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Mobile App Chapter 3: Problem 64
Monochromatic light of frequency \(5.5 \times 10^{14} \mathrm{Hz}\) falls on 10 slits separated by \(0.020 \mathrm{mm}\). What is the separation between the first and third maxima on a screen that is \(2.0 \mathrm{m}\) from the slits?
Short Answer
Expert verified
The separation between the first and third maxima on the screen is approximately \(1.09 \mathrm{mm}\).
Step by step solution
01
Identify the given values
The given values in the problem are:
1. Frequency of monochromatic light, \(f = 5.5 \times 10^{14} \mathrm{Hz}\)
2. Number of slits, \(n_s = 10\)
3. Separation between slits, \(d = 0.020 \mathrm{mm} = 0.020 \times 10^{-3} \mathrm{m}\)
4. Screen distance from slits, \(L = 2.0 \mathrm{m}\)
02
Calculate the wavelength of the light
To find the separation between the first and third maxima, we first need to calculate the wavelength \(\lambda\) of the monochromatic light. We can use the relationship between frequency, speed of light, and wavelength: \(c=f\lambda\), where \(c = 3 \times 10^8 \mathrm{m/s}\) is the speed of light. Solving for the wavelength, we get:
\[\lambda = \frac{c}{f}\]
Plugging in the given values, we get:
\[\lambda = \frac{3 \times 10^8 \mathrm{m/s}}{5.5 \times 10^{14} \mathrm{Hz}}\]
\[\lambda = 5.45 \times 10^{-7} \mathrm{m}\]
03
Calculate the angular position of the first and third maxima
Now, let's find the angular position of the first and third maxima (\(\theta_1\) and \(\theta_3\), respectively) using the formula for constructive interference in the multiple-slit interference pattern:
\[\theta_m = \arcsin{\frac{m\lambda}{nd}}\]
Where \(m\) is the order of the maximum, \(n\) is the number of slits, and \(d\) is the slit separation.
First maximum, \(m = 1\):
\[\theta_1 = \arcsin{\frac{1 \cdot 5.45 \times 10^{-7} \mathrm{m}}{10 \times 0.020 \times 10^{-3} \mathrm{m}}}\]
\[\theta_1 = \arcsin{2.725 \times 10^{-5} }\]
Similarly, for the third maximum, \(m = 3\):
\[\theta_3 = \arcsin{\frac{3 \cdot 5.45 \times 10^{-7} \mathrm{m}}{10 \times 0.020 \times 10^{-3} \mathrm{m}}}\]
\[\theta_3 = \arcsin{8.175 \times 10^{-5}}\]
04
Calculate the linear separation between the first and third maxima
Finally, we can calculate the linear separation \(\Delta y\) between the first and third maxima on the screen using the following formula:
\[\Delta y = L(\tan{\theta_3} - \tan{\theta_1})\]
Plugging in the known values, we find:
\[\Delta y = 2.0 \mathrm{m}(\tan{8.175 \times 10^{-5}} - \tan{2.725 \times 10^{-5}})\]
After evaluating the trigonometric functions, we get that the separation between the first and third maxima is approximately:
\[\Delta y \approx 0.00109 \mathrm{m}\]
So the separation between the first and third maxima on the screen is approximately \(1.09 \mathrm{mm}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Constructive Interference
Constructive interference is a critical concept in the study of light waves. It occurs when two or more waves overlap and combine to create a new wave with an amplitude larger than any of the individual contributing waves. Think of it as the combined strength of overlapping ripples in water—they add up to make a bigger wave. In the case of multiple-slit interference, such as in the given exercise, constructive interference produces patterns known as maxima on a screen. These maxima correspond to bright lines or spots, which are a result of light waves emerging from slits that are in phase or have a path difference that is an integer multiple of the wavelength. The condition for constructive interference in a multiple-slit setup is given by the equation \br \(m\text{λ} = n\text{d sin θ}\), \br where \(m\) is the order of the maximum, \(n\) represents the number of slits, \(d\) stands for the separation between slits, \(λ\) is the wavelength, and \(θ\) indicates the angle at which the maximum occurs. The result is a pattern with bright and dark regions indicative of where waves are reinforcing each other or canceling each other out.
Monochromatic Light Frequency
Monochromatic light refers to light of a single color, which means it has a single frequency and wavelength. In the exercise, the provided monochromatic light frequency is \(5.5 \times 10^{14} \mathrm{Hz}\). Frequency (\(f\)) is a key measure of how many wave cycles occur in one second and it's related to the wavelength (\(λ\)) and the speed of light (\(c\)) by the formula \br \(c = fλ\). \br The frequency of monochromatic light is critical for calculations involving interference patterns, as it allows us to determine the precise wavelength of light involved in the experiment. With light's speed being a constant \(3 \times 10^8 \mathrm{m/s}\), knowing the frequency gives a direct path to calculating the corresponding wavelength that is crucial for studying the resulting interference pattern generated by the interaction of these light waves at the slits and the screen.
Diffraction Grating
A diffraction grating is an optical component with a regular pattern of lines or apertures that diffract light into several beams. It is a tool widely used in physics to separate light into its component wavelengths, effectively spreading out monochromatic light into its spectrum.
When monochromatic light hits a diffraction grating, consisting of numerous closely spaced slits like in our example, it produces an interference pattern similar to the one mentioned in the problem. Each slit acts as a new source of wavefronts, and the interference of these wavefronts gives rise to the maxima and minima observed. The exercise utilizes the principles of a diffraction grating to find the separation between bright fringes on the screen. It's notable that the more slits there are in a grating, the sharper and more defined the maxima, because a greater number of interfering waves create more pronounced constructive and destructive interference effects.
When monochromatic light hits a diffraction grating, consisting of numerous closely spaced slits like in our example, it produces an interference pattern similar to the one mentioned in the problem. Each slit acts as a new source of wavefronts, and the interference of these wavefronts gives rise to the maxima and minima observed. The exercise utilizes the principles of a diffraction grating to find the separation between bright fringes on the screen. It's notable that the more slits there are in a grating, the sharper and more defined the maxima, because a greater number of interfering waves create more pronounced constructive and destructive interference effects.
Wavelength Calculation
Wavelength calculation is essential in understanding the behavior of waves, especially in applications involving interference and diffraction. In the context of the given exercise, the wavelength of monochromatic light is determined using the equation \br \(λ = \frac{c}{f}\), \br where \(λ\) represents the wavelength, \(c\) is the speed of light, and \(f\) is the frequency of the light. Upon computing the wavelength, it can be used to analyze the interference pattern created by light passing through multiple slits.
This wavelength data, combined with measurements from the experiment like slit separation and screen distance, enables us to predict the position of maxima and minima on the screen, as well as to calculate the distances between these bright and dark fringes, providing vital insight into the wave nature of light and the specific properties of the light used in the experiment.
This wavelength data, combined with measurements from the experiment like slit separation and screen distance, enables us to predict the position of maxima and minima on the screen, as well as to calculate the distances between these bright and dark fringes, providing vital insight into the wave nature of light and the specific properties of the light used in the experiment.
