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Laser light of wavelength \(500.0 \mathrm{nm}\) 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.

Short Answer

Expert verified
Slit width and separation are both approximately 45 μm (micrometers).

Step by step solution

01

Understand the Problem

We have laser light with a wavelength of \(500.0 \mathrm{nm}\), and it creates an interference pattern on a screen 90.0 cm away. The bright bands are 1.00 cm apart. The third-order bright band is missing. We need to find the slit width \(a\) and separation \(d\).
02

Use Young's Double-Slit Formula

The position \(y\) of the m-th bright fringe is given by: \[ y_m = \frac{m \cdot \lambda \cdot L}{d} \]where \(m\) is the order number, \(\lambda = 500 \times 10^{-9} \mathrm{m}\), and \(L = 0.9 \mathrm{m}\). The distance between consecutive bright fringes (fringe separation) is given by: \[ \Delta y = \frac{\lambda \cdot L}{d} \]We know \(\Delta y = 0.01 \mathrm{m} \).
03

Calculate the Slit Separation

From the fringe separation formula: \[ d = \frac{\lambda \cdot L}{\Delta y} \]Plug in the values: \[ d = \frac{500 \times 10^{-9} \cdot 0.9}{0.01} \approx 4.5 \times 10^{-5} \mathrm{m} \] This is the separation between the slits.
04

Understand Missing Order and its Implication

The third bright fringe being missing suggests destructive interference due to the single-slit diffraction condition: \[ a \cdot \sin \theta = m' \cdot \lambda \]For the third order, \(m'=3\), we test with small angles where \(\sin \theta \approx \tan \theta \approx \frac{y}{L}\).
05

Relate Missing Order to Slit Width

The position of the third interference fringe is: \[ y_3 = \frac{3 \cdot \lambda \cdot L}{d} \]From single-slit diffraction, the third order is at: \[ a \cdot \frac{3\cdot 0.01}{0.9} = 3 \cdot 500 \times 10^{-9} \]Solve for slit width \(a\): \[ a = \frac{0.9 \cdot 1500 \times 10^{-9}}{3\cdot 0.01} = 4.5 \times 10^{-5} \mathrm{m} \]
06

Conclusion

Both slit width \(a\) and separation \(d\) are approximately \(4.5 \times 10^{-5} \mathrm{m}\). They are equal due to the condition of the missing third-order fringe.

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Key Concepts

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

Wavelength
In the context of light and interference patterns, the wavelength is a fundamental concept. It refers to the distance between two corresponding points on consecutive cycles of a wave, such as from crest to crest. For visible light, wavelengths range from about 400 nm to 700 nm, with specific colors associated with particular wavelengths.

In the given problem, the laser light has a wavelength of 500 nm. This wavelength interacts with the slits to produce an interference pattern. Knowing the wavelength is essential, as it determines the spacing of the interference fringes on the detection screen. This pattern results from the wave nature of light experiencing constructive and destructive interference.
  • The wavelength helps determine fringe position in experiments like Young's double-slit.
  • Shorter wavelengths like ultraviolet light create narrower separation between fringes.
  • Wavelength is crucial for calculating optical properties and understanding wave behaviors.
Fringe Separation
Fringe separation refers to the distance between adjacent bright or dark bands in an interference pattern. It is a critical measurement in Young's double-slit experiment, helping to determine properties like slit separation.

In the problem, the fringe separation is 1.00 cm. This is the distance between successive bright fringes that appear on a screen due to interference of light waves passing through the double slits. The separation provides insight into actual physical properties of the slits.
  • The formula \( \Delta y = \frac{\lambda \cdot L}{d} \) describes fringe separation, where \( \Delta y \) is the fringe spacing.
  • Increasing screen distance \( L \) increases fringe separation.
  • Larger fringe separation indicates a larger distance between slits \( d \).
Destructive Interference
Destructive interference is a phenomenon where two waves combine to form a reduced amplitude. In the case of light waves, this results in dark bands on the interference pattern. It occurs when waves are out of phase by half of a wavelength, leading to cancellation.

In the exercise, the absence of the third bright fringe indicates destructive interference. The condition is imposed by single-slit effects interacting with the double-slit pattern, causing the third fringe to "disappear." This helps us find the slit dimensions.
  • Destructive interference results in missing fringes or dark areas in patterns.
  • Occurs at angles where the path difference between two waves is an odd multiple of half-wavelengths.
  • Can also indicate slit width characteristics when specific bright fringes are absent.
Young's Double-Slit Experiment
Young's double-slit experiment is a classic demonstration of the wave nature of light. It shows how light waves passing through two slits produce an interference pattern on a screen.

The experiment provides a way to measure wavelengths and explore interference effects. In the problem, it is utilized to calculate slit dimensions based on the resulting interference pattern. The appearance of bright and dark fringes is directly related to wave interference.
  • The experiment helps visualize wave phenomena like interference and diffraction.
  • It validates the principle of superposition of waves.
  • Can be adjusted to evaluate both light and particle behaviors with modern quantum experiments.

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Most popular questions from this chapter

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?

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 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 \(I_0\)?

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?

Monochromatic light with wavelength 620 nm passes through a circular aperture with diameter 7.4 \(\mu\)m. The resulting diffraction pattern is observed on a screen that is 4.5 m from the aperture. What is the diameter of the Airy disk on the screen?

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