Must Be Done In Microsoft Word And Must Follow The Structure
Must Be Done In Microsoft Word And Must Follow The Structured Problem
Must be done in Microsoft Word and must follow the Structured Problem Solving Format attached. Each problem must be done in separate documents. 38. Orange light with strikes a pair of slits separated by 0.580 mm. On a screen 1.20 m away, what is the distance between (a) the two second-order bright fringes; (b) the central maximum and one of the fourth dark fringes; (c) the third bright fringe on one side of the central maximum and the third dark fringe on the other? 42. You are doing a double-slit experiment with two different colors. You notice that the second-order bright fringe using 680-nm red light is in the same place as the third-order bright fringe of the other color. (a) What is the wavelength of the color? (b) What color is it?
Paper For Above instruction
The context of this assignment involves solving structured optics problems related to interference patterns produced by double-slit experiments. The deliberate focus on multiple specific problems encourages the application of wave optics principles, including the calculation of fringe separations, understanding of constructive and destructive interference, and the relationship between wavelength, slit separation, and fringe position within a double-slit interference setup. The problems require precise application of formulas governing interference fringes, as well as critical analysis to derive unknown wavelengths and identify observed colors based on interference pattern overlaps. This paper systematically addresses these problems by applying relevant physics equations, demonstrating problem-solving steps, and providing clear reasoning aligned with structured problem-solving methodologies.
Analysis and Solution of the Problems
Problem 1: Interference Pattern with Orange Light
This problem involves calculating fringe positions produced by a double-slit interference pattern using orange light, characterized by a known wavelength, slit separation, and screen distance.
a) Distance between the two second-order bright fringes
The fringe separation (distance between adjacent bright fringes) for double-slit interference is given by:
\[ y_m = \frac{\lambda L}{d} \]
where:
- \( y_m \) is the fringe position for order \( m \),
- \( \lambda \) is the wavelength of light,
- \( L \) is the distance from slits to screen,
- \( d \) is the slit separation.
Given \( \lambda \) (orange light) approximately 590 nm or 5.90 × 10\(^{-7}\) m, \( d = 0.580\,mm = 5.80 \times 10^{-4}\,m \), and \( L = 1.20\,m \), we first calculate the fringe spacing:
\[ y_m = \frac{(5.90 \times 10^{-7}\,m)(1.20\,m)}{5.80 \times 10^{-4}\,m} \approx 1.22 \times 10^{-3}\,m \text{ or } 1.22\,mm \]
The second-order bright fringe occurs at \( m=2 \):
\[
y_2 = 2 \times y_1
\]
The distance between the two second-order bright fringes is from \( + y_2 \) to \( - y_2 \), which is:
\[
\text{Distance} = 2 \times y_2 = 2 \times 1.22\,mm = 2.44\,mm
\]
Therefore, the distance between the two second-order bright fringes is approximately 2.44 mm.
b) Distance between the central maximum and the fourth dark fringe
The position of dark fringes occurs at points where destructive interference happens, specified by:
\[ y_{m}^{dark} = \left( m + \frac{1}{2} \right) \frac{\lambda L}{d} \]
where \( m \) is the fringe order for dark fringes.
For the fourth dark fringe, \( m=3 \):
\[ y_3^{dark} = \left( 3 + \frac{1}{2} \right) \times 1.22\,mm = 3.5 \times 1.22\,mm \approx 4.27\,mm \]
The central maximum is at \( y=0 \). The distance from the central maximum to this dark fringe is approximately 4.27 mm.
c) Distance between the third bright fringe and the third dark fringe on opposite sides
The third bright fringe (\( m=3 \)):
\[ y_3^{bright} = 3 \times 1.22\,mm = 3.66\,mm \]
The third dark fringe on the opposite side (\( m=3 \) as well but negative):
\[ y_3^{dark} = - 3.5 \times 1.22\,mm = -4.27\,mm \]
The total distance between these two fringes:
\[
\left| y_{bright} - y_{dark} \right| = 3.66\,mm + 4.27\,mm = 7.93\,mm
\]
So, the distance between the third bright fringe on one side and third dark fringe on the other is approximately 7.93 mm.
Problem 2: Wavelength and Color Identification in Double-Slit Experiment
This problem involves overlapping interference fringes of two different colors, with the known parameters: the second-order bright fringe of 680 nm red light coincides with the third-order bright fringe of another color. The goal is to find that other color's wavelength and identify it.
a) Calculating the wavelength of the other color
The position of interference fringes is given by:
\[ y_m = \frac{m \lambda L}{d} \]
Since fringes of different colors coincide at the same position:
\[ y_{m_1} = y_{m_2} \]
\[
m_1 \lambda_1 = m_2 \lambda_2
\]
where:
- \( m_1=2 \), \( \lambda_1=680\,nm \),
- \( m_2=3 \), \( \lambda_2= ? \),
and assuming the slit separation \( d \) and screen distance \( L \) are the same, the equality becomes:
\[
2 \times 680\,nm = 3 \times \lambda_2
\]
solving for \( \lambda_2 \):
\[
\lambda_2 = \frac{2 \times 680\,nm}{3} \approx \frac{1360\,nm}{3} \approx 453.33\,nm
\]
Hence, the wavelength of the other color is approximately 453.33 nm.
b) Identifying the color
A wavelength of roughly 453 nm corresponds to a blue color, specifically within the blue-cyan spectrum. Therefore, the other color is blue.
Conclusion
Through applying the fundamental equations of double-slit interference, this analysis calculated the fringe distances for orange light, clarifying the spatial pattern of bright and dark fringes. For the mixed-color interference problem, the overlap condition provided a method to determine the unknown wavelength, revealing it as blue. These solutions illustrate how interference principles help in understanding wave behavior, and how precise measurements and calculations turn observable patterns into quantitative insights about light's properties.
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