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Chapter 23: Problem 75

A beaker with a mirrored bottom is filled with a liquid whose index of refraction is 1.63. A light beam strikes the top surface of the liquid at an angle of \(42.5^{\circ}\) from the normal. At what angle from the normal will the beam exit from the liquid after traveling down through it, reflecting from the mirrored bottom, and returning to the surface?

Short Answer

Expert verified
The refraction angle in the liquid is calculated using Snell's Law.

Step by step solution

01

Understanding Snell's Law

To determine the angle at which the beam exits the liquid, we will first understand Snell's Law. Snell's Law relates the angles of incidence and refraction when light passes through different media. It is given by the formula: \( n_1 \sin \theta_1 = n_2 \sin \theta_2 \). Here, \( n_1 \) and \( n_2 \) are the refractive indices of medium 1 (air) and medium 2 (liquid), and \( \theta_1 \) and \( \theta_2 \) are the angles of incidence and refraction respectively.
02

Application of Snell's Law at the Liquid Surface

The light beam enters the liquid at an angle of \(42.5^{\circ}\). The index of refraction for air is approximately 1, and for the liquid, it is given as 1.63. Apply Snell's Law at the interface where the beam enters: \( 1 \times \sin 42.5^{\circ} = 1.63 \times \sin \theta_2 \). Solve this equation to find \( \theta_2 \), which is the angle of refraction inside the liquid.
03

Calculating the Refracted Angle Inside the Liquid

First, calculate \( \sin 42.5^{\circ} \). Then, rearrange the equation from Snell's Law to find \( \sin \theta_2 = \frac{\sin 42.5^{\circ}}{1.63} \). By calculating this, you can find \( \theta_2 \) by taking the inverse sine (arcsine): \( \theta_2 = \arcsin \left( \frac{\sin 42.5^{\circ}}{1.63} \right) \).

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Key Concepts

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

Index of Refraction
The index of refraction, often denoted as "n," is a measure of how much light slows down when it enters a substance. It is a crucial concept in optics that determines how light bends or refracts when moving between different materials. A higher index of refraction indicates that light travels more slowly in that medium compared to the speed of light in a vacuum. For instance:
  • A common index of refraction for air is approximately 1, meaning light travels nearly at its maximum speed.
  • A liquid in the exercise with an index of refraction of 1.63 implies light travels slower through it.
The value of the index is used in Snell's Law to calculate the amount of bending of light rays as they enter or exit a material.
Angle of Incidence
The angle of incidence is the angle formed between the incoming light ray and the normal (an imaginary line perpendicular to the surface at the point of contact). In optical problems, it is denoted by \( \theta_1 \). Understanding this angle is critical since it determines how light will interact with the surface it strikes.
In many cases, like in the exercise, the angle of incidence is provided. Here, the light beam enters the liquid at an angle of \( 42.5^{\circ} \). This is an essential part of applying Snell's Law because it helps predict the corresponding angle of refraction.
Angle of Refraction
The angle of refraction is the angle between the refracted ray and the normal inside the material into which the light has entered. In optics, it is represented by \( \theta_2 \). This angle is found using Snell's Law, helping to predict how light travels through a medium.
  • For instance, applying Snell's Law as in the exercise: \( n_1 \sin \theta_1 = n_2 \sin \theta_2 \).
  • The angle of refraction is significant because it dictates the path of the light within the new medium, such as a liquid.
After calculating this angle, it can also help in determining how the light exits the medium, potentially after reflecting off internal surfaces.
Mirror Reflection in Optics
Mirror reflection in optics refers to the bouncing back of a light ray after hitting a reflective surface. In many exercises, this involves light passing through a medium, reflecting off a mirrored surface, and then re-emerging back through the original medium. This reflection follows the law of reflection, stating the angle of incidence equals the angle of reflection.
In the context of the exercise, the light refracts as it enters a liquid, hits a mirrored bottom, and retraces its path back to the interface. This involves:
  • The initial angle of incidence impacting the path of the light through the liquid.
  • The reflective property of the bottom surface, allowing light to travel back upward.
This dual interaction - refraction and reflection - is key to solving complex optical problems like the one at hand.

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Most popular questions from this chapter

The intensity at a certain distance from a bright light source is \(6.00 \mathrm{~W} / \mathrm{m}^{2} .\) Find the radiation pressure (in pascals and in atmospheres) on (a) a totally absorbing surface and (b) a totally reflecting surface.

The microprocessor in a modern laptop computer runs on a 2.5 GHz clock. Assuming that the electrical signals in the computer travel at two-thirds of the speed of light, how far does a signal travel in one clock cycle?

The electric field of a sinusoidal electromagnetic wave obeys the equation \(E=-(375 \mathrm{~V} / \mathrm{m}) \sin \left[\left(5.97 \times 10^{15} \mathrm{rad} / \mathrm{s}\right) t+(1.99 \times\right.\) \(\left.\left.10^{7} \mathrm{rad} / \mathrm{m}\right) x\right] .\) (a) What are the amplitudes of the electric and magnetic fields of this wave? (b) What are the frequency, wavelength, and period of the wave? Is this light visible to humans? (c) What is the speed of the wave?

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 \(400 \mathrm{nm}\) (violet) to \(700 \mathrm{nm}\) (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?

We can reasonably model a \(75 \mathrm{~W}\) incandescent lightbulb as a sphere \(6.0 \mathrm{~cm}\) in diameter. Typically, only about \(5 \%\) of the energy goes to visible light; the rest goes largely to nonvisible infrared radiation. (a) What is the visible light intensity (in \(\mathrm{W} / \mathrm{m}^{2}\) ) at the surface of the bulb? (b) What are the amplitudes of the electric and magnetic fields at this surface, for a sinusoidal wave with this intensity?
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