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Red light \((n=1.520)\) and violet light \((n=1.538)\) traveling in air are incident on a slab of crown glass. Both colors enter the glass at the same angle of refraction. The red light has an angle of incidence of \(30.00^{\circ} .\) What is the angle of incidence of the violet light?

Short Answer

Expert verified
The angle of incidence for violet light is approximately 30.33°.

Step by step solution

01

Understand the Problem

We are given two different colors of light, red and violet, which both enter a crown glass slab at the same angle of refraction. Red light has a known angle of incidence of 30° in air. Our task is to find the angle of incidence for the violet light.
02

Use Snell's Law for Red Light

Snell's Law is written as \(n_1 \sin\theta_1 = n_2 \sin\theta_2\). For red light, we have \(n_1=1\) (refractive index of air), \(\theta_1=30^{\circ}\), and \(n_2=1.520\). Substitute these values into Snell's Law to find \(\theta_2\) (angle of refraction).
03

Calculate Angle of Refraction for Red Light

Using \(\sin(30^{\circ}) = 0.5\), we have:\[1 \times 0.5 = 1.520 \sin\theta_2\]Solve for \(\theta_2\):\[\sin\theta_2 = \frac{0.5}{1.520} \approx 0.3289\]\[\theta_2 \approx \sin^{-1}(0.3289) \approx 19.16^{\circ}\]
04

Express Snell's Law for Violet Light

Knowing the angle of refraction \(\theta_2\) for both red and violet lights is the same, use \(n_1 \sin\theta_1 = n_3 \sin\theta_2\) for violet light, where \(n_3 = 1.538\) and \(\sin\theta_2 = 0.3289\) from the previous step.
05

Solve for Angle of Incidence of Violet Light

With \(\theta_2 = 19.16^{\circ}\) for violet, calculate:\[1 \cdot \sin\theta_3 = 1.538 \cdot 0.3289\]\[\sin\theta_3 = 0.5057\]\[\theta_3 \approx \sin^{-1}(0.5057) \approx 30.33^{\circ}\]
06

Conclusion

The angle of incidence for violet light is approximately \(30.33^{\circ}\).

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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 crucial concept in understanding how light bends when it passes from one medium to another. It is denoted by the symbol "n" and is defined as the ratio of the speed of light in a vacuum to the speed of light in the medium. The refractive index determines how much the path of light is bent, or refracted, when entering another substance.

For example, in air, the refractive index is approximately 1, while in crown glass, it can vary slightly depending on the color of light. Red light, for instance, has a refractive index of 1.520 in crown glass, and violet light has a refractive index of 1.538. The difference in refractive indices for different colors is because light of different wavelengths travels at slightly different speeds in the same medium.
  • The higher the refractive index, the slower the light travels in the medium.
  • The refractive index enables calculation of how much light bends using Snell’s Law.
  • A medium with a higher refractive index bends light more sharply than one with a lower refractive index.
Understanding refractive indices is key to solving problems involving Snell's Law, which is used to determine the relationship between angles of incidence and refraction.
Angle of Incidence
The angle of incidence refers to the angle at which an incoming wave, such as a beam of light, strikes the surface of a medium. Using this, we can predict how the light will behave when it meets the boundary of a different medium.

In this scenario, red light is incident on crown glass at an angle of 30° from air. The angle of incidence is the angle between the incoming light ray and a perpendicular line to the surface, known as the normal.
  • The angle of incidence is critical in determining the angle of refraction when combined with the refractive indices of the involved media.
  • Using Snell’s Law, the refractive index and angle of incidence allows us to calculate the angle at which light exits or refracts within the new medium.
The solution to the exercise requires knowing the angle of incidence for both red and violet light, which is part of calculating how each color behaves differently when entering glass.
Angle of Refraction
The angle of refraction is the angle formed between the refracted ray and the normal to the surface at the point of refraction. It is calculated using Snell's Law, which relates the angle of incidence and the refractive indices of the two media.

For both red and violet light entering crown glass, the angle of refraction is a crucial element because it remains constant across different colors within the glass. In this example, both red and violet lights have an angle of refraction of approximately 19.16°. This angle helps determine how the light continues to travel through the second medium.
  • Once Snell's Law provides the angle of refraction, the path of light inside the medium can be accurately traced.
  • The consistency of the angle of refraction in different scenarios is vital for solving optical problems, especially with different light wavelengths.
In this exercise, using the same angle of refraction simplifies solving for the unknown angle of incidence for violet light since both colors penetrate at the same angle within the glass environment.

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Most popular questions from this chapter

A layer of oil \((n=1.45)\) floats on an unknown liquid. A ray of light originates in the oil and passes into the unknown liquid. The angle of incidence is \(64.0^{\circ},\) and the angle of refraction is \(53.0^{\circ} .\) What is the index of refraction of the unknown liquid?

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 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?

The distance between the lenses in a compound microscope is \(18 \mathrm{cm}\). The focal length of the objective is \(1.5 \mathrm{cm} .\) If the microscope is to provide an angular magnification of -83 when used by a person with a normal near point \((25 \mathrm{cm}\) from the eye), what must be the focal length of the eyepiece?

A nearsighted patient's far point is \(0.690 \mathrm{m}\) from her eyes. She is able to see distant objects in focus when wearing glasses with a refractive power of -1.50 diopters. What is the distance between her eyes and the glasses?

A flat sheet of ice has a thickness of \(2.0 \mathrm{cm} .\) It is on top of a flat sheet of crystalline quartz that has a thickness of \(1.1 \mathrm{cm} .\) Light strikes the ice perpendicularly and travels through it and then through the quartz. In the time it takes the light to travel through the two sheets, how far (in centimeters) would it have traveled in a vacuum?

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