Chapter 3: Problem 25
What is the smallest separation between two slits that will produce a second- order maximum for 720 -nm red light?
Short Answer
Expert verified
The smallest separation between the two slits that will produce a second-order maximum for 720-nm red light is \(1.44 \times 10^{-6} m\) or \(1.44 \mu m\).
Step by step solution
01
Identify the formula for double-slit interference pattern
We can use the formula for the double-slit interference pattern:
\[d sin(\theta) = m \lambda\]
where d is the distance between the slits, θ is the angle of the interference pattern maximum, m is the order of the maximum, and λ is the wavelength of the light.
In this problem, we are given the wavelength (λ = 720 nm) and the desired order (m = 2). We need to find the smallest distance between the slits (d) that will produce a second-order maximum.
02
Find the angle (θ) for the second-order maximum
In the second-order maximum (m=2), we need to find the angle (θ) for which the formula holds true. However, we want the smallest separation between the slits (minimum d). When d is minimum, the angle θ will be maximum. Let's use the maximum possible angle for interference, which is \(θ = 90^\circ\).
03
Convert angle to radians
Convert the angle to radians:
\(θ = 90^\circ\)
To convert to radians, use the formula:
\(radian = \frac{degree \times \pi}{180}\)
\(θ = \frac{90 \times \pi}{180} = \frac{\pi}{2}\)
04
Solve for the distance between the slits (d)
Now we can use the formula to find the smallest distance between the slits (d):
\[d sin(\theta) = m \lambda\]
\[d \sin(\frac{\pi}{2}) = 2(720 \times 10^{-9} m)\]
Since, \(\sin(\frac{\pi}{2}) = 1\)
\[d = 2(720 \times 10^{-9} m)\]
05
Calculate the smallest separation between the slits
Now, we can calculate the smallest separation between the slits:
\[d = 2(720 \times 10^{-9} m)\]
\[d = 1.44 \times 10^{-6} m\]
So, the smallest separation between the two slits that will produce a second-order maximum for 720-nm red light is \(1.44 \times 10^{-6} m\) or \(1.44 \mu m\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Interference Pattern Maximum
Understanding how an interference pattern maximum is formed is crucial when studying light diffraction through a double-slit experiment. As waves emanate from the two slits, they overlap and interfere with one another. This interference can be constructive or destructive, depending on the phase difference between the waves. Constructive interference occurs at specific angles where the path difference between waves is an integer multiple of the wavelength, leading to bright fringes known as interference pattern maxima.
In essence, the bright bands on the screen represent areas where waves from both slits arrive in phase and amplify each other, creating a visible 'maximum'. Accomplishing this requires precise alignment and conditions which is why performing such experiments with accuracy is essential for validating wave theories of light.
In essence, the bright bands on the screen represent areas where waves from both slits arrive in phase and amplify each other, creating a visible 'maximum'. Accomplishing this requires precise alignment and conditions which is why performing such experiments with accuracy is essential for validating wave theories of light.
Wavelength of Light
The wavelength of light is a fundamental property defining its color and place in the electromagnetic spectrum. Represented by the Greek letter lambda (λ), it measures the distance between successive peaks of a wave and is usually expressed in units of nanometers (nm) for visible light. When light of a specific wavelength passes through a double-slit, it creates an interference pattern because the light waves spread out and overlap past the slits—an effect described by wave optics.
Different wavelengths will create different patterns, and knowing the wavelength is crucial for calculating various aspects of these patterns, such as the position and spacing of the interference maxima. The wavelength is the key variable in the formula used to relate the slit separation to the interference pattern observed in a double-slit experiment.
Different wavelengths will create different patterns, and knowing the wavelength is crucial for calculating various aspects of these patterns, such as the position and spacing of the interference maxima. The wavelength is the key variable in the formula used to relate the slit separation to the interference pattern observed in a double-slit experiment.
Second-Order Maximum
In the context of a double-slit interference pattern, the term 'second-order maximum' refers to the second bright fringe from the central maximum on either side. These are the points where constructive interference occurs for the second time, meaning the path difference between the two sets of waves is twice the wavelength (2λ). This happens because the light waves from the two slits arrive at these points in phase after one wave has traveled exactly two wavelengths more than the other.
Calculating the position of this second-order maximum helps establish the necessary slit separation to accommodate the wavelength of light used, as seen in the exercise where we determined the smallest slit separation for a 720 nm wavelength required to observe such a maximum. The order of maximum, denominated by 'm', is fundamental in calculating and predicting the pattern outcome on the screen behind the slits.
Calculating the position of this second-order maximum helps establish the necessary slit separation to accommodate the wavelength of light used, as seen in the exercise where we determined the smallest slit separation for a 720 nm wavelength required to observe such a maximum. The order of maximum, denominated by 'm', is fundamental in calculating and predicting the pattern outcome on the screen behind the slits.