelength, and speed of this light (a) in vacuum and (b) in a material having refractive index \(n^{\prime} ?\) In each case, express your answers in terms of only \(f, \lambda, v, n,\) and \(n^{\prime} .\)

A light beam travels at \(1.94 \times 10^{8} \mathrm{m} / \mathrm{s}\) in quartz. The wavelength of the light in quartz is 355 \(\mathrm{nm}\) . (a) What is the index of refraction of quartz at this wavelength? (b) If this same light travels through air, what is its wavelength there?

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 75.0 \(\mathrm{W} / \mathrm{cm}^{2}\) after it passes through the stack. If the incident intensity is kept constant, what is the intensity of the light after it has passed through the stack if the second polarizer is removed?

The refractive index of a certain glass is \(1.66 .\) For what incident angle is light reflected from the surface of this glass completely polarized if the glass is immersed in (a) air and (b) water?

A ray of light is traveling in a glass cube that is totally immersed in water. You find that if the ray is incident on the glass-water interface at an angle to the normal larger than \(48.7^{\circ}\) , no light is refracted into the water. What is the refractive index of the glass?

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