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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 index of refraction for the oil is 1.288.

Step by step solution

01

Understanding Total Internal Reflection

Total internal reflection occurs when light traveling in a medium with a higher index of refraction cannot exit into a medium with a lower index of refraction because the angle of incidence is greater than the critical angle. To find the critical angle \( \theta_c \), we use \( \sin \theta_c = \frac{n_2}{n_1} \), where \( n_1 \) is the index of refraction of glass and \( n_2 \) is that of oil.
02

Calculate Critical Angle

Using \( n_1 = 1.52 \) for glass and \( n_2 = n_o \) for oil, the critical angle criterion becomes \( \sin \theta_c = \frac{n_o}{1.52} \). We know the angle of incidence \( 57.2^{\circ} \) must be greater than \( \theta_c \), thus \( 57.2^{\circ} > \theta_c \). Calculating \( \theta_c \) using this condition gives us \( \sin(57.2^{\circ}) = 0.848 \).
03

Determine Maximum Index of Refraction for Oil

Since \( \sin \theta_c = \frac{n_o}{1.52} \) and \( \sin(57.2^{\circ}) = 0.848 \), the oil's maximum index of refraction \( n_o \) can be determined as \( n_o < 0.848 \times 1.52 \). Calculating this gives \( n_o < 1.288 \). Therefore, the maximum possible index of refraction for the oil for total internal reflection to occur is \( n_o = 1.288 \).

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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 passes through a medium. It is a dimensionless number that represents the ratio of the speed of light in a vacuum to its speed in the medium.
  • If \( n = 1 \), the medium is a vacuum, where light travels at its maximum speed.
  • A higher \( n \) indicates that light travels slower in the medium.
  • For instance, in the given exercise, the glass block has an index of refraction of \( n = 1.52 \), meaning light travels slower in glass compared to a vacuum.
The index of refraction is crucial in determining how light bends, or refracts, when transitioning between different materials.
By comparing the indices of refraction of two materials, you can predict the bending angle of the light at the interface. This is especially important when analyzing phenomena like total internal reflection, where knowing the indices helps to understand under what conditions light will be totally reflected back into the original medium.
Critical Angle
The critical angle is a specific angle of incidence at which light traveling from a medium with a higher index of refraction to a medium with a lower index of refraction is refracted along the boundary. Beyond this angle, light does not exit the medium, leading to total internal reflection.
To find the critical angle \( \theta_c \), use the formula:\[\sin \theta_c = \frac{n_2}{n_1}\]where:
  • \( n_1 \) is the index of refraction of the initial medium (glass in this case, with \( n_1 = 1.52 \)).
  • \( n_2 \) is the index of refraction of the second medium (oil in the exercise).
The key is that the angle of incidence must be greater than the critical angle for total internal reflection to occur.
In the exercise, the angle of incidence is \( 57.2^{\circ} \), which must be greater than the critical angle to ensure total internal reflection. Calculating the critical angle involves using the known indices and confirming it is surpassed, as was done with \( \sin(57.2^{\circ}) = 0.848 \).
Snell's Law
Snell's Law is the fundamental principle that describes the behavior of light as it passes from one medium into another, including how it bends.
The law is stated as:\[n_1 \sin \theta_1 = n_2 \sin \theta_2\]where:
  • \( n_1 \) and \( n_2 \) are the indices of refraction for the media the light travels through.
  • \( \theta_1 \) and \( \theta_2 \) are the angles of incidence and refraction, respectively, measured from the normal (an imaginary line perpendicular to the surface).
Using Snell's Law allows you to find unknown angles or indices of refraction when certain parameters are known.
In situations of total internal reflection as in the exercise, the concept is a bit inverted. We use the condition that the light does not exit the first medium because the angle of incidence exceeds the critical angle calculated using the indices of refraction as explained previously. Thus, Snell's Law helps bridge understanding between refraction and total reflection behaviors based on existing conditions.

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Most popular questions from this chapter

The critical angle for total internal reflection at a liquid-air interface is \(42.5^{\circ} .\) (a) If a ray of light traveling in the liquid has an angle of incidence of \(35.0^{\circ}\) at the interface, what angle does the refracted ray in the air make with the normal? (b) If a ray of light traveling in air has an angle of incidence of \(35.0^{\circ}\) at the interface, what angle does the refracted ray in the liquid make with the normal?

A parallel-sided plate of glass having a refractive index of 1.60 is in contact with the surface of water in a tank. A ray coming from above makes an angle of incidence of \(32.0^{\circ}\) with the normal to the top surface of the glass. What angle does this ray make with the normal in the water?

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?

A sinusoidal electromagnetic wave from a radio station passes perpendicularly through an open window that has area \(0.500 \mathrm{~m}^{2}\). At the window, the electric field of the wave has rms value \(0.0200 \mathrm{~V} / \mathrm{m} .\) How much energy does this wave carry through the window during a 30.0 s commercial?

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