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35.16. Coherent light of frequency \(6.32 \times 10^{14} \mathrm{Hz}\) passes through two thin slits and falls on a screen 85.0 \(\mathrm{cm}\) away. You observe that the third bright fringe occurs at \(\pm 3.11 \mathrm{cm}\) on either side of the central bright fringe. (a) How far apart are the two slits? (b) At what distance from the central bright fringe will the third dark fringe occur?

Short Answer

Expert verified
(a) 0.388 mm; (b) 2.59 cm from the central fringe.

Step by step solution

01

Understand the Problem

We are working with a double-slit interference pattern where coherent light of frequency \(6.32 \times 10^{14} \mathrm{Hz}\) creates bright and dark fringes on a screen 85.0 cm away. We need to find the distance between the slits and the position of the third dark fringe from the central bright fringe.
02

Calculate Wavelength

First, calculate the wavelength \(\lambda\) of the light using the speed of light \(c = 3 \times 10^{8} \mathrm{m/s}\) and the frequency \(f = 6.32 \times 10^{14} \mathrm{Hz}\).\[\lambda = \frac{c}{f} = \frac{3 \times 10^8 \mathrm{m/s}}{6.32 \times 10^{14} \mathrm{Hz}}\]Calculating this gives \(\lambda = 4.75 \times 10^{-7} \mathrm{m}\).
03

Use Bright Fringe Formula

For the third-order bright fringe (\(m = 3\)), the position on the screen is given as \(x_m = \pm 3.11 \mathrm{cm}\). Use the equation for bright fringe: \[d \cdot \sin(\theta) = m \cdot \lambda\]Convert \(x_m\) to meters: \(x_m = 0.0311 \mathrm{m}\). Using small angle approximation (\(\sin(\theta) \approx \tan(\theta) \approx \frac{x}{L}\)), substitute and solve for \(d\): \[d = \frac{m \cdot \lambda \cdot L}{x_m}\]\[d = \frac{3 \cdot 4.75 \times 10^{-7} \times 0.85}{0.0311}\]This gives \(d \approx 3.88 \times 10^{-4} \mathrm{m}\) or 0.388 mm.
04

Use Dark Fringe Formula

For the third dark fringe, use the equation for the dark fringe:\[d \cdot \sin(\theta) = (m + 0.5) \cdot \lambda\]Substitute \(m = 2.5\) for the third dark fringe (halfway between second and third bright fringes), convert fringes similarly, and solve:\[x_{2.5} = \frac{(2 + 0.5) \cdot \lambda \cdot L}{d}\]\[x_{2.5} = \frac{2.5 \cdot 4.75 \times 10^{-7} \times 0.85}{3.88 \times 10^{-4}}\]This gives \(x_{2.5} \approx 0.0259 \mathrm{m}\) or 2.59 cm.

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Key Concepts

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

Coherent Light
Coherent light is a fundamental concept in understanding many optical phenomena, such as the double-slit interference experiment. Coherent light refers to light waves that have a constant phase difference and the same frequency. This is essential for creating predictable interference patterns, like the alternating bright and dark fringes seen when coherent light passes through two slits.
The best example of coherent light is that produced by a laser. A laser emits a beam of light where the waves are in phase and have the same frequency and wavelength. This coherence is crucial for precise scientific experiments and for the creation of stable interference patterns. Without coherence, the interference pattern would be unpredictable or non-existent.
In this exercise, coherent light is used to ensure that the light waves from the two slits interfere constructively or destructively at consistent points on the screen, forming clear, predictable patterns of bright and dark fringes.
Wavelength Calculation
The wavelength of light is another critical concept for understanding interference patterns. In the given problem, the wavelength \[\lambda\]is calculated using the formula:\[\lambda = \frac{c}{f}\]where \(c\) is the speed of light \[3 \times 10^8\, \mathrm{m/s}\]and \(f\) is the frequency \[6.32 \times 10^{14}\, \mathrm{Hz}\].
Plugging these values into the formula gives a wavelength of \[4.75 \times 10^{-7}\, \mathrm{m}\]. This wavelength allows us to use the interference formulas to find the positions of the bright and dark fringes.
It's important to understand that the wavelength directly affects the spacing of these fringes on the screen. Longer wavelengths result in more spread-out fringes, while shorter wavelengths bring fringes closer together.
Bright Fringe
In a double-slit experiment, bright fringes are points of constructive interference where the light waves from each slit arrive in phase and reinforce each other. The position of these bright fringes can be calculated using the formula for bright fringes:\[d \cdot \sin(\theta) = m \cdot \lambda\]where \(d\) is the slit separation, \(\theta\) is the angle of the fringe, and \(m\) is the order of the fringe (in this case, 3 for the third bright fringe).
By using the small angle approximation, where \[\sin(\theta) \approx \tan(\theta) \approx \frac{x}{L}\], we can find the separation of the slits \(d\) as \[0.388\, \mathrm{mm}\]. This shows how the position and order of a bright fringe on the screen are directly tied to the wavelength and slit separation.
Bright fringes indicate that the light waves have traveled path lengths differing by an integer number of wavelengths, resulting in constructive interference.
Dark Fringe
Dark fringes are the regions of destructive interference in a double-slit interference pattern. These appear between the bright fringes and are points where the light waves from the slits arrive out of phase, effectively canceling each other out. In this way, they create a band of darkness on the screen.
The position of dark fringes can be determined using the formula:\[d \cdot \sin(\theta) = (m + 0.5) \cdot \lambda\]where \((m + 0.5)\) accounts for the half-wavelength path difference that results in cancellation of the light waves.
For the third dark fringe, by using values of \(m = 2.5\), \[d\], \[\lambda\], and the distance \[L\], we find that the third dark fringe occurs at roughly \[2.59\, \mathrm{cm}\] from the central bright fringe.
These positions of dark fringes provide insight into the path difference and aid in understanding wave interference in optics.

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Most popular questions from this chapter

35.29. Two rectangular pieces of plane glass are laid one upon the other on a table. A thin strip of paper is placed between them at one edge so that a very thin wedge of air is formed The plates are illuminated at normal incidence by 546 -nm light from a mercury-vapor lamp. Interference fringes are formed, with 15.0 fringes per centimeter. Find the angle of the wedge.

35.38. Jan first uses a Michelson interferometer with the 606 -nm light from a krypton-86 lamp. He displaces the movable mirror away from him, counting 818 fringes moving across a line in his field of view. Then Linda replaces the krypton lamp with filtered 502 -nm light from a helium lamp and displaces the movable mirror toward her. She also counts 818 fringes, but they move across the line in her field of view opposite to the direction they moved for Jan. Assume that both Jan and Linda counted to 818 correctly. (a) What distance did each person move the mirror? (b) What is the resultant displacement of the mirror?

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.

35.57. Two thin parallel slits are made in an opaque sheet of film. When a monochromatic beam of light is shone through them at normal incidence, the first bright fringes in the transmitted light occur in air at \(\pm 18.0^{\circ}\) with the original direction of the light beam on a distant screen when the apparatus is in air. When the apparatus is immersed in a liquid, the same bright fringes now occur at \(\pm 12.6^{\circ} .\) Find the index of refraction of the liquid.

35.19. 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 angle is \(60.0^{\circ} ?\)

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