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A single-slit diffraction pattern is formed by monochromatic electromagnetic radiation from a distant source passing through a slit 0.105 mm wide. At the point in the pattern 3.25\(^\circ\) from the center of the central maximum, the total phase difference between wavelets from the top and bottom of the slit is 56.0 rad. (a) What is the wavelength of the radiation? (b) What is the intensity at this point, if the intensity at the center of the central maximum is \(I_0\)?

Short Answer

Expert verified
(a) Using the formula with given values, \(\lambda = 365.4 \text{ nm}\). (b) Intensity \(I(\theta) = 0.19 I_0\).

Step by step solution

01

Relate phase difference to path difference

The path difference across the slit is related to the phase difference by the formula:\[ \Delta \phi = \frac{2\pi}{\lambda} \Delta x \], where \(\Delta \phi\) is the phase difference, \(\lambda\) is the wavelength, and \(\Delta x\) is the path difference across the slit. We are given that \(\Delta \phi = 56.0\, \text{rad}\).
02

Calculate path difference using angle

The path difference \(\Delta x\) for a point \(\theta\) from the central maximum is given by \(\Delta x = a \sin(\theta)\), where \(a\) is the slit width (0.105 mm). The angle \(\theta\) is given as 3.25° or 0.0567 radians (after converting from degrees to radians).
03

Substitute values to find wavelength

Substitute the values into the phase difference equation to find the wavelength: \(56.0 = \frac{2\pi}{\lambda} \times 0.105 \times 10^{-3} \times \sin(0.0567)\). Solve for \(\lambda\) to find the wavelength.
04

Calculate the intensity using the Fraunhofer diffraction formula

The intensity \(I(\theta)\) at a given angle \(\theta\) compared to the central maximum \(I_0\) is given by the formula: \( I(\theta) = I_0 \left( \frac{\sin(\beta)}{\beta} \right)^2 \), where \(\beta = \frac{\pi a \sin(\theta)}{\lambda}\).
05

Substitute wavelength to find intensity at the point

Use the previously calculated wavelength and substitute it in the expression for \(\beta\). Then calculate: \( \beta = \frac{\pi \times 0.105 \times 10^{-3} \times \sin(0.0567)}{\lambda}\). Find \(\sin(\beta)\) and use the Fraunhofer equation to find the intensity \(I(\theta)\).

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Key Concepts

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

Monochromatic Radiation
Monochromatic radiation refers to light that consists of a single wavelength. This concept is essential in understanding single-slit diffraction patterns, as the light source's uniformity directly impacts the pattern's clarity and predictability. This kind of radiation is often used in experiments because it creates distinct, measurable patterns. Monochromatic sources, like lasers, ensure that the resulting diffraction pattern is clean and uncomplicated by multiple wavelengths interacting and overlapping.
This simplicity allows for easier calculations and a deeper understanding of the wave properties involved. In a single-slit experiment, knowing that the light is monochromatic helps us predict and calculate the pattern using precise mathematical models.
Phase Difference
In physics, the phase difference between waves refers to the amount by which one wave lags or leads another. It is measured in radians, where a complete cycle corresponds to a phase difference of 2\( \pi \) radians. In the context of single-slit diffraction, the phase difference is crucial for understanding how waves arriving from different parts of the slit interfere with each other.
  • The formula \( \Delta \phi = \frac{2\pi}{\lambda} \Delta x \) relates phase difference \( \Delta \phi \) to wavelength \( \lambda \) and path difference \( \Delta x \).
  • A larger phase difference results in more significant interference effects, influencing the position and size of dark and bright fringes in the diffraction pattern.
Understanding phase differences allows for predicting and calculating the fringe pattern in detail, giving insights into factors like light frequency and slit width.
Path Difference
Path difference is a term that describes the difference in distance traveled by two waves from their origin to a common point of observation. In the single-slit diffraction pattern, the path difference determines how the waves interfere when they merge on a surface.
The relationship between path difference and the diffraction pattern is governed by the equation \( \Delta x = a \sin(\theta) \), where \( a \) is the width of the slit and \( \theta \) is the angle from the central maximum. Calculating the path difference helps in understanding whether constructive or destructive interference occurs at a particular point in the diffraction pattern.
This understanding is critical in predicting where bright and dark bands will form, based on the geometry and properties of the light and slit.
Fraunhofer Diffraction Formula
The Fraunhofer diffraction formula is a paramount tool for calculating the intensity patterns in single-slit diffraction. It helps to quantify how light intensity varies at different angles from the slit.
  • The formula \( I(\theta) = I_0 \left( \frac{\sin(\beta)}{\beta} \right)^2 \) relates the intensity \( I(\theta) \) of light at an angle \( \theta \) to the maximum intensity \( I_0 \).
  • The parameter \( \beta \) is given by \( \beta = \frac{\pi a \sin(\theta)}{\lambda} \), where \( a \) is the slit width and \( \lambda \) is the wavelength.
This formula empowers physicists to predict the precise distribution of light and dark bands in the diffraction pattern, providing detailed insights into wave behaviors, based on fundamental wave properties. By employing the Fraunhofer formula, scientists and engineers can accurately model light behavior, enhancing applications ranging from optical instrument design to precise scientific measurements.

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Most popular questions from this chapter

A wildlife photographer uses a moderate telephoto lens of focal length 135 mm and maximum aperture \(f/\)4.00 to photograph a bear that is 11.5 m away. Assume the wavelength is 550 nm. (a) What is the width of the smallest feature on the bear that this lens can resolve if it is opened to its maximum aperture? (b) If, to gain depth of field, the photographer stops the lens down to \(f/\)22.0, what would be the width of the smallest resolvable feature on the bear?

Diffraction occurs for all types of waves, including sound waves. High- frequency sound from a distant source with wavelength 9.00 cm passes through a slit 12.0 cm wide. A microphone is placed 8.00 m directly in front of the center of the slit, corresponding to point \(O\) in Fig. 36.5a. The microphone is then moved in a direction perpendicular to the line from the center of the slit to point \(O\). At what distances from \(O\) will the intensity detected by the microphone be zero?

Monochromatic light of wavelength 592 nm from a distant source passes through a slit that is 0.0290 mm wide. In the resulting diffraction pattern, the intensity at the center of the central maximum \((\theta = 0^\circ)\) is 4.00 \(\times\) 10\(^{-5}\) W/m\(^2\). What is the intensity at a point on the screen that corresponds to \(\theta\) = 1.20\(^\circ\)?

Monochromatic light of wavelength \(\lambda\) = 620 nm from a distant source passes through a slit 0.450 mm wide. The diffraction pattern is observed on a screen 3.00 m from the slit. In terms of the intensity \(I_0\) at the peak of the central maximum, what is the intensity of the light at the screen the following distances from the center of the central maximum: (a) 1.00 mm; (b) 3.00 mm; (c) 5.00 mm?

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?

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