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Parallel rays of green mercury light with a wavelength of 546 nm pass through a slit covering a lens with a focal length of 60.0 cm. In the focal plane of the lens, the distance from the central maximum to the first minimum is 8.65 mm. What is the width of the slit?

Short Answer

Expert verified
The width of the slit is 37.9 micrometers.

Step by step solution

01

Understand the Problem

We are given the wavelength \( \lambda = 546 \text{ nm} \), the focal length of the lens \( f = 60.0 \text{ cm} \), and the distance from the central maximum to the first minimum \( x_1 = 8.65 \text{ mm} \). Our task is to determine the width of the slit \( a \).
02

Use the Single-Slit Diffraction Formula

In single-slit diffraction, the position of the first minimum is given by the formula: \( a \sin \theta = m \lambda \) for \( m = 1 \).
03

Relate the Angle to the Geometry

From geometry, the angle \( \theta \) can be approximated using \( \tan \theta \approx \sin \theta = \frac{x_1}{f} \), since \( \theta \) is small. Thus, we have \( \sin \theta = \frac{8.65 \times 10^{-3} \text{ m}}{0.60 \text{ m}} \).
04

Solve for the Width of the Slit \( a \)

Substitute \( \sin \theta \) from Step 3 into the diffraction formula to find \( a \):\[a = \frac{m \lambda}{\sin \theta} = \frac{1 \times 546 \times 10^{-9} \text{ m}}{\frac{8.65 \times 10^{-3} \text{ m}}{0.60 \text{ m}}}\approx 3.79 \times 10^{-5} \text{ m}.\]
05

Finalize the Answer

Convert the slit width to micrometers: \( a \approx 37.9 \mu\text{m} \). Thus, the width of the slit is 37.9 micrometers.

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Key Concepts

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

Wavelength of Light
The wavelength of light is a critical concept in understanding many optical phenomena. It refers to the distance between successive peaks of the light wave. For visible light, this distance is typically measured in nanometers (nm), with 1 nm equalling one billionth of a meter.
For example, in the exercise, the given wavelength is 546 nm, which is in the range of green light on the electromagnetic spectrum. Knowing the wavelength of light is essential when calculating diffraction patterns, as it directly affects the location of diffraction minima and maxima.
When light waves encounter obstacles or openings, such as a slit, their path is altered in a manner dependent on the wavelength. This alteration gives rise to diffraction, a core concept of wave optics. Understanding wavelength helps us calculate other related phenomena like interference, a key mechanism in various optical devices.
Focal Length
Focal length is an important characteristic of lenses, determining how strongly they converge or diverge light. It is the distance from the lens to its focus, the point where parallel light rays coming into the lens are brought together.
In the exercise, the focal length provided is 60.0 cm. This distance affects the geometry of diffraction patterns formed through the lens.
The focal length influences the position and size of diffraction patterns in the lens's focal plane.
  • A shorter focal length lens will converge light more sharply, forming smaller patterns.
  • A longer focal length lens creates larger, more spread-out patterns.
This principle is crucial for designing optical systems, helping us predict how light will behave when passing through different lenses.
Diffraction Minimum
Diffraction in wave optics refers to the bending of light as it encounters an obstacle or passes through an opening. The points where the light waves destructively interfere, resulting in minimal light intensity, define the diffraction minimum.
In the single-slit diffraction formula, the first minimum occurs when the path difference is equal to one wavelength, mathematically expressed as \( a \sin \theta = m \lambda \) for \( m = 1 \).
This exercise involves finding the slit width by measuring the distance from the central maximum to the first diffraction minimum. Using the relationship of angles and distances, we calculate the slit width, which controls how light diffracts through the slit. Such understanding helps optimize devices like optical filters where specific wavelengths need to be minimized.
Wave Optics
Wave optics, also known as physical optics, is the study of light's interaction with objects in terms of waves, rather than particles. Key phenomena under wave optics include interference, diffraction, and polarization.
In the exercise, single-slit diffraction showcases wave optics principles, as light behaves according to the wave nature of light. Through slits, light waves spread out and form patterns due to their uniform wavelength and interaction.
This behavior contrasts with ray optics, which considers light as straight rays. Wave optics provides explanations for aspects like the diffraction pattern seen in slits or the coloring of oil slicks due to interference.
  • Diffraction details how waves encounter and navigate obstacles.
  • Interference leads to patterns of light and dark bands due to the superposition of waves.
Wave optics is essential to understanding optical technology and natural phenomena.

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Most popular questions from this chapter

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?

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?

A laser beam of wavelength \(\lambda\) = 632.8 nm shines at normal incidence on the reflective side of a compact disc. (a) The tracks of tiny pits in which information is coded onto the CD are 1.60 \(\mu\)m apart. For what angles of reflection (measured from the normal) will the intensity of light be maximum? (b) On a DVD, the tracks are only 0.740 \(\mu\)m apart. Repeat the calculation of part (a) for the DVD.

Laser light of wavelength 632.8 nm falls normally on a slit that is 0.0250 mm wide. The transmitted light is viewed on a distant screen where the intensity at the center of the central bright fringe is 8.50 W/m\(^2\). (a) Find the maximum number of totally dark fringes on the screen, assuming the screen is large enough to show them all. (b) At what angle does the dark fringe that is most distant from the center occur? (c) What is the maximum intensity of the bright fringe that occurs immediately before the dark fringe in part (b)? Approximate the angle at which this fringe occurs by assuming it is midway between the angles to the dark fringes on either side of it.

If the planes of a crystal are 3.50 \(\AA\) (1 \(\AA\) = 10\(^{-10}\) m = 1 \(\AA\)ngstrom unit) apart, (a) what wavelength of electromagnetic waves is needed so that the first strong interference maximum in the Bragg reflection occurs when the waves strike the planes at an angle of 22.0\(^\circ\), and in what part of the electromagnetic spectrum do these waves lie? (See Fig. 32.4.) (b) At what other angles will strong interference maxima occur?

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