/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 47 Light in vacuum is incident on t... [FREE SOLUTION] | 91Ó°ÊÓ

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Light in vacuum is incident on the surface of a glass slab. In the vacuum the beam makes an angle of \(32.0^{\circ}\) with the normal to the surface, while in the glass it makes an angle of \(21.0^{\circ}\) with the normal. What is the index of refraction of the glass?

Short Answer

Expert verified
The index of refraction of the glass is approximately 1.478.

Step by step solution

01

Understand the Problem

We are given the angles of incidence and refraction of a light beam as it passes from vacuum into glass. We need to find the index of refraction of the glass.
02

Apply Snell's Law

Snell's Law is given by the formula: \(n_1 \sin(\theta_1) = n_2 \sin(\theta_2)\), where \(n_1\) and \(n_2\) are the indices of refraction for the first and second medium respectively, and \(\theta_1\) and \(\theta_2\) are the angles of incidence and refraction respectively. In this case, the first medium is a vacuum, where \(n_1 = 1\), and the second medium is glass, where \(n_2\) is what we need to find.
03

Substitute Known Values

Substitute the known values into Snell's Law: \(1 \cdot \sin(32.0^{\circ}) = n_2 \cdot \sin(21.0^{\circ})\).
04

Solve for the Index of Refraction

Rearrange the previous equation to solve for \(n_2\): \(n_2 = \frac{\sin(32.0^{\circ})}{\sin(21.0^{\circ})}\).
05

Calculate the Index of Refraction

Using a calculator, find \(\sin(32.0^{\circ}) \approx 0.5299\) and \(\sin(21.0^{\circ}) \approx 0.3584\). Thus, \(n_2 \approx \frac{0.5299}{0.3584} \approx 1.478\).

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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 is a measure of how much a material slows down light compared to its speed in a vacuum. It's usually denoted by the symbol \( n \). In a vacuum, the speed of light is approximately \( 3.00 \times 10^8 \) meters per second, and this is the fastest that light can travel. The index of refraction of a medium quantifies how slower light travels in that medium.
For example, if the index of refraction of a medium is 1.5, light travels 1.5 times slower in that medium than in a vacuum. The index of refraction can also be understood as how much the light bends when entering the medium. Each material has a unique index, with water typically around 1.33, and glass can vary between 1.5 to 1.9 depending on its composition.
Angles of Incidence and Refraction
Angles in optics are measured with respect to a line called the "normal." The normal is an imaginary line perpendicular to the surface where the light enters or exits. - **Angle of Incidence** is the angle between the incoming light beam and the normal.- **Angle of Refraction** is the angle between the refracted light beam and the normal after it has passed into the second medium.
In our original exercise, the light hits the glass at an angle of \(32.0^{\circ}\) with respect to the normal; this is the angle of incidence. Once it enters the glass, it changes direction and makes an angle of \(21.0^{\circ}\) with the normal - this is the angle of refraction. Understanding these angles is crucial in applying Snell's Law correctly.
Light in Vacuum
Light behaves uniquely when traveling through a vacuum. In such an environment, there are no particles or medium to slow it down, so it travels at its maximum speed, often referred to as the "speed of light." This speed is constant and thus light does not change speed when continuing through another vacuum. The refractive index of a vacuum is defined as 1 by convention, which serves as a baseline when calculating the refractive indices of other media.
When light transitions from a vacuum to any medium with a different refractive index, it changes speed and direction. This change is described by Snell's Law, which helps us understand and predict the behavior of light in diverse environments.
Calculating Refraction
Calculating refraction involves using mathematical tools to determine how light bends when it passes from one medium to another. Snell's Law is the key formula here:\[ n_1 \sin(\theta_1) = n_2 \sin(\theta_2) \]Here:\- \( n_1 \) and \( n_2 \) are the indices of refraction of the first and second medium respectively.\- \( \theta_1 \) is the angle of incidence.\- \( \theta_2 \) is the angle of refraction.
In the provided exercise, the light is moving from a vacuum (where \( n_1 = 1 \)) into glass. By rearranging the formula to solve for \( n_2 \), the index of refraction of the glass can be calculated. This process shows us how the physics of light traveling through different substances can be predicted and understood using mathematical equations.

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Most popular questions from this chapter

Sunlight just outside Earth's atmosphere has an intensity of \(1.40 \mathrm{~kW} / \mathrm{m}^{2} .\) Calculate (a) \(E_{m}\) and (b) \(B_{m}\) for sunlight there, assuming it to be a plane wave.

A ray of white light traveling through fused quartz is incident at a quartz- air interface at angle \(\theta_{1}\). Assume that the index of refraction of quartz is \(n=1.456\) at the red end of the visible range and \(n=\) \(1.470\) at the blue end. If \(\theta_{1}\) is (a) \(42.00^{\circ}\), (b) \(43.10^{\circ}\), and (c) \(44.00^{\circ}\), is the refracted light white, white dominated by the red end of the visible range, or white dominated by the blue end of the visible range, or is there no refracted light?

What is the radiation pressure \(1.5 \mathrm{~m}\) away from a \(500 \mathrm{~W}\) lightbulb? Assume that the surface on which the pressure is exerted faces the bulb and is perfectly absorbing and that the bulb radiates uniformly in all directions.

When red light in vacuum is incident at the Brewster angle on a certain glass slab, the angle of refraction is \(32.0^{\circ}\). What are (a) the index of refraction of the glass and (b) the Brewster angle?

A helium-neon laser, radiating at \(632.8 \mathrm{~nm}\), has a power output of \(3.0 \mathrm{~mW}\). The beam diverges (spreads) at angle \(\theta=0.17\) mrad (Fig. \(33-72\) ). (a) What is the intensity of the beam \(40 \mathrm{~m}\) from the laser? (b) What is the power of a point source providing that intensity at that distance?

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