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Chapter 32: Problem 58

The critical angle for a certain liquid-air surface is \(49.6^{\circ}\) What is the index of refraction of the liquid?

Short Answer

Expert verified
The index of refraction of the liquid is approximately 1.315.

Step by step solution

01

Understand the Concept of Critical Angle

The critical angle is the angle of incidence beyond which light cannot pass through the boundary between two media and instead is totally internally reflected. The critical angle only occurs when moving from a more optically dense medium to a less dense one.
02

Use Snell's Law

Snell's Law states \( n_1 \sin(\theta_1) = n_2 \sin(\theta_2) \), where \( n_1 \) and \( n_2 \) are the refractive indices of the two media. At the critical angle (\( \theta_c \)), \( \theta_2 = 90^{\circ} \) and \( \sin(\theta_2) = 1 \). This simplifies to \( n_1 \sin(\theta_c) = n_2 \).
03

Set Up the Equation for Critical Angle

For a liquid-air surface, the index of refraction of air \( n_2 \) is approximately 1 (since air has an index of refraction close to 1). The equation becomes \( n_{liquid} \cdot \sin(49.6^{\circ}) = 1 \).
04

Calculate the Index of Refraction

Rearrange the equation to solve for the index of refraction of the liquid: \( n_{liquid} = \frac{1}{\sin(49.6^{\circ})} \). Use a calculator to find \( \sin(49.6^{\circ}) \), which is approximately 0.7604. Then, compute: \( n_{liquid} \approx \frac{1}{0.7604} \approx 1.315 \).

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Key Concepts

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

Optical Density
Optical density plays a critical role in understanding how light interacts with materials. When we talk about optical density, we refer to the ability of a material to slow down light passing through it. Materials with a higher optical density slow down light more significantly than materials with a lower optical density. This effect occurs because of the interactions at the atomic level, where light waves are absorbed and re-emitted as they pass through the material. It鈥檚 important to note that optical density is not the same as mass density. A material might be very dense in terms of mass but still be low in optical density if light passes through it easily. The key takeaway is that optical density is related to the refractive index: the greater the optical density, the higher the refractive index. Therefore, when light travels from a medium with higher optical density to one with lower optical density, it speeds up.
Snell's Law
Snell's Law is a fundamental principle that explains how light bends, or refracts, as it passes from one medium into another. Snell's Law is expressed with the equation:
  • \( n_1 \sin(\theta_1) = n_2 \sin(\theta_2) \)
This formula shows the relationship between the angles of incidence and refraction and the refractive indices of two different media. In practical terms, if light passes from air into water, the angle formed with the normal by the light's entry (incidence) and its path within the water (refraction) are influenced by the media鈥檚 refractive indices. When the angle of refraction becomes 90 degrees, we reach the critical angle, and the light cannot enter the second medium but instead reflects entirely within the first medium. At this critical angle point, Snell鈥檚 Law simplifies to \( n_1 \sin(\theta_c) = n_2 \), which is crucial for calculating the refractive index when dealing with critical angles.
Index of Refraction
The index of refraction, also called the refractive index, measures how much light slows down as it enters a material. It is calculated by the ratio of the speed of light in a vacuum to the speed of light in the medium. Mathematically, it's represented by:
  • \( n = \frac{c}{v} \)
where \( c \) is the speed of light in a vacuum, and \( v \) is the speed of light in the medium. The refractive index also determines how much the light bends as it enters a new medium. A higher index of refraction means that light will bend more upon entering the medium. For example, water has a refractive index of about 1.33, meaning light travels more slowly in water compared to air. In the exercise, by knowing the critical angle, we can rearrange Snell's Law to solve for the unknown refractive index of the liquid. This shows how these concepts are interconnected and underscore the importance of understanding the refractive index in studying optics.

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Most popular questions from this chapter

(I) A light beam strikes a piece of glass at a \(60.00^{\circ}\) incident angle. The beam contains two wavelengths, 450.0 \(\mathrm{nm}\) and \(700.0 \mathrm{nm},\) for which the index of refraction of the glass is 1.4831 and 1.4754 , respectively. What is the angle between the two refracted beams?

Some rearview mirrors produce images of cars to your rear that are smaller than they would be if the mirror were flat. Are the mirrors concave or convex? What is a mirror's radius of curvature if cars \(18.0 \mathrm{~m}\) away appear 0.33 their normal size?

(II) A concave mirror has focal length \(f .\) When an object is placed a distance \(d_{\mathrm{o}}>f\) from this mirror, a real image with magnification \(m\) is formed. \((a)\) Show that \(m=f /\left(f-d_{\mathrm{o}}\right)\) (b) Sketch \(m\) vs. \(d_{0}\) over the range \(f < d_{\mathrm{o}} < +\infty\) where \(f=0.45 \mathrm{m} .(c)\) For what value of \(d_{\mathrm{o}}\) will the real image have the same (lateral) size as the object? \((d)\) To obtain a real image that is much larger than the object, in what general region should the object be placed relative to the mirror?

(II) The index of refraction, \(n,\) of crown flint glass at different wavelengths \((\lambda)\) of light are given in the Table below. $$\begin{array}{|c|c|c|c|c|}\hline \lambda(\mathrm{nm}) & {1060} & {546.1} & {365.0} & {312.5} \\ \hline n & {1.50586} & {1.51978} & {1.54251} & {1.5600} \\\ \hline\end{array}$$ Make a graph of \(n\) versus \(\lambda\) . The variation in index of refraction with wavelength is given by the Cauchy equation \(n=A+B / \lambda^{2} .\) Make another graph of \(n\) versus 1\(/ \lambda^{2}\) and determine the constants \(A\) and \(B\) for the glass by fitting the data with a straight line.

(1) Suppose that you want to take a photograph of yourself as you look at your image in a mirror 2.8 \(\mathrm{m}\) away. For what distance should the camera lens be focused?
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