24.1 Waves Versus Particles; Huygens’ Principle and Diffraction Huygens’ principle: Every point on a wave front acts as a point source; the wavefront as it develops is tangent to their envelope
24.1 Waves Versus Particles; Huygens’ Principle and Diffraction Huygens’ principle is consistent with diffraction:
24.3 Interference – Young’s Double-Slit Experiment If light is a wave, there should be an interference pattern.
Interference Two coherent sources producing waves of the same frequency and amplitude produce an interference pattern. Sum of 2 waves in phase
Constructive Interference: waves add up constructively Sum of 2 waves in phase
Destructive Interference: waves add up destructively Interference Sum of 2 waves out phase
24.3 Interference – Young’s Double-Slit Experiment Between the maxima and the minima, the interference varies smoothly.
24.3 Interference – Young’s Double-Slit Experiment The interference occurs because each point on the screen is not the same distance from both slits. Depending on the path length difference, the wave can interfere constructively (bright spot) or destructively (dark spot).
24.3 Interference – Young’s Double-Slit Experiment We can use geometry to find the conditions for constructive and destructive interference:
Example: Monochromatic light falling on two slits 0.042 mm apart produces the fifth-order fringe at a 7.8° angle. What is the wavelength of the light used? For constructive interference, the path difference is a multiple of the wavelength: d sin = m , m = 0, 1, 2, 3, …. For the fifth order, we have (4.2 x10 –5 m) sin 7.8° = (5), = 1.1 x10 –6 m = 1.1 µm.
Example:. 3. (I) The third-order fringe of 650 nm light is observed at an angle of 15° when the light falls on two narrow slits. How far apart are the slits? For constructive interference, the path difference is a multiple of the wavelength: d sin = m, m = 0, 1, 2, 3, …. For the third order, we have d sin 15° = (3)(650 x10 –9 m), d = 7.5 x10 –6 m = 7.5 µm.
4.(II) A parallel beam of light from a He-Ne laser, with a wavelength 656 nm, falls on two very narrow slits 0.060 mm apart. How far apart are the fringes in the center of the pattern on a screen 3.6 m away? 4.For constructive interference, the path difference is a multiple of the wavelength: We find the location on the screen from For small angles, we have For adjacent fringes, so we have
24.3 Interference – Young’s Double-Slit Experiment Since the position of the maxima (except the central one) depends on wavelength, the first- and higher-order fringes contain a spectrum of colors.
24.4 The Visible Spectrum and Dispersion Wavelengths of visible light: 400 nm to 750 nm Shorter wavelengths are ultraviolet; longer are infrared
24.4 The Visible Spectrum and Dispersion The index of refraction of a material varies somewhat with the wavelength of the light.
24.4 The Visible Spectrum and Dispersion This variation in refractive index is why a prism will split visible light into a rainbow of colors.
24.4 The Visible Spectrum and Dispersion Actual rainbows are created by dispersion in tiny drops of water.
Single slit diffraction D sin = m m = 1,2,3,... destructive interference
24.5 Diffraction by a Single Slit or Disk The resulting pattern of light and dark stripes is called a diffraction pattern. This pattern arises because different points along a slit create wavelets that interfere with each other just as a double slit would.
24.5 Diffraction by a Single Slit or Disk The minima of the single-slit diffraction pattern occur when (24-3b)
Single slit diffraction Example: Monochromatic light falls on a slit that is 3.00 X 10 -3 mm wide. If the angle between the first dark fringes on either side of the central maximum is 37.0°, what is the wavelength of the light used? The angle from the central maximum to the first minimum is 18.5°. We find the wavelength from D sin = m ; (3.00 x 10 –6 m) sin (18.5°) = (1), = 9.52 x10 –7 m = 952 nm.
20.(II) A single slit 1.0 mm wide is illuminated by 450-nm light. What is the width of the central maximum (in cm) in the diffraction pattern on a screen 5.0 m away? We find the angle to the first minimum from We find the distance on the screen from For small angles, we have Thus the width of the central maximum is
24.6 Diffraction Grating A diffraction grating consists of a large number of equally spaced narrow slits or lines. The more lines or slits there are, the narrower the peaks.
24.6 Diffraction Grating The maxima of the diffraction pattern are defined by (24-4)
Example: At what angle will 650-nm light produce a second order maximum when falling on a grating whose slits are 1.15 X 10 -3 cm apart? We find the angle for the second order from d sin = m ; (1.15 x10 –5 m) sin = (2)(650 x 10 –9 m), sin = 0.113, so = 6.49°.
Example: A 3500-line/cm grating produces a third-order fringe at a 22.0° angle. What wavelength of light is being used? We find the wavelength from d sin = m ; [1/(3500 lines/cm)](10 –2 m/cm) sin 22.0° = 3, = 3.57 x10 –7 m = 357 nm.
24.7 The Spectrometer and Spectroscopy A spectrometer makes accurate measurements of wavelengths using a diffraction grating or prism.
24.7 The Spectrometer and Spectroscopy The wavelength can be determined to high accuracy by measuring the angle at which the light is diffracted. Atoms and molecules can be identified when they are in a thin gas through their characteristic emission lines.
24.8 Interference by Thin Films Another way path lengths can differ, and waves interfere, is if the travel through different media. If there is a very thin film of material – a few wavelengths thick – light will reflect from both the bottom and the top of the layer, causing interference. This can be seen in soap bubbles and oil slicks, for example.
24.8 Interference by Thin Films The wavelength of the light will be different in the oil and the air, and the reflections at points A and B may or may not involve reflection.
24.8 Interference by Thin Films A similar effect takes place when a shallowly curved piece of glass is placed on a flat one. When viewed from above, concentric circles appear that are called Newton’s rings.