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Parallel rays of green mercury light with a wavelength of 546 \(\mathrm{nm}\) pass through a slit covering a lens with a focal length of \(60.0 \mathrm{cm} .\) In the focal plane of the lens the distance from the central maximum to the first minimum is 10.2 \(\mathrm{mm}\) . What is the width of the slit?

Short Answer

Expert verified
The width of the slit is approximately 0.0321 mm.

Step by step solution

01

Identify the formula

For a single-slit diffraction pattern, the distance from the central maximum to the first minimum is given by the formula: \( a \sin \theta = m\lambda \) where \( a \) is the slit width, \( \theta \) is the angle, \( m \) is the order number (\( m = 1 \) for the first minimum), and \( \lambda \) is the wavelength. Since \( \theta \) is small, \( \sin \theta \approx \tan \theta \approx \frac{y}{f} \), where \( y \) is the distance from the central maximum to the first minimum on the focal plane, and \( f \) is the focal length.
02

Apply the formula

Substitute \( \lambda = 546 \mathrm{nm} = 546 \times 10^{-9} \mathrm{m} \), \( y = 10.2 \mathrm{mm} = 10.2 \times 10^{-3} \mathrm{m} \), \( f = 60.0 \mathrm{cm} = 0.6 \mathrm{m} \), and \( m = 1 \) into the equation \( a \sin \theta = m\lambda \). Since \( \sin \theta \approx \frac{y}{f} \), we have \( a \frac{y}{f} = \lambda \).
03

Solve for the slit width (a)

Rearrange the equation \( a \frac{y}{f} = \lambda \) to find \( a \). This gives: \( a = \frac{\lambda f}{y} \). Substitute the known values to calculate \( a \): \( a = \frac{546 \times 10^{-9} \mathrm{m} \times 0.6 \mathrm{m}}{10.2 \times 10^{-3} \mathrm{m}} \approx 3.21 \times 10^{-5} \mathrm{m} \).
04

Convert to a convenient unit

Convert the result from meters to millimeters for a more practical unit of measurement: \( a \approx 3.21 \times 10^{-5} \mathrm{m} = 0.0321 \mathrm{mm} \).

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Key Concepts

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

Single-slit diffraction
Single-slit diffraction occurs when light waves pass through a narrow slit and spread out, creating a diffraction pattern consisting of bright and dark fringes. This phenomenon is described by the equation \( a \sin \theta = m\lambda \), where:
  • \(a\) is the width of the slit,
  • \(\theta\) is the diffraction angle,
  • \(m\) is the order of the minimum (e.g., \(m = 1\) for the first minimum), and
  • \(\lambda\) is the wavelength of the light.
Understanding the Equation Components:- **Slit Width \(a\):** Indicates how narrow the slit is. A smaller slit width results in a larger and more noticeable diffraction pattern.- **Diffraction Angle \(\theta\):** The angle between the light's original path and its direction after passing through the slit.- **Order \(m\):** Represents the sequence of dark (minimum) fringes with \(m = 0\) for the central maximum. The first minimum is usually focused on in problems.
The single-slit diffraction pattern is a fundamental concept in wave optics and helps in understanding how waves interact with obstacles.
Mercury light wavelength
Mercury light is commonly used in physics experiments due to its sharp emission lines. One significant wavelength of mercury light is 546 nanometers (nm). This green light is part of the visible spectrum and is often used in diffraction experiments.- **Nanometer (nm):** A unit of length in the metric system, equal to one billionth of a meter (\(10^{-9} m\)).- **Green Light:** The wavelength 546 nm places it in the green portion of the visible spectrum, which ranges from approximately 495 to 570 nm.
Understanding the wavelength is crucial because it affects the diffraction pattern produced. In the diffraction equation \( a \sin \theta = m\lambda \), the wavelength determines the distance between the fringes for a given slit width. This dependency is why accurately knowing the wavelength is necessary for calculating diffraction effects accurately.
Focal length
The focal length of a lens is the distance between the lens and its focal point, where it converges the light rays passing through it. It's a key parameter in optics, dictating how a lens focuses light.- **Measuring Focal Length:** Typically measured in centimeters (cm) or meters (m). In the given exercise, the focal length is 60.0 cm.- **Role in Diffraction Experiments:** In a single-slit diffraction setup, the lens focuses the diffracted light rays onto a screen or sensor, forming the diffraction pattern. The focal length helps determine the size of the pattern observed.
Using the relation \( \sin \theta \approx \frac{y}{f} \), we can connect the focal length \(f\) with other parameters like the distance from the central maximum to the first minimum \(y\). This approximation helps in simplifying calculations, particularly when the angles involved are small. Understanding focal length ensures accurate predictions of where points like the central maximum will appear in relation to the lens.

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Most popular questions from this chapter

The intensity of light in the Fraunhofer diffraction pattern of a single slit is $$ I=I_{0}\left(\frac{\sin \gamma}{\gamma}\right)^{2} $$ where $$ \gamma=\frac{\pi a \sin \theta}{\lambda} $$ (a) Show that the equation for the values of \(\gamma\) at which \(I\) is a maximum is \(\tan \gamma=\gamma\) (b) Determine the three smallest positive values of \(\gamma\) that are solutions of this equation (Hint: You can use a trial-and-error procedure. Guess a value of \(\gamma\) and adjust your guess to bring tan \(\gamma\) closer to \(\gamma\) A graphical solution of the equation is very helpful in locating the solutions approximately, to get good initial guesses.)

A series of parallel linear water wave fronts are traveling directly toward the shore at 15.0 \(\mathrm{cm} / \mathrm{s}\) on an otherwise placid lake. A long concrete barrier that runs parallel to the shore at a distance of 3.20 \(\mathrm{m}\) away has a hole in it. You count the wave crests and observe that 75.0 of them pass by each minute, and you also observe that no waves reach the shore at \(\pm 61.3 \mathrm{cm}\) from the point directly opposite the hole, but waves do reach the shore everywhere within this distance. (a) How wide is the hole in the barrier? (b) At what other angles do you find no waves hitting the shore?

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.)

An astronaut in orbit can just resolve two point sources on the earth that are 75.0 \(\mathrm{m}\) apart. Assume that the resolution is diffraction limited, and use Rayleigh's criterion. What is the astronaut's altitude above the earth? Treat her eye as a circular aperture with a diameter of 4.00 \(\mathrm{mm}\) (the diameter of her pupil), and take the wavelength of the light to be 500 \(\mathrm{nm}\) .

Monochromatic light is at normal incidence on a plane transmission grating. The first-order maximum in the interference pattern is at an angle of \(8.94^{\circ} .\) What is the angular position of the fourth-order maximum?

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