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Orange light has a wavelength of \(600 \mathrm{nm},\) and green light has a wavelength of \(510 \mathrm{nm}\). (a) What is the difference in frequency between the two types of light? (b) If you doubled the wavelength of both, what type of light would thev become?

Short Answer

Expert verified
(a) The difference in frequency is \(0.88 \times 10^{14} \text{ Hz}\). (b) Both become infrared light.

Step by step solution

01

Understand the Relationship between Wavelength and Frequency

The frequency \( f \) of light is related to its wavelength \( \lambda \) and the speed of light \( c \) by the equation \( f = \frac{c}{\lambda} \), where \( c = 3 \times 10^8 \) meters per second.
02

Calculate the Frequency of Orange Light

The wavelength of orange light is \( \lambda_o = 600 \text{ nm} = 600 \times 10^{-9} \text{ meters} \). Using the formula \( f = \frac{c}{\lambda} \), compute the frequency: \[ f_o = \frac{3 \times 10^8}{600 \times 10^{-9}} = 5 \times 10^{14} \text{ Hz} \].
03

Calculate the Frequency of Green Light

The wavelength of green light is \( \lambda_g = 510 \text{ nm} = 510 \times 10^{-9} \text{ meters} \). Using the same formula, compute the frequency: \[ f_g = \frac{3 \times 10^8}{510 \times 10^{-9}} = 5.88 \times 10^{14} \text{ Hz} \].
04

Calculate the Difference in Frequency

Subtract the frequency of orange light from the frequency of green light: \( \Delta f = f_g - f_o = 5.88 \times 10^{14} - 5 \times 10^{14} = 0.88 \times 10^{14} \text{ Hz} \).
05

Determine the New Wavelengths When Doubled

If the wavelengths of both lights are doubled, the new wavelengths are \( \lambda'_o = 2 \times 600 = 1200 \text{ nm} \) and \( \lambda'_g = 2 \times 510 = 1020 \text{ nm} \).
06

Identify the Type of Light Based on New Wavelengths

1200 nm is in the infrared region, and 1020 nm is also in the infrared region. Therefore, both wavelengths become types of infrared light.

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Key Concepts

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

Orange Light
Orange light sits in a fascinating part of the visible spectrum. Specifically, it has a wavelength of about 600 nanometers (nm). The visible light spectrum ranges approximately between 400 nm and 700 nm.
What makes orange light special is how it interacts with our eyes. Shorter wavelengths generally have higher frequencies. This explains why the orange glow is quite distinct.
  • Wavelength: 600 nm
  • Frequency: Calculated as \(5 \times 10^{14}\) Hz using the formula \(f = \frac{c}{\lambda}\).
The frequency calculation is crucial. It helps us understand how often the light waves hit a point in a second. Knowing this can help characterize different light types.
When you see a radiant orange sunset, you are seeing light waves mainly around this wavelength.
Green Light
Green light is another major player in the visible spectrum. Its wavelength measures around 510 nanometers (nm). It is situated between blue and yellow in the spectrum, and this impacts how we perceive colors.
  • Wavelength: 510 nm
  • Frequency: Roughly \(5.88 \times 10^{14}\) Hz as calculated using \(f = \frac{c}{\lambda}\).
The frequency indicates how rapid the wave cycles are. This is higher than that of orange light due to its shorter wavelength. Green light is abundant in nature, especially in plants. The chlorophyll in leaves strongly reflects it.
That's why forests are usually rich in green hues. Understanding this light helps us grasp the role light plays in our environment.
Infrared Light
Infrared light is a type of electromagnetic radiation with longer wavelengths than visible light. In many cases, it's associated with heat. Devices like remote controls use it to send signals.
  • Wavelengths: Typically range from 700 nm to 1 mm.
When the wavelengths of orange and green light are doubled, they transition into the infrared region. Both lights become infrared, specifically:
  • New Orange Wavelength: 1200 nm
  • New Green Wavelength: 1020 nm
Infrared light can't be seen by the naked eye, but we can feel it as heat. This is why when you place your hand near a warm object, you're sensing infrared radiation.
In summary, the shift of visible light to infrared demonstrates how altering wavelengths impacts the light's properties.
Speed of Light
The speed of light is a fundamental constant, crucial to understanding electromagnetic waves. Light travels at a staggering speed of approximately 300,000,000 meters per second (or \(3 \times 10^8\) m/s).
  • Relation: Speed of light \( (c) = \text{Wavelength} \times \text{Frequency} \).
This speed is not just a number; it is critical to equations that describe light behavior, like determining frequencies and wavelengths.
Remember, light speed in a vacuum is the ultimate speed limit in the universe. Understanding this helps when examining the behavior of different light waves, such as how orange and green light differ in frequency.
Ultimately, the speed of light underpins much of modern physics and gives insight into the behaviors and characteristics of different forms of light.

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Most popular questions from this chapter

An ideal transformer steps \(8.0 \mathrm{~V}\) up to \(2000 \mathrm{~V}\), and the 4000-turn secondary coil carries 2.0 A. (a) Find the number of turns in the primary coil. (b) Find the current in the primary coil.

A circular loop with an area of \(0.015 \mathrm{~m}^{2}\) is in a uniform magnetic field of \(0.30 \mathrm{~T}\). What is the flux through the loop's plane if it is (a) parallel to the field, (b) at an angle of \(37^{\circ}\) to the field, and (c) perpendicular to the field?

A boy is traveling due north at a constant speed while carrying a metal rod. The rod's length is oriented in the east-west direction and is parallel to the ground. (a) There will be no induced emf when the rod is (1) at the equator, (2) near the Earth's magnetic poles, (3) somewhere between the equator and the poles. Why? (b) Assume that the Earth's magnetic field is \(1.0 \times 10^{-4} \mathrm{~T}\) near the North Pole and \(1.0 \times 10^{-5} \mathrm{~T}\) near the equator. If the boy runs with a speed of \(1.3 \mathrm{~m} / \mathrm{s}\) northward near each location, and the rod is \(1.5 \mathrm{~m}\) long, calculate the induced emf in the rod in each location.

The plane of a conductive loop with an area of \(0.020 \mathrm{~m}^{2}\) is perpendicular to a uniform magnetic field of \(0.30 \mathrm{~T}\). If the field drops to zero in \(0.0045 \mathrm{~s}\), what is the magnitude of the average emf induced in the loop?

An ideal solenoid with a current of 1.5 A has a radius of \(3.0 \mathrm{~cm}\) and a turn density of 250 turns \(/ \mathrm{m}\). (a) What is the magnetic flux (due to its own field) through only one of its loops at its center? (b) What current would be required to double the flux value in part (a)?

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