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Chapter 36: Problem 43

(A) How many bright fringes appear between the first diffraction-envelope minima to either side of the central maximum in a double-slit pattern if \(\lambda=550 \mathrm{nm}, d=0.150 \mathrm{~mm}\) and \(a=30.0 \mu \mathrm{m} ?\) (b) What is the ratio of the intensity of the third bright fringe to the intensity of the central fringe?

Short Answer

Expert verified
1. 9 bright fringes 2. Intensity ratio is approximately 0.993.

Step by step solution

01

Identify the Formula for Diffraction Minima

To solve this, we first need the formula for the diffraction envelope minima of a single slit, which occurs when \( a \sin \theta = m \lambda \), where \( a \) is the slit width, \( \lambda \) is the wavelength, and \( m \) is the order of the minimum (\( m = \pm 1, \pm 2, \pm 3, \ldots \)). We use this to find the angular positions of the minima that define the envelope.
02

Calculate Angular Position for First Minima

Calculate the angular position of the first minimum: \[ a \sin \theta_1 = \lambda \]Substitute the values: \( a = 30 \times 10^{-6} \text{m} \), \( \lambda = 550 \times 10^{-9} \text{m} \).\[ \sin \theta_1 = \frac{550 \times 10^{-9}}{30 \times 10^{-6}} \approx 0.01833 \]Thus, \( \theta_1 \approx \sin^{-1}(0.01833) \).
03

Determine Angular Positions of Bright Fringes

For a double-slit interference, bright fringes occur at: \[ d \sin \theta = m \lambda \]where \( d \) is the distance between the slits. Here, \( d = 0.150 \text{ mm} = 0.150 \times 10^{-3} \text{ m} \).Calculate \( m \) for which \( \theta \) fits between \(-\theta_1\) and \(\theta_1\).
04

Quantify Number of Bright Fringes

Substitute \( \theta_1 \) to find the maximum order \( m \) possible:\[ \sin \theta = \frac{m \lambda}{d} \leq \sin \theta_1 \]\[ m \leq \frac{d \sin \theta_1}{\lambda} = \frac{0.150 \times 10^{-3} \times 0.01833}{550 \times 10^{-9}} \approx 5 \]Thus, there are a total of \(2 \times 5 - 1 = 9 \) bright fringes between the first minima on either side of the central maximum.
05

Find Intensity Ratio of Third Bright Fringe to Central Fringe

Intensity of an interference pattern fringe is given by: \[ I = I_0 \cos^2 \left(\frac{\pi d \sin \theta}{\lambda}\right) \]For the central fringe (maximum):\[ \sin \theta = 0 \Rightarrow I_{central} = I_0 \]For the third bright fringe:\[ m = 3 \Rightarrow I_3 = I_0 \cos^2 \left(\frac{3 \pi \lambda}{d}\right) \]Substitute the values:\[ \cos^2 \left(\frac{3 \pi \times 550 \times 10^{-9}}{0.150 \times 10^{-3}}\right) \approx \cos^2 (0.0346 \pi) \approx 0.993 \]So, the ratio is approximately 0.993.

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Key Concepts

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

Diffraction Minima
In the context of double-slit interference, the term "diffraction minima" refers to the positions along the diffraction pattern where the intensity of light approaches zero. These are critical points that help define the shape and behavior of the overall interference pattern. In a single-slit setup, these minima occur because of the condition:\[ a \sin \theta = m \lambda \]Here, \(a\) is the width of the slit, \(\lambda\) is the wavelength of the light, and \(m\) represents the order of the minimum (where \(m = \pm 1, \pm 2, \ldots\)). This equation indicates that the path difference leads to destructive interference.
For practical calculations like in our problem, we focus on the first order of minima as it establishes the boundary for the main or central diffraction peak. Identifying these angles helps to determine where bright fringes will appear in relation to these diminishing points of intensity.
Bright Fringes
Bright fringes are the regions of maximum intensity visible in an interference pattern resulting from the constructive interference of light waves passing through the slits. For double-slit interference, these bright fringes occur according to:\[ d \sin \theta = m \lambda \]Here, \(d\) is the slit separation and \(m\) (\(m = 0, \pm 1, \pm 2, \ldots\)) specifies the order number of the fringe. The central bright fringe corresponds to \(m = 0\), which is the most intense and located at angle \(\theta = 0\).
In our exercise, we need to find out how many bright fringes exist between the first diffraction envelope minima on either side of the central maximum. The maxima repetition pattern falls between these minima due to constructive interference within the defined angle. The total number of such fringes can be determined by calculating the corresponding \(m\) values, ensuring they lie inside the range set by the first diffraction minima boundaries.
Intensity Ratio
In an interference pattern, the intensity of the bright fringes is not uniform. It diminishes as we move away from the center. The intensity for any of these fringes can be calculated using:\[ I = I_0 \cos^2 \left(\frac{\pi d \sin \theta}{\lambda}\right) \]Here, \(I_0\) stands for the intensity at the central maximum (or central fringe), which is the maximum intensity possible. The term \(\cos^2(\dots)\) determines the reduction in intensity for non-central fringes.
Using this formula, one can find the intensity ratio between any two fringes by comparing their respective \(m\) values for the angles. In our case, we calculate the ratio between the third order bright fringe \((m=3)\) and the central fringe. This helps us understand how the intensity changes within the pattern, providing insight into how bright these subsequent fringes will appear relative to the center.

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Most popular questions from this chapter

The pupil of a person's eye has a diameter of \(5.00 \mathrm{~mm}\). According to Rayleigh's criterion, what distance apart must two small objects be if their images are just barely resolved when they are \(250 \mathrm{~mm}\) from the eye? Assume they are illuminated with light of wavelength \(500 \mathrm{nm}\).

A \(0.10-\mathrm{mm}\) -wide slit is illuminated by light of wavelength \(589 \mathrm{nm}\). Consider a point \(P\) on a viewing screen on which the diffraction pattern of the slit is viewed; the point is at \(30^{\circ}\) from the central axis of the slit. What is the phase difference between the Huygens wavelets arriving at point \(P\) from the top and midpoint of the slit?

Two yellow flowers are separated by \(60 \mathrm{~cm}\) along a line perpendicular to your line of sight to the flowers. How far are you from the flowers when they are at the limit of resolution according to the Rayleigh criterion? Assume the light from the flowers has a single wavelength of \(550 \mathrm{nm}\) and that your pupil has a diameter of \(5.5 \mathrm{~mm}\)

Find the separation of two points on the Moon's surface that can just be resolved by the 200 in. \((=5.1 \mathrm{~m})\) telescope at Mount Palomar, assuming that this separation is determined by diffraction effects. The distance from Earth to the Moon is \(3.8 \times 10^{5} \mathrm{~km}\). Assume a wavelength of \(550 \mathrm{nm}\) for the light.

Prove that it is not possible to determine both wavelength of incident radiation and spacing of reflecting planes in a crystal by measuring the Bragg angles for several orders.
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