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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?

Short Answer

Expert verified
The refractive index of glass is approximately 1.77.

Step by step solution

01

Understand the Critical Angle Concept

When a ray of light passes from a medium with higher refractive index to a medium with a lower refractive index, total internal reflection occurs if the angle of incidence is greater than the critical angle. In this case, since no light is refracted into water beyond 48.7°, this is the critical angle ( abla_c").
02

Apply Snell’s Law

Snell's Law relates the refractive indices of two media and the sine of the angle of incidence and refraction: \[ n_1 imes \sin(\theta_1) = n_2 imes \sin(\theta_2) \]For the critical angle, \( \theta_2 = 90^\circ \) because the refracted ray travels along the boundary. So \( \sin(\theta_2) = 1 \).
03

Rearrange Snell’s Law for Critical Angle

Using the critical angle, Snell's Law becomes \[ n_1 \times \sin(\theta_c) = n_2 \]where \( n_1 \) is the refractive index of glass, \( \theta_c = 48.7^\circ \), and \( n_2 = 1.333 \) (the refractive index of water).
04

Calculate Refractive Index of Glass

Rearrange the equation to find \( n_1 \) (the refractive index of the glass):\[ n_1 = \frac{n_2}{\sin(\theta_c)} \]Substitute \( n_2 = 1.333 \) and \( \sin(48.7^\circ) \) into the equation:\[ n_1 = \frac{1.333}{\sin(48.7^\circ)} \approx 1.77 \].
05

Final Step: Verify Calculation

Calculate \( \sin(48.7^\circ) \) which is approximately 0.752, and then verify the refractive index calculation:\[ n_1 \approx \frac{1.333}{0.752} \approx 1.77 \]This confirms the refractive index of glass is approximately 1.77.

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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 that governs the behavior of light as it travels between different media. Imagine a ray of light moving from one medium into another. Snell's Law helps us calculate how much the light ray bends, changing its direction, at the boundary between the two media.

The law is mathematically expressed as: \[ n_1 \sin(\theta_1) = n_2 \sin(\theta_2) \] where:
  • \( n_1 \) and \( n_2 \) are the refractive indices of the first and second medium, respectively.
  • \( \theta_1 \) is the angle of incidence, the angle between the incoming ray and the normal to the surface.
  • \( \theta_2 \) is the angle of refraction, the angle between the refracted ray and the normal.
When light passes from a denser medium like glass into a less dense medium like water, it bends away from the normal if \(n_1 > n_2\). Snell's Law is essential for understanding phenomena like lenses, prisms, and fiber optics.

Always remember, when applying Snell's Law, the sine of the angles is used, and this law only applies when light passes from one medium into another, not when it reflects.
critical angle
The critical angle is a special angle of incidence that occurs when light travels from a medium with a higher refractive index to one with a lower refractive index. Above this angle, total internal reflection occurs, meaning that all the light is reflected back into the initial medium rather than passing through the interface.

Understanding this is fundamental for applications in fiber optic cables where light needs to be confined inside a core.
  • The critical angle \( \theta_c \) can be found when the angle of refraction \( \theta_2 \) is \( 90^{\circ} \). At this point, the refracted ray runs along the surface of the two media.
  • Using Snell's Law, when the refracted angle is \( 90^{\circ} \), we have: \[ n_1 \sin(\theta_c) = n_2 \sin(90^{\circ}) \] which simplifies to \[ n_1 \sin(\theta_c) = n_2 \]
  • Thus, \( \theta_c = \arcsin\left( \frac{n_2}{n_1} \right) \).
In scenarios like the exercise, when no light is refracted beyond the critical angle, it results in total internal reflection, a concept crucial in the design of many optical devices.
total internal reflection
Total internal reflection is an optical phenomenon that happens when light traveling through a medium hits the boundary with another medium at an angle greater than the critical angle. When this condition is met, none of the light crosses the boundary; instead, all of it reflects back into the original medium.

Let's explore the key factors:
  • Occurs when light moves from a more dense medium, like glass, into a less dense medium, like air or water.
  • Requires the angle of incidence to be greater than the critical angle \( \theta_c \).
  • Applications of total internal reflection include optical fibers, where light signals are confined within the core to transmit data efficiently over long distances without loss.
This principle is beneficial in directing light as it prevents light from escaping, allowing it to "bounce" along the walls of the medium. Understanding total internal reflection helps in designing efficient lighting systems, enhancing the performance of optical instruments, and crucially in enhancing technologies like endoscopes and telecommunications.

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Most popular questions from this chapter

The critical angle for total internal reflection at a liquid-air interface is \(42.5^{\circ} .\) (a) If a ray of light traveling in the liquid has an angle of incidence at the interface of \(35.0^{\circ},\) what angle does the refracted ray in the air make with the normal? (b) If a ray of light traveling in air has an angle of incidence at the interface of \(35.0^{\circ}\) , what angle does the refracted ray in the liquid make with the normal?

Light traveling in air is incident on the surface of a block of plastic at an angle of \(62.7^{\circ}\) to the normal and is bent so that it makes a \(48.1^{\circ}\) angle with the normal in the plastic. Find the speed of light in the plastic.

Heart Sonogram. Physicians use high-frequency \((f=1-5 \mathrm{MHz})\) sound waves, called ultrasound, to image internal organs. The speed of these ultrasound waves is 1480 \(\mathrm{m} / \mathrm{s}\) in muscle and 344 \(\mathrm{m} / \mathrm{s}\) in air. We define the index of refraction of a material for sound waves to be the ratio of the speed of sound in air to the speed of sound in the material. Snell's law then applies to the refraction of sound waves. (a) At what angle from the normal does an ultrasound beam enter the heart if it leaves the lungs at an angle of \(9.73^{\circ}\) from the normal to the heart wall? (Assume that the speed of sound in the lungs is 344 \(\mathrm{m} / \mathrm{s} .\) ) (b) What is the critical angle for sound waves in air incident on muscle?

A light beam is directed parallel to the axis of a hollow cylindrical tube. When the tube contains only air, it takes the light 8.72 ns to travel the length of the tube, but when the tube is filled with a transparent jelly, it takes the light 2.04 ns longer to travel its length. What is the refractive index of this jelly?

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 \(n^{\prime} .\)

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