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Light of two different wavelengths is used in a double-slit experiment. The location of the third-order maximum for the first light, yellow-orange light \((\lambda=600 \mathrm{nm})\) coincides with the location of the fourth-order maximum for the other color's light. What is the wavelength of the other light?

Short Answer

Expert verified
The wavelength of the other light is 450 nm.

Step by step solution

01

Understanding the Problem

In a double-slit experiment, the location of a maximum for light of a certain order is given by the path difference formula: \( m_1 \lambda_1 = m_2 \lambda_2 \). Here, \( \lambda_1 = 600 \text{ nm} \), \( m_1 = 3 \) for the yellow-orange light, and \( m_2 = 4 \) for the other light. We need to find \( \lambda_2 \).
02

Setting up the Equation

We know that the third-order maximum of the yellow-orange light coincides with the fourth-order maximum of the other light. We use the equation \( m_1 \lambda_1 = m_2 \lambda_2 \), which becomes \( 3 \times 600 = 4 \times \lambda_2 \).
03

Solving the Equation

Rearrange the equation to solve for \( \lambda_2 \): \( \lambda_2 = \frac{3 \times 600}{4} \). Divide to find \( \lambda_2 = 450 \text{ nm} \).
04

Conclusion

The wavelength of the other light is \( \lambda_2 = 450 \text{ nm} \).

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Key Concepts

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

Light Wavelength
Light travels in waves and a crucial property of these waves is called the wavelength, symbolized by \( \lambda \). The wavelength is the distance between two consecutive peaks or troughs of a wave.
This concept is important in many scientific phenomena, including the double-slit experiment, where different wavelengths can result in different patterns.
Here are some important points about light wavelength you need to remember:
  • The wavelength determines the color of the light we see. For instance, yellow-orange light has a wavelength of about 600 nm (nanometers).
  • Shorter wavelengths are typically found in the blue/violet end of the spectrum, while longer wavelengths reside in the red/orange end.
  • In experiments, knowing the wavelength allows us to predict how light will behave as it moves through different materials or encounters obstacles.
Interference Pattern
When light passes through two slits close to each other, it creates an interference pattern on a screen behind the slits. This pattern is a series of bright and dark lines.
  • The bright lines are called maxima, where light waves from the slits meet in phase, amplifying each other.
  • The dark lines, or minima, occur where light waves meet out of phase, cancelling each other out.
These patterns occur because of the wave nature of light, showcasing one of the most celebrated properties of waves: they can interfere with each other.
This interference is central to experiments such as the double-slit experiment, helping us understand fundamental properties of light and quantum mechanics.
The pattern observed provides clues to the wavelength of the light being used, which can be crucial for calculations and understanding material properties.
Path Difference Formula
The path difference formula is a fundamental equation when studying the interference of light, especially in experiments like the double-slit experiment.
This formula helps us calculate the position of maxima and minima.
Here's the core equation and details to note:
  • The formula is expressed as \( m \lambda = d \sin \theta \), where \( m \) represents the order of the maximum, \( \lambda \) is the wavelength, \( d \) is the distance between the slits, and \( \theta \) is the angle of diffraction.
  • In the exercise you encountered, the equation \( m_1 \lambda_1 = m_2 \lambda_2 \) was adapted to find the wavelength of an unknown light source.
  • This formula helps determine where bright and dark fringes appear on the screen, and is vital in understanding and predicting light patterns.
Knowing and using the path difference formula is essential for solving problems involving interference patterns in double-slit experiments.
Order of Maximum
The order of maximum is a key term used when examining interference patterns from a double-slit experiment. It basically refers to the number in the sequence of bright spots where interference is constructive.
For example:
  • An order of 1 would be the first bright fringe from the center.
  • An order of 2 is the second bright fringe, and so on.
In the original exercise, you learned that the third-order maximum of one light coincides with the fourth-order maximum of another. This tells us that the same spot on the fringe pattern can correspond to different orders for different wavelengths, due to the relationship \( m_1 \lambda_1 = m_2 \lambda_2 \).
This concept is important because it allows us to calculate unknown variables in experiments by understanding how different light wavelengths relate to each other within the interference pattern.

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Most popular questions from this chapter

A teacher standing in a doorway \(1.0 \mathrm{~m}\) wide blows a whistle with a frequency of \(1000 \mathrm{~Hz}\) to summon children from the playground (v Fig. 24.31 ). Two boys are playing on the swings \(20 \mathrm{~m}\) away from the school building. One boy is at an angle of \(0^{\circ}\) and another one at \(19.6^{\circ}\) from a line normal to the doorway. Taking the speed of sound in air to be \(335 \mathrm{~m} / \mathrm{s}\), which boy may not hear the whisle? Prove your answer.

A single slit is illuminated with monochromatic light, and a screen is placed behind the slit to observe the diffraction pattern. (a) If the width of the slit is increased, will the width of the central maximum (1) increase, (2) remain the same, or (3) decrease? Why? (b) If the width of the slit is \(0.50 \mathrm{~mm}\), the wavelength is \(680 \mathrm{nm},\) and the screen is \(1.80 \mathrm{~m}\) from the slit, what would be the width of the central maximum? (c) What if the width of the slit is \(0.60 \mathrm{~mm}\) ?

Unpolarized light of intensity \(I_{\mathrm{o}}\) is incident on a polarizer- analyzer pair. (a) If the angle between the polarizer and analyzer increases in the range of \(0^{\circ}\) to \(90^{\circ}\) the transmitted light intensity will (1) also increase, (2) decrease, (3) remain the same. Explain. (b) If the angle between the polarizer and analyzer is \(30^{\circ},\) what light intensity would be transmitted through the polarizer and the analyzer, respectively? (c) What if the angle is \(60^{\circ}\) ?

In the development of Young's double-slit experiment, a small-angle approximation \((\tan \theta \approx \sin \theta)\) was used to find the lateral displacements of the maxima (bright) and minima (dark) positions. How good is this approximation? For example, what is the percentage error for \(\theta=10^{\circ} ?\)

(a) Derive a relationship that gives the locations of the minima in a Young's double-slit experiment. What is the distance between adjacent minima? (b) For a thirdorder minimum (the third side dark position from the central maximum), what is the path length difference between that location and the two slits?

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