Chapter 4: Problem 38
A monochromatic light of wavelength 589 nm incident on a double slit with slit width \(2.5 \mu \mathrm{m}\) and unknown separation results in a diffraction pattem containing nine interference peaks inside the central maximum. Find the separation of the slits.
Short Answer
Expert verified
The separation of the slits in the double-slit experiment is approximately 1.2 μm.
Step by step solution
01
Interpret the given information
We're given the following information:
- Monochromatic light wavelength λ = 589 nm
- Slit width a = 2.5 μm
- Number of interference peaks n = 9 (inside the central maximum)
- Separation of the slits d (that's what we need to find)
02
Calculate the angular width of the central maximum
To find the angular width (Δθ) of the central maximum in the diffraction pattern, we use the formula for single-slit diffraction:
\[\Delta \theta = 2 \times \arcsin(\frac{m \lambda}{2a})\]
where m = 1 for the central maximum.
Plugging in the given values, we get:
\[\Delta \theta = 2 \times \arcsin(\frac{1 \times 589 \times 10^{-9}}{2 \times 2.5 \times 10^{-6}})\]
Calculate the Δθ:
\[\Delta \theta = 2 \times \arcsin(0.1178)\]
\[\Delta \theta \approx 13.6^\circ\]
03
Calculate the total angular separation between interference peaks
Since there are 9 interference peaks within the central maximum, the total angular separation between these peaks (Δθ') can be calculated as:
\[\Delta \theta' = \frac{\Delta \theta}{n}\]
Plugging in the values, we get:
\[\Delta \theta' = \frac{13.6^\circ}{9}\]
\[\Delta \theta' \approx 1.51^\circ\]
04
Calculate the separation of the slits using the double-slit formula
Now that we have the total angular separation between the interference peaks, we can use the formula for double-slit interference to find the separation of the slits:
\[\Delta \theta' \approx \frac{\lambda}{d}\]
Rearranging the formula to solve for d:
\[d \approx \frac{\lambda}{\Delta \theta'}\]
Plugging in the given values, we get:
\[d \approx \frac{589 \times 10^{-9}}{1.51^\circ \times (\pi/180)}\]
Calculate the separation of the slits:
\[d \approx 1.2 \times 10^{-6}\]
Hence, the separation of the slits is approximately 1.2 μm.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Monochromatic Light
When we talk about monochromatic light, we are referring to light that consists of a single color, or technically, a single wavelength. In experiments like the double slit interference, monochromatic light is crucial because it ensures that all the light waves are coherent and in phase with each other. This consistency is essential for creating a clear and predictable interference pattern.
Imagine you have a beam of pure yellow light—that would be an example of monochromatic light. In the problem provided, the monochromatic light has a wavelength of 589 nm, which is in the yellow region of the visible spectrum. This single wavelength property means that when the light passes through the double slit apparatus, it will produce interference patterns that can be analyzed to reveal properties of the slits, like their separation.
Imagine you have a beam of pure yellow light—that would be an example of monochromatic light. In the problem provided, the monochromatic light has a wavelength of 589 nm, which is in the yellow region of the visible spectrum. This single wavelength property means that when the light passes through the double slit apparatus, it will produce interference patterns that can be analyzed to reveal properties of the slits, like their separation.
Slit Separation Calculation
Determining the slit separation in a double slit experiment is vital for analyzing the diffraction and interference patterns produced. The separation of slits, often denoted as 'd', affects the spacing between the bright and dark bands, also known as interference peaks and troughs.
To calculate slit separation from the information given in the problem, you must first interpret the observed interference pattern. Here, we know the number of peaks within the central maximum and the wavelength of the monochromatic light used. Using these data points, together with known formulas from wave optics, we can work out the separation between the slits. It's essential to correctly apply the formulas, remembering to adjust units where necessary, as calculating the slit separation is a fundamental step in understanding the underlying physics of the double slit experiment.
To calculate slit separation from the information given in the problem, you must first interpret the observed interference pattern. Here, we know the number of peaks within the central maximum and the wavelength of the monochromatic light used. Using these data points, together with known formulas from wave optics, we can work out the separation between the slits. It's essential to correctly apply the formulas, remembering to adjust units where necessary, as calculating the slit separation is a fundamental step in understanding the underlying physics of the double slit experiment.
Diffraction Pattern
A diffraction pattern arises when waves encounter an obstacle or slit that is comparable in size to their wavelength. It's characterized by a series of dark and bright regions called 'fringes'. With a double slit, these patterns result from both diffraction and interference.
In our scenario, light waves spread out after passing through the slits and overlap, creating an interference pattern on a screen. The central maximum is the brightest part and contains the most interference peaks — the number of these peaks is directly related to the slits' properties. Analyzing this diffraction pattern allows us to delve deeper into the wave nature of light; it's the visual evidence of both constructive and destructive interference happening due to the wave properties of light.
In our scenario, light waves spread out after passing through the slits and overlap, creating an interference pattern on a screen. The central maximum is the brightest part and contains the most interference peaks — the number of these peaks is directly related to the slits' properties. Analyzing this diffraction pattern allows us to delve deeper into the wave nature of light; it's the visual evidence of both constructive and destructive interference happening due to the wave properties of light.
Interference Peaks
Interference peaks are the bright bands observed on a screen in a double slit experiment. They occur due to the constructive interference of light waves—when the crests of two waves add up, enhancing the light intensity. The number of interference peaks within the central maximum is a key piece of information and is used to calculate the slit separation.
The peaks' angular separation helps us find out the distance between the slits. As the number of peaks increases, we could infer that the slits are closer together. Each peak represents a point where light from both slits arrive in phase, causing reinforcement. This concept of interference peaks ties together the behavior of light as waves with the geometric constraints of the experiment's setup, leading to a better understanding of wave optics.
The peaks' angular separation helps us find out the distance between the slits. As the number of peaks increases, we could infer that the slits are closer together. Each peak represents a point where light from both slits arrive in phase, causing reinforcement. This concept of interference peaks ties together the behavior of light as waves with the geometric constraints of the experiment's setup, leading to a better understanding of wave optics.