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Chapter 26: Problem 107

A glass block \((n=1.56)\) is immersed in a liquid. A ray of light within the glass hits a glass-liquid surface at a \(75.0^{\circ}\) angle of incidence. Some of the light enters the liquid. What is the smallest possible refractive index for the liquid?

Short Answer

Expert verified
The smallest refractive index for the liquid is approximately 1.51.

Step by step solution

01

Understand Snell's Law

Snell's Law describes how light bends when passing through different media. It is given by the formula \( n_1 \sin\theta_1 = n_2 \sin\theta_2 \), where \( n_1 \) and \( \theta_1 \) are the refractive index and angle of incidence in the first medium, and \( n_2 \) and \( \theta_2 \) are the refractive index and angle of refraction in the second medium.
02

Identify Given Values

From the problem statement, we know: - The refractive index for the glass is \( n_1 = 1.56 \).- The angle of incidence is \( \theta_1 = 75.0^\circ \).- We need to find the smallest possible refractive index for the liquid, \( n_2 \).
03

Apply Critical Angle Concept

The critical angle occurs when the angle of refraction is \(90^\circ\), meaning the light travels along the boundary. For the smallest refractive index in the liquid, we are considering this scenario. At critical angle, \( \theta_2 = 90^\circ \), and \( \sin\theta_2 = 1 \).
04

Use Snell's Law for the Critical Angle

Using Snell's Law for the scenario when the light barely refracts, set up the equation: \[ n_1 \sin\theta_1 = n_2 \sin 90^\circ \]Simplifying, we get \[ n_1 \sin\theta_1 = n_2 \].
05

Calculate the Smallest Refractive Index

Insert the known values into the equation: \[ 1.56 \sin 75.0^\circ = n_2 \]Calculate \( \,\sin 75.0^\circ \, \approx \, 0.9659 \):\[ 1.56 \times 0.9659 = n_2 \].Thus, the smallest possible refractive index for the liquid is approximately \( n_2 \approx 1.51 \).

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Key Concepts

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

Refractive Index
The refractive index is a measure of how much light bends, or refracts, when it enters a different medium. Think of it like a speed limit sign for light traveling through various materials. When light moves from one medium into another鈥攍ike from air into glass鈥攊t slow downs or speeds up.
This change in speed causes the light to bend. The refractive index, denoted as \(n\), is calculated by comparing the speed of light in a vacuum to its speed in the material:
  • Air, which barely slows light down, has a refractive index close to 1.
  • For glass, it is typically around 1.56, much like the situation in this problem.
High refractive indices mean light bends more sharply whereas low values result in gentler bending. Understanding this helps us predict how light will behave at boundaries between different materials.
Angle of Incidence
The angle of incidence is the angle at which incoming light strikes a surface. Imagine shining a flashlight directly onto a mirror; the light hits the surface at an angle, and this angle is what we call the angle of incidence. It is measured from the normal, an imaginary line perpendicular to the surface, not from the surface itself.
In our exercise, the angle of incidence is given as \(75.0^{\circ}\), which means the light is hitting the glass-liquid interface quite steeply. This angle is crucial because it influences how much the light will bend when it enters another medium, as per Snell's Law.
Understanding the angle of incidence aids in calculating how light travels across different surfaces and is pivotal for designing things like lenses and optical fibers.
Critical Angle
The critical angle is a special angle of incidence when light moves from a denser medium to a less dense one, like from glass to air or liquid. It's the "just right" angle where light skims along the boundary rather than passing into the second medium.
At this angle, the refracted ray is exactly \(90^{\circ}\), meaning it doesn't enter the new medium but runs parallel to the boundary surface.
  • Below this angle, light will refract into the second medium.
  • At and beyond, it reflects back into the original medium鈥攁 phenomenon known as total internal reflection.
In the exercise, determining the critical angle allows us to calculate the smallest possible refractive index for the liquid. Knowing the critical angle is vital for technologies like fiber optics, which use it to trap light within glass or plastic fibers, guiding it over long distances.

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Most popular questions from this chapter

Violet light and red light travel through air and strike a block of plastic at the same angle of incidence. The angle of refraction is \(30.400^{\circ}\) for the violet light and \(31.200^{\circ}\) for the red light. The index of refraction for violet light in plastic is greater than that for red light by \(0.0400 .\) Delaying any rounding off of calculations until the very end, find the index of refraction for violet light in plastic.

An object is located \(9.0 \mathrm{cm}\) in front of a converging lens \((f=\) \(6.0 \mathrm{cm}$$ )\). Using an accurately drawn ray diagram, determine where the image is located.

A ray of sunlight is passing from diamond into crown glass; the angle of incidence is \(35.00^{\circ} .\) The indices of refraction for the blue and red components of the ray are: blue \(\left(n_{\text {diamond }}=2.444, n_{\text {crown glass }}=1.531\right),\) and red \(\left(n_{\text {diamond }}=2.410, n_{\text {crown glass}}=1.520\right) .\) Determine the angle between the refracted blue and red rays in the crown glass.

In a compound microscope, the focal length of the objective is \(3.50 \mathrm{cm}\) and that of the eyepiece is \(6.50 \mathrm{cm}\). The distance between the lenses is \(26.0 \mathrm{cm} .\) (a) What is the angular magnification of the microscope if the person using it has a near point of \(35.0 \mathrm{cm} ?\) (b) If, as usual, the first image lies just inside the focal point of the eyepiece (see Figure 26.32 ), how far is the object from the objective? (c) What is the magnification (not the angular magnification) of the objective?

A ray of light impinges from air onto a block of ice \((n=1.309)\) at a \(60.0^{\circ}\) angle of incidence. Assuming that this angle remains the same, find the difference \(\theta_{2, \text { ice }}-\theta_{2, \text { water }}\) in the angles of refraction when the ice turns to water \((n=1.333)\)
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