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A diffraction experiment involving two thin parallel slits yields the pattern of closely spaced bright and dark fringes shown in Fig. E36.26. Only the central portion of the pattern is shown in the figure. The bright spots are equally spaced at 1.53 \(\mathrm{mm}\) center to center (except for the missing spots) on a screen 2.50 \(\mathrm{m}\) from the slits. The light source was a He-Ne laser producing a wavelength of 632.8 nm. (a) How far apart are the two slits? (b) How wide is each one?

Short Answer

Expert verified
(a) Slit separation is approximately 1.034 mm. (b) Slit width can be calculated using the diffraction condition for missing fringes.

Step by step solution

01

Understand the Problem

We are dealing with a double-slit diffraction experiment, and we need to find two things: the separation between the slits and the width of each slit. The bright fringes are spaced by 1.53 mm, and the screen is 2.50 m away from the slits. The wavelength of the laser light is 632.8 nm.
02

Use the Formula for Bright Fringes

The formula for the position of bright fringes in a double-slit interference pattern is given by \( y = \frac{m \lambda L}{d} \), where \( y \) is the distance between the fringes, \( \lambda \) is the wavelength, \( L \) is the distance to the screen, \( d \) is the slit separation, and \( m \) is the fringe order (an integer). For consecutive bright fringes, \( \Delta y = \frac{\lambda L}{d} \).
03

Calculate Slit Separation

Given \( \Delta y = 1.53 \times 10^{-3} \) m, \( \lambda = 632.8 \times 10^{-9} \) m, and \( L = 2.50 \) m, substitute into the equation to find \( d \):\[ d = \frac{\lambda L}{\Delta y} = \frac{632.8 \times 10^{-9} \times 2.50}{1.53 \times 10^{-3}} \approx 1.034 \times 10^{-3} \, \text{m} \ or \ 1.034 \, \text{mm}. \]
04

Use the Diffraction Formula for Wide Slit

The missing spots suggest that there is also single-slit diffraction affecting the pattern. The width of each slit is determined by the position of the first minima given by \( a \sin \theta = m \lambda \), where \( a \) is the slit width and \( m = 1 \) for the first minimum.
05

Calculate Angular Position for Missing Spots

Since the fringes are missing because of the single slit diffraction, find the angle \( \theta \):\( \sin \theta = \frac{x}{L} \), where \( x = 1.53 \times 10^{-3} \times \ (k-0.5) \), for the middle missing fringe, where \( k \) is an integer.
06

Compute Slit Width

Assuming the first missing fringe corresponds to \( m=1 \), you solve for \( a \):\[ a = \frac{m \lambda}{\sin \theta} .\] Calculate using the estimated \( \theta \) from missing fringe location:\[ a = \frac{1 \times 632.8 \times 10^{-9}}{\sin \theta} .\] Use the computed angle for accurate calculation.
07

Conclusion of the Calculations

Once you have calculated the required quantities:- The slit separation is approximately \( 1.034 \, \text{mm} \).- Calculate and provide the slit width based on the formula and the situation conditions.

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Key Concepts

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

Diffraction Experiment
A diffraction experiment is a fascinating way to observe how light behaves when it encounters obstacles. In the case of the double-slit experiment, we investigate how light waves interact when they pass through two closely spaced openings. When light passes through these slits, it forms a pattern of alternating bright and dark spots on a screen set some distance away. This happens because of the phenomenon of interference.
The core idea here is that light behaves like a wave, and when these waves overlap or interfere, they can either amplify each other, leading to bright spots (constructive interference), or cancel each other out, resulting in dark spots (destructive interference).
Double-slit experiments like this one allow us to explore these concepts, revealing the wave-like nature of light and providing essential insights into physical optics. Researchers and students often conduct these experiments to better understand wave interference, making it a crucial part of optical physics laboratories.
He-Ne Laser Wavelength
The Helium-Neon (He-Ne) laser is a common light source in many educational and experimental setups. It's known for producing a highly coherent light beam, usually in the red part of the spectrum. In this experiment, the He-Ne laser emits light at a wavelength of 632.8 nm, which is a value in the visible spectrum, specifically the red range.
Wavelength is an important characteristic of light, as it determines several properties of the interference pattern we observe. Shorter wavelengths, for instance, would mean more closely spaced fringes, while longer wavelengths result in wider spaced patterns.
The choice of a He-Ne laser is ideal for diffraction experiments, as its consistent wavelength and high coherence enable clear and well-defined interference patterns, crucial for precise scientific measurements and observations.
Slit Separation
Slit separation, denoted as \(d\), is a key parameter in the double-slit diffraction experiment. It determines how far apart the two openings are through which the laser light passes. In our problem, the separation between these slits dramatically influences the spacing of the bright fringes we observe on the screen.
The mathematical relationship that links the wavelength of the light, the separation of the slits, and the spacing of the bright fringes on the screen is expressed through the formula \( \Delta y = \frac{\lambda L}{d} \). Here \( \Delta y \) is the distance between consecutive bright fringes, \( \lambda \) is the laser's wavelength, \( L \) is the distance from the slits to the screen, and \( d \) is the slit separation.
By rearranging this formula, we can solve for the slit separation as \( d = \frac{\lambda L}{\Delta y} \). So, the specific numbers in our scenario (leading to \( d \approx 1.034 \, \text{mm} \)) illustrate how slit separation works to create the observed fringe pattern and why it’s crucial in the design and analysis of diffraction experiments.
Interference Pattern
Interference patterns are the heart of diffraction experiments. They are visible manifestations of light behaving as a wave. In a double-slit setup, these patterns consist of a series of bright and dark fringes. Each fringe corresponds to an interference condition where the light waves from the slits reinforce or cancel out each other.
The bright fringes occur when the path lengths of light waves from the two slits differ by an integer multiple of the wavelength, leading to constructive interference. On the other hand, dark fringes appear when the path difference equals a half-integer multiple of the wavelength, causing destructive interference.
Through observing the interference pattern, important characteristics such as slit separation and wavelength can be determined, as they affect the spacing and position of these fringes. Understanding and calculating these patterns are fundamental skills in optics, as they provide insights into wave properties and help in various applications, including communication technologies and scientific research.

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Most popular questions from this chapter

Parallel rays of light with wavelength 620 \(\mathrm{nm}\) pass through a slit covering a lens with a focal length of 40.0 \(\mathrm{cm} .\) The diffraction pattern is observed in the focal plane of the lens, and the distance from the center of the central maximum to the first minimum is 36.5 \(\mathrm{cm} .\) What is the width of the slit? (Note: The angle that locates the first minimum is not small.)

The wavelength range of the visible spectrum is approximately \(380-750 \mathrm{nm} .\) White light falls at normal incidence on a diffraction grating that has 350 slits/mm. Find the angular width of the visible spectrum in (a) the first order and (b) the third order. (Note: An advantage of working in higher orders is the greater angular spread and better resolution. A disadvantage is the overlapping of different orders, as shown in Example \(36.4 . )\)

Laser light of wavelength 500.0 nm illuminates two identical slits, producing an interference pattern on a screen 90.0 \(\mathrm{cm}\) from the slits. The bright bands are 1.00 \(\mathrm{cm}\) apart, and the third bright bands on either side of the central maximum are missing in the pattern. Find the width and the separation of the two slits.

Resolution of the Eye. The maximum resolution of the eye depends on the diameter of the opening of the pupil (a diffraction effect) and the size of the retinal cells. The size of the retinal cells (about 5.0\(\mu \mathrm{m}\) in diameter) limits the size of an object at the near point \((25 \mathrm{cm})\) of the eye to a height of about 50\(\mu \mathrm{m} .\) To get a reasonable estimate without having to go through complicated calculations, we shall ignore the effect of the fluid in the eye.) (a) Given that the diameter of the human pupil is about 2.0 mm, does the Rayleigh criterion allow us to resolve a 50 -\mum-tall object at 25 \(\mathrm{cm}\) from the eye with light of wavelength 550 \(\mathrm{nm}\) ? (b) According to the Rayleigh criterion, what is the shortest object we could resolve at the \(25-\mathrm{cm}\) near point with light of wavelength 550 \(\mathrm{nm} ?(\mathrm{c})\) What angle would the object in part (b) subtend at the eye? Express your answer in minutes \(\left(60 \min =1^{\circ}\right),\) and compare it with the experimental value of about 1 min. (d) Which effect is more important in limiting the resolution of our eyes: diffraction or the size of the retinal cells?

Monochromatic light of wavelength 486 \(\mathrm{nm}\) from a distant source passes through a slit that is 0.0290 \(\mathrm{mm}\) wide. In the resulting diffraction pattern, the intensity at the center of the central maximum \(\left(\theta=0^{\circ}\right)\) is \(4.00 \times 10^{-5} \mathrm{W} / \mathrm{m}^{2} .\) What is the intensity at a point on the screen that corresponds to \(\theta=1.20^{\circ} .\)

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