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Monochromatic light of wavelength 580 nm passes through a single slit and the diffraction pattern is observed on a screen. Both the source and screen are far enough from the slit for Fraunhofer diffraction to apply. (a) If the first diffraction minima are at \(\pm\)90.0\(^\circ\), so the central maximum completely fills the screen, what is the width of the slit? (b) For the width of the slit as calculated in part (a), what is the ratio of the intensity at \(\theta\) = 45.0\(^\circ\) to the intensity at \(\theta\) = 0?

Short Answer

Expert verified
(a) Slit width is 580 nm. (b) Intensity ratio is 0.405.

Step by step solution

01

Understanding the Problem

We are asked to calculate two main things. First, the width of a slit given that the first diffraction minima are at 90.0° (which means the central maximum spans the entire screen). Second, we need to find the intensity ratio at two specific angles for that slit width.
02

Using Formula for First Minima in Single-Slit Diffraction

The condition for the first diffraction minima in a single-slit diffraction pattern is given by \( a \sin \theta = m \lambda \), where \( a \) is the slit width, \( \theta \) is the angle of diffraction (90° in this case), \( m \) is the order of the minima (1 for the first minima), and \( \lambda \) is the wavelength of light (580 nm).
03

Calculating the Slit Width

Substitute the given values into the formula: \( a \sin 90.0° = 1 \times 580 \times 10^{-9} \), where \( \sin 90.0° = 1 \). Therefore, \( a = 580 \times 10^{-9} \text{ m} = 580 \, \text{nm}\). The width of the slit is 580 nm.
04

Apply Intensity Formula for Single-Slit Diffraction

The relative intensity for a single slit is given by \( I(\theta) = I_0 \left(\frac{\sin(\beta)}{\beta}\right)^2 \), where \( \beta = \frac{\pi a \sin \theta}{\lambda} \) and \( I_0 \) is the intensity at \( \theta = 0 \).
05

Calculate Intensity at \( \theta = 45.0° \)

Substitute \( \theta = 45.0° \) into \( \beta \):\[ \beta = \frac{\pi \times 580 \times 10^{-9} \times \sin 45.0°}{580 \times 10^{-9}} = \frac{\pi}{\sqrt{2}} \]Next, calculate \( \left(\frac{\sin(\beta)}{\beta}\right)^2 \) at \( \beta = \frac{\pi}{\sqrt{2}} \).
06

Calculate Intensity Ratio

Calculate \( \left(\frac{\sin(\pi/\sqrt{2})}{\pi/\sqrt{2}}\right)^2 \) to find the ratio. The term simplifies to approximately 0.405. Therefore, the intensity at \( \theta = 45.0° \) is 40.5% of the intensity at \( \theta = 0\). The ratio is 0.405.

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Key Concepts

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

Single-slit diffraction
When light passes through a narrow slit, it doesn't just continue straight. Instead, it spreads out in a phenomenon known as diffraction. This is more noticeable when the slit is comparable in size to the wavelength of light. One of the simplest setups to observe this is in single-slit diffraction. Here, a single narrow slit is used to diffract light onto a screen, creating a pattern of light and dark areas due to the interference of light waves.

Single-slit diffraction is a key example of Fraunhofer diffraction. This type of diffraction occurs when both the light source and the observation screen are far enough from the slit. In such cases, the light waves can be treated as parallel rays. The central bright area on the screen is flanked by a series of dark and bright fringes.

These patterns arise because parts of the light waves overlap and interfere. Where they overlap constructively, we see brightness. Where they interfere destructively, darkness appears. Understanding single-slit diffraction is fundamental in optics and helps in explaining various concepts like the resolving power of optical instruments.
Diffraction pattern
The diffraction pattern generated by a single slit is a classic example of light displaying wave-like behavior. These patterns are characterized by the central maximum and numerous surrounding minima and maxima. The central part of the pattern is the brightest and broadest, known as the central maximum. Around it are a series of smaller and dimmer spots. These are the result of light waves bending around the edges of the slit and interfering with one another.

The position of these dark and bright fringes depends on the wavelength of the light and the width of the slit. For instance, the condition for the first minima occurs when the path difference between light from the edges of the slit is exactly one wavelength. This is described by the formula: \( a \sin \theta = m \lambda \), where \( a \) is the slit width, \( \theta \) is the angle from the center of the pattern, \( m \) is the order of the minima, and \( \lambda \) is the wavelength. This equation helps determine the angles where darkness appears, thus mapping out the diffraction pattern.
Wavelength of light
The wavelength of light, denoted by \( \lambda \), is a critical factor in diffraction phenomena. It represents the distance between two successive peaks of a wave. In the context of diffraction, the wavelength determines how and where the light will spread out and interfere after passing through a slit.

For light of wavelength 580 nm, as given in the example, the diffraction pattern's features—such as how the light and dark bands (minima and maxima) are spaced—are also dictated by this wavelength. A longer wavelength results in a more spread-out pattern, with bands further apart. A shorter wavelength causes a more compressed pattern.

Understanding the wavelength's role helps in predicting and analyzing optical phenomena. It also aids in designing experiments and interpreting the results. Wavelength is not only crucial for visible light but also for understanding various applications ranging from telecommunications to spectroscopy.
Intensity distribution
Intensity distribution refers to how the light's brightness varies across the diffraction pattern. In single-slit diffraction, the intensity, or brightness, of light does not remain constant but changes depending on the angle \( \theta \) relative to the central maximum.

For a single slit, the intensity at any angle \( \theta \) is given by the formula: \( I(\theta) = I_0 \left(\frac{\sin(\beta)}{\beta}\right)^2 \), where \( \beta = \frac{\pi a \sin \theta}{\lambda} \) and \( I_0 \) is the maximum intensity at \( \theta = 0 \). This means that the light is brightest at the center and gradually diminishes as you move away, forming the distinct pattern of peaks and troughs.

At specific angles such as 45 degrees, the formula allows us to calculate the exact intensity relative to the central maximum. For example, in the provided exercise, at \( \theta = 45.0^\circ \), the intensity is about 40.5% of the center. These calculations are crucial for applications that rely on precise control and manipulation of light, such as optical engineering and beam shaping.

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Most popular questions from this chapter

When the light is passed through the bottom of the sample container, the interference maximum is observed to be at 41\(^\circ\); when it is passed through the top, the corresponding maximum is at 37\(^\circ\). What is the best explanation for this observation? (a) The microspheres are more tightly packed at the bottom, because they tend to settle in the suspension. (b) The microspheres aremore tightly packed at the top, because they tend to float to the top of the suspension. (c) The increased pressure at the bottom makes the microspheres smaller there. (d) The maximum at the bottom corresponds to \(m\) = 2, whereas the maximum at the top corresponds to \(m\) = 1.

A diffraction grating has 650 slits>mm. What is the highest order that contains the entire visible spectrum? (The wavelength range of the visible spectrum is approximately 380-750 nm.)

Monochromatic x rays are incident on a crystal for which the spacing of the atomic planes is 0.440 nm. The first-order maximum in the Bragg reflection occurs when the incident and reflected x rays make an angle of 39.4\(^\circ\) with the crystal planes. What is the wavelength of the x rays?

A loudspeaker with a diaphragm that vibrates at 960 Hz is traveling at 80.0 m/s directly toward a pair of holes in a very large wall. The speed of sound in the region is 344 m/s. Far from the wall, you observe that the sound coming through the openings first cancels at \(\pm11.4^\circ\) with respect to the direction in which the speaker is moving. (a) How far apart are the two openings? (b) At what angles would the sound first cancel if the source stopped moving?

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 \(\mu\)m in diameter) limits the size of an object at the near point (25 cm) of the eye to a height of about 50 \(\mu\)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-\(\mu\)m- tall object at 25 cm from the eye with light of wavelength 550 nm? (b) According to the Rayleigh criterion, what is the shortest object we could resolve at the 25-cm near point with light of wavelength 550 nm? (c) What angle would the object in part (b) subtend at the eye? Express your answer in minutes (60 min = 1\(^\circ\)), 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?

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