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35.21. Coherent light with wavelength 500 \(\mathrm{nm}\) passes through narrow slits separated by 0.340 \(\mathrm{mm}\) . At a distance from the slits large compared to their separation, what is the phase difference (in radians) in the light from the two slits at an angle of \(23.0^{\circ}\) from the centerline?

Short Answer

Expert verified
The phase difference is approximately 4.99 radians.

Step by step solution

01

Identify Key Variables

We'll first identify the key variables in the problem. The wavelength \( \lambda \) of the light is 500 nm (or \( 500 \times 10^{-9} \) m), and the slit separation \( d \) is 0.340 mm (or \( 0.340 \times 10^{-3} \) m). The angle \( \theta \) is given as 23.0 degrees.
02

Convert Angle to Radians

Convert the angle \( \theta \) from degrees to radians, as the phase difference needs to be in radians. Use the formula:\[\theta_{\text{radians}} = \theta_{\text{degrees}} \times \frac{\pi}{180}\]Substitute \( \theta = 23.0^{\circ} \):\[\theta_{\text{radians}} = 23.0 \times \frac{\pi}{180} \approx 0.401\, \text{radians}.\]
03

Calculate the Path Difference

Using the formula for path difference \( \Delta x \), which is the additional distance light from one slit travels compared to the other, calculated as:\[\Delta x = d \sin \theta\]Substitute in the known values:\[\Delta x = 0.340 \times 10^{-3} \times \sin(0.401) \approx 0.340 \times 10^{-3} \times 0.3907 \approx 1.329 \times 10^{-4} \, \text{m}.\]
04

Calculate the Phase Difference

Calculate the phase difference \( \Delta \phi \) using the formula:\[\Delta \phi = \frac{2\pi}{\lambda} \Delta x\]Substitute \( \lambda = 500 \times 10^{-9} \) m and \( \Delta x = 1.329 \times 10^{-4} \, \text{m} \):\[\Delta \phi = \frac{2\pi}{500 \times 10^{-9}} \times 1.329 \times 10^{-4} \approx 1670.53 \, \text{radians}.\]
05

Simplify the Phase Difference

Since phase difference in a wave is generally considered modulo \( 2\pi \), find the remainder of \( 1670.53 \, \text{radians} \) divided by \( 2\pi \) to express it within standard limits:\[\Delta \phi \mod 2\pi \approx 4.99 \, \text{radians}.\]

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Key Concepts

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

Interference
Interference is a fascinating phenomenon that occurs when waves, such as light , overlap each other. When waves meet, they can combine to create new wave patterns through constructive or destructive interference. Constructive interference happens when the waves are in phase, meaning their peaks align. This creates a wave with greater amplitude—a bright spot in terms of light. On the other hand, destructive interference occurs when the waves are out of phase, with peaks aligning with troughs. This results in reduced amplitude or even complete cancellation of the wave, appearing as a dark spot for light waves.

For light waves passing through two slits, the interference pattern is a series of alternating bright and dark fringes. This pattern forms because light from each slit travels slightly different paths, leading to different phases which can either construct or destruct when they meet. This principle is the foundation of many concepts in optics, including understanding light diffraction and calculating phase differences.
Diffraction
Diffraction is the bending of waves around obstacles or when they pass through narrow openings. This bending results in a spreading of the wave, particularly noticeable when the opening or obstacle is comparable in size to the wave's wavelength. When light passes through a small slit or a series of slits, it spreads out, creating a diffraction pattern. This pattern typically has a central bright fringe with successive dark and bright fringes to either side.

In the context of the exercise, when light passes through narrow slits, diffraction causes it to spread and merge, forming an interference pattern observed as bands of light and dark. This phenomenon highlights the wave nature of light and is why diffraction and interference are often studied together in optics. Understanding diffraction is crucial to interpreting and analyzing wave behaviors in various applications like optics and acoustics.
Wavelength
Wavelength is a fundamental property of a wave, defined as the distance between consecutive points such as peaks on a wave. For light, this distance is typically measured in nanometers (nm). In optics, the wavelength determines the color of visible light and plays a critical role in phenomena like interference and diffraction.

For a laser or coherent light source, each stream of light maintains a consistent wavelength. In the problem, the wavelength is given as 500 nm. This characteristic is essential for calculating path differences and phase shifts, especially when light interacts with narrow slits or other waves. Having a clear understanding of wavelength helps us analyze patterns formed due to interference and diffraction, as the wavelength will affect the spacing and intensity of the fringes observed.
Phase Difference
Phase difference in wave optics refers to the difference in phase between two points on a wave or between two waves. It is a measure of how "in step" or "out of step" two waves are. This concept is critical in the study of wave interference, as phase difference will determine whether two waves interfere constructively or destructively.

The phase difference can be calculated using the path difference, which is the physical distance one wave has traveled compared to another. In the exercise, the path difference was calculated using the slit separation, the angle of light, and trigonometric functions. By knowing the wavelength, the phase difference (\(\Delta\phi\)in radians) can be determined, telling us exactly how the waves will interfere when they meet. Understanding phase difference helps in accurately predicting the resulting wave patterns of overlapping waves, making it an essential concept in optical and wave physics.

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Most popular questions from this chapter

35.34. Light with wavelength 648 \(\mathrm{nm}\) in air is incident perpendicularly from air on a film 8.76\(\mu \mathrm{m}\) thick and with refractive index \(1.35 .\) Part of the light is reflected from the first surface of the film, and part enters the film and is reflected back at the second surface, where the film is again in contact with air. (a) How many waves are contained along the path of this second part of the light in its round trip through the film? (b) What is the phase difference between these two parts of the light as they leave the film?

35.39. The radius of curvature of the convex surface of a planoconvex lens is 95.2 \(\mathrm{cm}\) . The lens is placed convex side down on a perfectly flat glass plate that is illuminated from above with red light having a wavelength of 580 \(\mathrm{nm}\) . Find the diameter of the second bright ring in the interference pattern.

35.11. Coherent light from a sodium-vapor lamp is passed through a filter that blocks everything except light of a single wavelength. It then falls on two slits separated by 0.460 \(\mathrm{mm}\) . In the resulting interference pattern on a screen 2.20 \(\mathrm{m}\) away, adjacent bright fringes are separated by 2.82 \(\mathrm{mm}\) . What is the wavelength?

35.54. White light reflects at normal incidence from the top and bottom surfaces of a glass plate \((n=1.52) .\) There is air above and below the plate. Constructive interference is observed for light whose wavelength in air is 477.0 \(\mathrm{nm}\) . What is the thickness of the plate if the next longer wavelength for which there is constructive interference is 540.6 \(\mathrm{nm}\) ?

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?

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