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Two slits spaced 0.450 mm apart are placed 75.0 cm from a screen. What is the distance between the second and third dark lines of the interference pattern on the screen when the slits are illuminated with coherent light with a wavelength of 500 nm?

Short Answer

Expert verified
The distance is approximately 0.083 mm.

Step by step solution

01

Understand the Concepts

Interference patterns involve alternating light and dark bands. Dark bands (minima) occur at positions where the path difference between light from the two slits is a half-integer multiple of the wavelength, i.e., \[ d \sin \theta = (m + \frac{1}{2})\lambda \]where \(d\) is the slit separation, \(\theta\) is the angle of the dark fringe, \(m\) is the order of the minimum (1, 2 for second and third dark lines), and \(\lambda\) is the wavelength of the light.
02

Calculate the Angle for the Second Dark Line

For the second dark line, \(m = 1\). Using the formula for the dark fringes, \[ 0.450 \times 10^{-3} \sin \theta_2 = (1 + \frac{1}{2})(500 \times 10^{-9}) \]Solve for \(\theta_2\):\[ \sin \theta_2 = \frac{1.5 \times 500 \times 10^{-9}}{0.450 \times 10^{-3}} \approx 1.67 \times 10^{-3} \]Thus, \(\theta_2 \approx \sin^{-1}(1.67 \times 10^{-3})\).
03

Calculate the Angle for the Third Dark Line

For the third dark line, \(m = 2\). Use the dark fringe pattern formula:\[ 0.450 \times 10^{-3} \sin \theta_3 = (2 + \frac{1}{2})(500 \times 10^{-9}) \]Solve for \(\theta_3\):\[ \sin \theta_3 = \frac{2.5 \times 500 \times 10^{-9}}{0.450 \times 10^{-3}} \approx 2.78 \times 10^{-3} \]Thus, \(\theta_3 \approx \sin^{-1}(2.78 \times 10^{-3})\).
04

Find the Positions on the Screen

Use the small angle approximation, where \(\sin \theta \approx \tan \theta \approx \theta\) and the formula for distance from the center, \[ y = L \tan \theta \approx L \theta \]For \(\theta_2\) and \(\theta_3\) calculated earlier, with \(L = 75.0 \times 10^{-2} \) m:\[ y_2 = 75.0 \times 10^{-2} \cdot 1.67 \times 10^{-3} \approx 1.2525 \times 10^{-1} \text{ mm} \]\[ y_3 = 75.0 \times 10^{-2} \cdot 2.78 \times 10^{-3} \approx 2.085 \times 10^{-1} \text{ mm} \]
05

Determine the Distance Between Second and Third Dark Lines

Subtract the positions of the second and third dark lines:\[ \Delta y = y_3 - y_2 = 0.2085 - 0.12525 \approx 0.08325 \text{ mm} \]
06

Conclusion

The distance between the second and third dark lines on the screen is approximately 0.083 mm.

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Key Concepts

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

Young's Double-Slit Experiment
Young's Double-Slit Experiment is a classic physics experiment demonstrating the wave nature of light. When light passes through two closely spaced slits, it creates an interference pattern on a screen placed behind the slits. This pattern consists of alternating bright and dark areas called fringes.
If the light coming through the slits was considered as particles, you would expect two distinct lines on the screen because particles travel straight. However, by observing a pattern of multiple lines, it suggests that light behaves like a wave, experiencing constructive and destructive interference.
Understanding this collaboration of light waves through the slits remains a cornerstone of optical and quantum physics.
Optical Path Difference
The Optical Path Difference (OPD) is crucial in analyzing interference patterns. It refers to the difference in the travel paths of light waves coming from the two slits to a common point on the screen.
As waves travel different distances, the OPD leads to either amplification or cancellation of the waves. This interference is mathematically expressed as
\[ d \sin \theta = m\lambda \quad \text{for bright fringes} \]
and
\[ d \sin \theta = \left(m + \frac{1}{2}\right)\lambda \quad \text{for dark fringes} \]
Here, \(d\) is the slit separation, \(\lambda\) is the wavelength of the light, and \(m\) is the fringe order.
The OPD concept helps explain why you see specific patterns of light and dark bands.
Wavelength and Slit Separation
Understanding the relationship between wavelength and slit separation is vital in Young's Double-Slit Experiment. The wavelength \(\lambda\) is the distance over which the wave's shape repeats. In this experiment, knowing the wavelength of the incoming light is essential to calculate where the interference fringes will appear.
Slit separation \(d\) is the physical distance between the two slits. This separation plays a critical role in determining the spacing of the interference pattern. As the slit separation increases, the fringes move closer together, and vice versa.
The slit separation and wavelength together shape the entire interference pattern according to the formula:
\[ d \sin \theta = \left(m + \frac{1}{2}\right)\lambda \]
where \(\theta\) is the angle at which a fringe occurs.
Dark Fringe Calculation
Dark fringes, which are the areas of minima in an interference pattern, occur where the path difference leads to destructive interference. To calculate these dark fringes, the formula \[ d \sin \theta = \left(m + \frac{1}{2}\right)\lambda \] is used.
Here, \(m\) denotes the order of the dark fringe being calculated. For example, \(m = 1\) for the second dark line and \(m = 2\) for the third, and so forth.
Once the angle \(\theta\) for these dark fringes is found using the above formula, you can convert it to a linear distance \(y\) on the screen using \[ y = L \tan \theta \approx L \theta \].
This conversion is possible due to the small angle approximation where \(\sin \theta \approx \theta\), allowing you to find the position where the light waves cancel out, leading to the dark fringes.

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Most popular questions from this chapter

When viewing a piece of art that is behind glass, one often is affected by the light that is reflected off the front of the glass (called \(glare\)), which can make it difficult to see the art clearly. One solution is to coat the outer surface of the glass with a film to cancel part of the glare. (a) If the glass has a refractive index of 1.62 and you use Ti\(O^2\) , which has an index of refraction of 2.62, as the coating, what is the minimum film thickness that will cancel light of wavelength 505 nm? (b) If this coating is too thin to stand up to wear, what other thickness would also work? Find only the three thinnest ones.

A plastic film with index of refraction 1.70 is applied to the surface of a car window to increase the reflectivity and thus to keep the car's interior cooler. The window glass has index of refraction 1.52. (a) What minimum thickness is required if light of wavelength 550 nm in air reflected from the two sides of the film is to interfere constructively? (b) Coatings as thin as that calculated in part (a) are difficult to manufacture and install. What is the next greater thickness for which constructive interference will also occur?

In a two-slit interference pattern, the intensity at the peak of the central maximum is \(I_0\) . (a) At a point in the pattern where the phase difference between the waves from the two slits is 60.0\(^\circ\), what is the intensity? (b) What is the path difference for 480-nm light from the two slits at a point where the phase difference is 60.0\(^\circ\)?

Eyeglass lenses can be coated on the \(inner\) surfaces to reduce the reflection of stray light to the eye. If the lenses are medium flint glass of refractive index 1.62 and the coating is fluorite of refractive index 1.432, (a) what minimum thickness of film is needed on the lenses to cancel light of wavelength 550 nm reflected toward the eye at normal incidence? (b) Will any other wavelengths of visible light be cancelled or enhanced in the reflected light?

Coherent light with wavelength 400 nm passes through two very narrow slits that are separated by 0.200 mm, and the interference pattern is observed on a screen 4.00 m from the slits. (a) What is the width (in mm) of the central interference maximum? (b) What is the width of the first-order bright fringe?

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