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Calculate the wavelength of "(a) the sodium line at \(589 \mathrm{~nm}\) in an aqueous solution with a refractive index of \(1.35\). (b) the output of a ruby laser at \(694.3 \mathrm{~nm}\) when it is pasing through a piece of quartz that has a refractive index of \(1.55 .\)

Short Answer

Expert verified
In water: ~436.3 nm, in quartz: ~447.3 nm.

Step by step solution

01

Understand the Problem

We need to calculate the wavelength of light in a medium given its original wavelength in a vacuum and the refractive index of the medium. This requires us to use the formula for the wavelength in a medium: \( \lambda_{\text{medium}} = \frac{\lambda_{\text{vacuum}}}{n} \), where \( \lambda_{\text{medium}} \) is the wavelength in the medium, \( \lambda_{\text{vacuum}} \) is the wavelength in vacuum, and \( n \) is the refractive index.
02

Calculate Wavelength for Sodium Line in Water

Given \( \lambda_{\text{vacuum}} = 589 \ \text{nm} \) and \( n = 1.35 \) for water, use the formula: \[ \lambda_{\text{water}} = \frac{589 \ \text{nm}}{1.35} \approx 436.3 \ \text{nm} \]
03

Calculate Wavelength for Ruby Laser in Quartz

Given \( \lambda_{\text{vacuum}} = 694.3 \ \text{nm} \) and \( n = 1.55 \) for quartz, use the formula: \[ \lambda_{\text{quartz}} = \frac{694.3 \ \text{nm}}{1.55} \approx 447.3 \ \text{nm} \]
04

Conclusion

The wavelength of the sodium line in water is approximately 436.3 nm, and the wavelength of the ruby laser in quartz is approximately 447.3 nm.

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Key Concepts

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

Refractive Index
Refractive index is an important concept when we talk about how light travels through different materials. It is a dimensionless number that indicates how much a material reduces the speed of light compared to the speed of light in a vacuum. The formula for refractive index, denoted as \( n \), is given by:
  • \( n = \frac{c}{v} \)
where \( c \) is the speed of light in a vacuum, and \( v \) is the speed of light in the material. When light enters a new medium at an angle, it bends, which is called refraction. The refractive index tells us how much bending happens.
For example, water has a refractive index of about 1.33. This means light travels 1.33 times slower in water than in a vacuum. Knowing the refractive index allows us to determine how the wavelength of light changes as it moves through different materials.

Factors Affecting Refractive Index

Several factors can influence the refractive index of a material:
  • Temperature: Generally, as temperature increases, the refractive index decreases.
  • Wavelength of Light: The refractive index varies with the wavelength, causing dispersion.
  • Material Composition: Different materials have different refractive indexes.
Light Propagation in Medium
When light enters a medium other than a vacuum, it slows down due to the material's refractive index. This slowing down affects the wavelength of light because the frequency remains constant but the speed decreases. To find the new wavelength of light in a medium, we use the formula:
  • \( \lambda_{\text{medium}} = \frac{\lambda_{\text{vacuum}}}{n} \)
where \( \lambda_{\text{medium}} \) is the wavelength in the new medium, \( \lambda_{\text{vacuum}} \) is the original wavelength in vacuum, and \( n \) is the refractive index of the medium.
The direction and speed of light change when it moves from one medium to another. This is due to the change in wavelength and is why objects appear distorted when observed through different substances, like water or glass. This distortion is often what causes optical illusions.

Practical Applications

Understanding how light propagates through different media is crucial for several practical applications:
  • Designing Optical Instruments: Lenses and mirrors need precise calculations for focus and clarity.
  • Fiber Optic Communication: Knowledge of light propagation is essential for efficient data transfer.
  • Medical Imaging: Techniques such as endoscopy and optical coherence tomography rely on light interaction with tissues.
Electromagnetic Spectrum
The electromagnetic spectrum encompasses all types of electromagnetic radiation. From radio waves to gamma rays, different light forms are defined by their wavelengths and frequencies. Visible light, a small part of this spectrum, ranges from about 400 nm (violet) to about 700 nm (red). Wavelength and frequency are inversely related, meaning if one increases, the other decreases.
In various media, the wavelength of light changes, while its frequency remains constant. This shift can affect how we perceive color and is crucial for the many technologies that rely on different wavelengths of light, such as infrared cameras or ultraviolet lamps.

Components of the Electromagnetic Spectrum

The electromagnetic spectrum includes several types of radiation, each with its specific characteristics:
  • Radio Waves: Low frequency and long wavelength used in communication.
  • Microwaves: Slightly higher frequency; used for cooking and certain communication technologies.
  • Infrared: Emitted by warm objects; used for thermal imaging.
  • Visible Light: The only part of the spectrum visible to the human eye.
  • Ultraviolet: Higher energy than visible light; can cause skin tanning or sunburn.
  • X-rays: Used in medical imaging due to their ability to penetrate tissues.
  • Gamma Rays: Highest energy; used in cancer treatment.
Understanding the electromagnetic spectrum helps us harness these different types of radiation for various practical and technological applications.

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Most popular questions from this chapter

Calculate the wavelength in centimeters of *(a) an airport tower transmitting at \(118.6 \mathrm{MHz}\) (b) a VOR (radio navigation aid) transmitting at \(114.10 \mathrm{kHz}\) *(c) an NMR signal at \(105 \mathrm{MHz}\) (d) an infrared absorption peak with a wavenumber of \(1210 \mathrm{~cm}^{-1}\).

At \(580 \mathrm{~nm}\), the wavelength of its maximum absorption, the complex \(\mathrm{Fe}(\mathrm{SCN})^{2+}\) has a molar absorptivity of \(7.00 \times 10^{5} \mathrm{~L} \mathrm{~cm}^{-1} \mathrm{~mol}^{-1}\). Calculate (a) the absorbance of a \(3.40 \times 10^{-5} \mathrm{M}\) solution of the complex at \(580 \mathrm{~nm}\) in a \(1.00-\mathrm{cm}\) cell. (b) the absorbance of a solution in which the concentration of the complex is twice that in (a). (c) the transmittance of the solutions described in (a) and (b). (d) the absorbance of a solution that has half the transmittance of that described in (a).

Convert the accompanying transmittance data to absorbances. "(a) \(27.290\) (b) \(0.579\) *(c) \(30.6 \%\) (d) \(3.98 \%\) "(e) \(0.093\) (f) \(63.7 \%\)

Calculate the frequency in hertz of *(a) an X-ray beam with a wavelength of \(2.65 \mathrm{~A}\). (b) an emission line for copper at \(211.0 \mathrm{~nm}\). * (c) the line at \(694.3 \mathrm{~nm}\) produced by a ruby laser. (d) the output of a \(\mathrm{CO}_{2}\) laser at \(10.6 \mu \mathrm{m}\). *(e) an infrared absorption peak at \(19.6 \mu \mathrm{m}\). (f) a microwave beam at \(1.86 \mathrm{~cm}\).

Beryllium(II) forms a complex with acetylacetone (166.2 g \(/ \mathrm{mol})\). Calculate the molar absorptivity of the complex, given that a \(2.25\) ppm solution has a transmittance of \(37.5 \%\) when measured in a \(1.00-\mathrm{cm}\) cell at \(295 \mathrm{~nm}\), the wavelength of maximum absorption.

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