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35.14. Coherent light that contains two wavelengths, 660 \(\mathrm{nm}\) ( red) and 470 \(\mathrm{nm}\) (blue), passes through two narrow slits separated by \(0.300 \mathrm{mm},\) and the interference pattern is observed on a screen 5.00 \(\mathrm{m}\) from the slits. What is the distance on the screen between the first-order bright fringes for the two wavelengths?

Short Answer

Expert verified
The distance between the first-order bright fringes is 3.17 mm.

Step by step solution

01

Identify the Known Values

We have two wavelengths: \(\lambda_1 = 660\, \mathrm{nm} = 660 \times 10^{-9}\, \mathrm{m}\) (red) and \(\lambda_2 = 470\, \mathrm{nm} = 470 \times 10^{-9}\, \mathrm{m}\) (blue). The slit separation \(d = 0.300\, \mathrm{mm} = 0.300 \times 10^{-3}\, \mathrm{m}\), and the distance from the slits to the screen \(L = 5.00\, \mathrm{m}\).
02

Use the Formula for Fringe Position

The position \(y\) of the \(m\)-th order bright fringe for a given wavelength in a double-slit interference pattern is calculated using the formula: \[ y_m = \frac{m \lambda L}{d} \]For the first-order bright fringes \(m = 1\).
03

Calculate Position of First-Order Bright Fringe for 660 nm

Substitute \(m = 1\), \(\lambda = 660 \times 10^{-9}\, \mathrm{m}\), \(L = 5.00\, \mathrm{m}\), and \(d = 0.300 \times 10^{-3}\, \mathrm{m}\) into the formula: \[ y_1^{660} = \frac{(1)(660 \times 10^{-9})(5.00)}{0.300 \times 10^{-3}} = \frac{3300 \times 10^{-9}}{0.300 \times 10^{-3}} = 0.011 \, \mathrm{m} = 11 \, \mathrm{mm} \]
04

Calculate Position of First-Order Bright Fringe for 470 nm

Substitute \(m = 1\), \(\lambda = 470 \times 10^{-9}\, \mathrm{m}\), \(L = 5.00\, \mathrm{m}\), and \(d = 0.300 \times 10^{-3}\, \mathrm{m}\) into the formula: \[ y_1^{470} = \frac{(1)(470 \times 10^{-9})(5.00)}{0.300 \times 10^{-3}} = \frac{2350 \times 10^{-9}}{0.300 \times 10^{-3}} = 0.00783 \, \mathrm{m} = 7.83 \, \mathrm{mm} \]
05

Find the Distance Between the Two Fringes

Subtract the position of the blue fringe from the red fringe: \[ \Delta y = y_1^{660} - y_1^{470} = 11 \mathrm{mm} - 7.83 \mathrm{mm} = 3.17 \mathrm{mm} \]

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Key Concepts

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

Wavelength
The wavelength of light is a fundamental concept in understanding double-slit interference. It is the distance between consecutive peaks of a wave and is typically measured in nanometers (nm) for light. In this exercise, we have two wavelengths: 660 nm for red light and 470 nm for blue light. These wavelengths determine the color of the light and play a crucial role in the interference pattern observed.
Wavelengths dictate how the light waves interfere with each other when they pass through slits. Longer wavelengths, like the red light in our exercise, will spread out more and produce a broader fringe pattern. Conversely, shorter wavelengths, like blue light, are more closely spaced, leading to more narrowly spaced fringes. Thus, understanding the concept of wavelength helps explain why different colors of light form distinctive patterns when they pass through slits.
Fringe Position
In double-slit interference, 'fringe position' refers to the specific locations on the observation screen where bright and dark bands appear. These bands, known as fringes, are the result of constructive and destructive interference of light waves. Constructive interference happens when the path difference between the two waves equals an integer multiple of the wavelength, resulting in bright fringes.
The position of these fringes is calculated using the formula: \[ y_m = \frac{m \lambda L}{d} \]where \(y_m\) is the fringe position, \(m\) is the order number of the fringe (1, 2, 3,...), \(\lambda\) is the wavelength of light, \(L\) is the distance from the slits to the screen, and \(d\) is the separation between the slits. In our problem, the first-order fringe (\(m = 1\)) positions are calculated for two different wavelengths, demonstrating how different wavelengths result in fringes appearing at different positions on the screen.
Coherent Light
Coherent light is a key requirement for creating an interference pattern, such as in the double-slit experiment. Coherent light consists of waves that are in phase or have a constant phase difference. In simpler terms, coherent light waves travel together in a uniform manner, allowing them to consistently interfere with each other—crucial for forming clear and stable patterns of bright and dark fringes.
For the exercise, the coherent source ensures that light maintains its wave characteristics after passing through the slits. This coherence allows us to apply our formula for fringe positions confidently, as the waves maintain a predictable relationship, leading to the formation of observable interference patterns. Without coherence, the light waves would not interact in a consistent way, leading to a lack of structured patterns on the screen.

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Most popular questions from this chapter

35.10. Coherent light with wavelength 450 \(\mathrm{nm}\) falls on a double slit. On a screen 1.80 \(\mathrm{m}\) away, the distance between dark fringes is 4.20 \(\mathrm{mm}\) . What is the separation of the slits?

35\. 41. Suppose you illuminate two thin slits by monochromatic coherent light in air and find that they produce their first interference minima at \(\pm 35.20^{\circ}\) on either side of the central bright spot. You then immerse these slits in a transparent liquid and illuminate them with the same light, Now you find that the first minima occur at \(\pm 19.46^{\circ}\) instead. What is the index of refraction of this liquid?

35.55. A source \(S\) of monochrounatic light and a detector \(D\) are both located in air a distance \(h\) above a horizontal plane sheet of glass, and are separated by a horizontal distance \(x\) . Waves reaching \(D\) directly from \(S\) interfere with waves that reflect off the glass. The distance \(x\) is small compared to \(h\) so that the reflection is at close to normal incidence. (a) Show that the condition for constructive interference is \(\sqrt{x^{2}+4 h^{2}}-x=\left(m+\frac{1}{2}\right) \lambda,\) and the condition for destructive interference is \(\sqrt{x^{2}+4 h^{2}}-x=m \lambda\) (Hint: Take into account the phase change on reflection.) (b) Let \(h=24 \mathrm{cm}\) and \(x=14 \mathrm{cm} .\) What is the longest wavelength for which there will be constructive interference?

35.31. A uniform film of \(\mathrm{TiO}_{2}, 1036 \mathrm{nm}\) thick and having index of refraction \(2.62,\) is spread uniformly over the surface of crown glass of refractive index \(1.52 .\) Light of wavelength 520.0 \(\mathrm{nm}\) falls at normal incidence onto the film from air. You want to increase the thickness of this film so that the reflected light cancels. (a) What is the minimum thickness of \(\mathrm{TiO}_{2}\) that you must add so the reflected light cancels as desired? (b) After you make the adjustment in part (a), what is the path difference between the light reflected off the top of the film and the light that cancels it after traveling through the film? Express your answer in (i) nanometers and (ii) wave-lengths of the light in the TiO_ 2 film.

35.22. GPSTransmission. The GPS (Global Positioning System) satellites are approximately 5.18 \(\mathrm{m}\) across and transmit two low-power signals, one of which is at 1575.42 \(\mathrm{MHz}\) (in the UHF band). In a series of laboratory tests on the satellite, you put two 1575.42-MHz UHF transmitters at opposite ends of the satellite. These broadcast in phase uniformly in all directions. You measure the intensity at points on a circle that is several hundred meters in radius and centered on the satellite. You measure angles on this circle relative to a point that lies along the centerline of the satellite (that is, the perpendicular bisector of a line which extends from one transmitter to the other). At this point on the circle, the measured intensity is \(2.00 \mathrm{W} / \mathrm{m}^{2} .\) (a) At how many other angles in the range \(0^{\circ}<\theta<90^{\circ}\) is the intensity also 2.00 \(\mathrm{W} / \mathrm{m}^{2} ?\) (b) Find the four smallest angles in the range \(0^{\circ}<\theta<90^{\circ}\) for which the intensity is \(2.00 \mathrm{W} / \mathrm{m}^{2} .\) (c) What is the intensity at a point on the circle at an angle of \(4.65^{\circ}\) from the centerline?

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