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\(\bullet$$\bullet\) A ray of light traveling in a block of glass \((n=1.52)\) is incident on the top surface at an angle of \(57.2^{\circ}\) with respect to the normal in the glass. If a layer of oil is placed on the top surface of the glass, the ray is totally reflected. What is the maximum possible index of refraction of the oil?

Short Answer

Expert verified
The maximum possible index of refraction of the oil is approximately 1.28.

Step by step solution

01

Understand the Problem

A ray of light inside a glass block with an index of refraction \(n_{glass} = 1.52\) is incident at an angle \(\theta_{glass} = 57.2^{\circ}\) to the normal of the surface. An oil layer is added on top and the light is totally internally reflected. We must find the maximum index of refraction \(n_{oil}\) for this scenario.
02

Apply Snell's Law for Total Internal Reflection

For total internal reflection at the interface between the glass and the oil, the critical angle \(\theta_c\) must satisfy:\[\sin(\theta_c) = \frac{n_{oil}}{n_{glass}}\]Since the light is totally internally reflected, an incident angle equal to or greater than \(\theta_c\) means \(\theta_{glass} = \theta_c\).
03

Solve for the Critical Angle

The critical condition for total internal reflection is:\[sin(\theta_{glass}) = \frac{n_{oil}}{n_{glass}}\]Hence:\[\sin(57.2^{\circ}) = \frac{n_{oil}}{1.52}\]Calculate \(\sin(57.2^{\circ})\).
04

Calculate \(\sin(57.2^{\circ})\)

Using a calculator, find:\[sin(57.2^{\circ}) \approx 0.8415\]This value will be used in the next step to find \(n_{oil}\).
05

Solve for Maximum Index of Refraction of Oil

Substitute \(\sin(57.2^{\circ})\) into the equation:\[0.8415 = \frac{n_{oil}}{1.52}\]Rearrange to solve for \(n_{oil}\):\[n_{oil} = 0.8415 \times 1.52\]Calculate this value.
06

Final Calculation

Perform the multiplication:\[n_{oil} \approx 1.2791\]Hence, the maximum possible index of refraction of the oil is approximately \(1.28\).

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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 describes how light behaves when it moves between different media. It states that the ratio of the sine of the angles of incidence and refraction is equal to the ratio of the indices of refraction of those two media. Mathematically, it is expressed as:
  • \( n_1 \sin(\theta_1) = n_2 \sin(\theta_2) \\)
Here, \( n_1 \) and \( n_2 \) are the indices of refraction, and \( \theta_1 \) and \( \theta_2 \) are the angles of incidence and refraction, respectively.
This law helps us predict how light will bend at the boundary between two different materials. For example, if light travels from air (\( n = 1.00 \)) into water (\( n = 1.33 \)), Snell's Law can determine how much it will bend. Understanding this principle is crucial for tackling problems where light behaves in unusual ways, such as through lenses or over transitions into different media.
In our exercise, Snell's Law helps to establish the conditions under which total internal reflection occurs. When light transitions from glass to oil, using this law will ensure that the conditions for total internal reflection are correctly identified by balancing the indices.
Critical Angle
The critical angle is a specific angle of incidence at which light traveling from one medium to another less dense medium ends up being reflected entirely back into the original medium. This phenomenon only occurs if the light is moving from a medium with a higher index of refraction to one with a lower index.
Here's how it works:
  • If \( \theta > \theta_c \): total internal reflection happens.
The critical angle \( \theta_c \) can be found using Snell’s Law in scenarios of total internal reflection, like so:
  • \( \sin(\theta_c) = \frac{n_{low}}{n_{high}} \)
where \( n_{high} \) and \( n_{low} \) are the indices of refraction for the more and less dense media, respectively.
In our original exercise, the angle exceeds the critical angle, causing total internal reflection at the boundary of glass and oil. Knowing this critical angle helps us determine the maximum possible index of refraction for the oil, setting up the scenario where this reflection condition holds true.
Index of Refraction
The index of refraction, sometimes called the refractive index, is a measure of how much light slows down as it enters a material. Each material has its own unique refractive index, and this value is crucial for explaining many optical phenomena.
Here’s what you should know:
  • A higher index means light travels slower in that medium and bends more when it enters the material.
  • It’s a dimensionless number, typically greater than 1, signifying that light always travels slower in a medium compared to a vacuum.
In our exercise, glass has an index of 1.52. This tells us that light travels significantly slower in glass than in air.
Understanding the refractive indices of different materials enables us to apply Snell's Law correctly and analyze conditions like total internal reflection. For instance, in the optical setup with glass and oil, the task is to find the maximum refractive index of the oil, ensuring that light does not just incidentally pass through but is reflected inwardly. Using steps from the solution, we realize the calculation is key to optimizing how the light behaves at these interfaces.

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Most popular questions from this chapter

\(\bullet$$\bullet\) Two sources of sinusoidal electromagnetic waves have average powers of 75 \(\mathrm{W}\) and 150 \(\mathrm{W}\) and emit uniformly in all directions. At the same distance from each source, what is the ratio of the maximum electric field for the 150 \(\mathrm{W}\) source to that of the 75 \(\mathrm{W}\) source?

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\(\bullet\) A glass plate having parallel faces and a refractive index of 1.58 lies at the bottom of a liquid of refractive index \(1.70 . \mathrm{A}\) ray of light in the liquid strikes the top of the glass at an angle of incidence of \(62.0^{\circ} .\) Compute the angle of refraction of this light in the glass.

\(\bullet\) Two ideal polarizing filters are oriented so that they transmit the maximum amount of light when unpolarized light is shone on them. To what fraction of its maximum value \(I_{0}\) is the intensity of the transmitted light reduced when the second filter is rotated through (a) \(22.5^{\circ},\) (b) \(45.0^{\circ},\) and (c) \(67.5^{\circ} ?\)

\(\bullet\) Fields from a lightbulb. 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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