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**Diffraction and Polarization**

Chapter 35 opener. Parallel coherent light from a laser, which acts as nearly a point source, illuminates these shears. Instead of a clean shadow, there is a dramatic diffraction pattern, which is a strong confirmation of the wave theory of light. Diffraction patterns are washed out when typical extended sources of light are used, and hence are not seen, although a careful examination of shadows will reveal fuzziness. We will examine diffraction by a single slit, and how it affects the double-slit pattern. We also discuss diffraction gratings and diffraction of X-rays by crystals. We will see how diffraction affects the resolution of optical instruments, and that the ultimate resolution can never be greater than the wavelength of the radiation used. Finally we study the polarization of light.

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**Diffraction by a Single Slit or Disk**

Intensity in Single-Slit Diffraction Pattern Diffraction in the Double-Slit Experiment Limits of Resolution; Circular Apertures Resolution of Telescopes and Microscopes; the λ Limit Resolution of the Human Eye and Useful Magnification Diffraction Grating

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**The Spectrometer and Spectroscopy**

Peak Widths and Resolving Power for a Diffraction Grating X-Rays and X-Ray Diffraction Polarization Liquid Crystal Displays (LCD) Scattering of Light by the Atmosphere

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**Diffraction by a Single Slit or Disk**

If light is a wave, it will diffract around a single slit or obstacle. Figure If light is a wave, a bright spot will appear at the center of the shadow of a solid disk illuminated by a point source of monochromatic light.

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**Diffraction by a Single Slit or Disk**

The resulting pattern of light and dark stripes is called a diffraction pattern. Figure Diffraction pattern of (a) a circular disk (a coin), (b) razor, (c) a single slit, each illuminated by a coherent point source of monochromatic light, such as a laser.

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**Diffraction by a Single Slit or Disk**

This pattern arises because different points along a slit create wavelets that interfere with each other just as a double slit would. Figure Analysis of diffraction pattern formed by light passing through a narrow slit of width D.

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**Diffraction by a Single Slit or Disk**

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**Diffraction by a Single Slit or Disk**

The minima of the single-slit diffraction pattern occur when Figure Intensity in the diffraction pattern of a single slit as a function of sin θ. Note that the central maximum is not only much higher than the maxima to each side, but it is also twice as wide (2λ/D wide) as any of the others (only λ/D wide each).

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**Diffraction by a Single Slit or Disk**

Single-slit diffraction maximum. Light of wavelength 750 nm passes through a slit 1.0 x 10-3 mm wide. How wide is the central maximum (a) in degrees, and (b) in centimeters, on a screen 20 cm away? Solution: a. The first minimum occurs at sin θ = λ/D = 0.75, or θ = 49°. The full width is twice this, or 98°. b. The width is 46 cm.

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**Diffraction by a Single Slit or Disk**

Diffraction spreads. Light shines through a rectangular hole that is narrower in the vertical direction than the horizontal. (a) Would you expect the diffraction pattern to be more spread out in the vertical direction or in the horizontal direction? (b) Should a rectangular loudspeaker horn at a stadium be high and narrow, or wide and flat? Solution: a. The pattern will be more spread out in the vertical direction, as the slit is narrower there. b. A pattern that is wider than it is high is desired, so the speaker should be high and narrow. A B

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**Intensity in Single-Slit Diffraction Pattern**

Light passing through a single slit can be divided into a series of narrower strips; each contributes the same amplitude to the total intensity on the screen, but the phases differ due to the differing path lengths: Figure Slit of width D divided into N strips of width Δy. .

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**Slit of width D divided into N strips of width Δy**

Slit of width D divided into N strips of width Δy. Each strip is a wave with intensity of I0/N. Path difference between two adjacent strips is and the corresponding phase angle difference is The intensity of the diffraction is, by superposition, the vector sum of the N strips of light with N approaching infinity.

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**Intensity in Single-Slit Diffraction Pattern**

Phasor diagrams give us the intensity as a function of angle. Figure Phasor diagram for single-slit diffraction, giving the total amplitude Eθ at four different angles θ.

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**Intensity in Single-Slit Diffraction Pattern**

Taking the limit as the width becomes infinitesimally small gives the field as a function of angle: Figure Determining amplitude Eθ as a function of θ for single-slit diffraction.

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**Intensity in Single-Slit Diffraction Pattern**

Finally, we have the phase difference and the intensity as a function of angle: and .

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**Intensity in Single-Slit Diffraction Pattern**

Intensity at secondary maxima. Estimate the intensities of the first two secondary maxima to either side of the central maximum. Solution: The secondary maxima occur approximately halfway between the minima. Equation 35-7 then gives, for m = 1, I = I0, and for m = 2, I = I0.

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**Diffraction in the Double-Slit Experiment**

The double-slit experiment also exhibits diffraction effects, as the slits have a finite width. This means the amplitude at an angle θ will be modified by the same factor as in the single-slit experiment: The intensity is, as usual, proportional to the square of the field.

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**Diffraction in the Double-Slit Experiment**

The diffraction factor (depends on β) appears as an “envelope” modifying the more rapidly varying interference factor (depends on δ). Figure (a) Diffraction factor, (b) interference factor, and (c) the resultant intensity plotted as a function of θ for d = 6D = 60λ.

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**Diffraction in the Double-Slit Experiment**

Diffraction plus interference. Show why the central diffraction peak shown, plotted for the case where d = 6D = 60λ, contains 11 interference fringes. Solution: The first minimum in the diffraction pattern occurs where sin θ = λ/D. Since d = 6D, d sin θ = 6λ. This means that the m=6 interference fringes will not appear; the central diffraction peak contains the fringes for m = 0 (one) and m = 1 through 5 (two each for a total of 10), making 11 fringes.

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**Limits of Resolution; Circular Apertures**

Resolution is the distance at which a lens can barely distinguish two separate objects. Resolution is limited by aberrations and by diffraction. Aberrations can be minimized, but diffraction is unavoidable; it is due to the size of the lens compared to the wavelength of the light.

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**Limits of Resolution; Circular Apertures**

For a circular aperture of diameter D, the central maximum has an angular width: Figure Intensity of light across the diffraction pattern of a circular hole.

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**Limits of Resolution; Circular Apertures**

The Rayleigh criterion states that two images are just resolvable when the center of one peak is over the first minimum of the other. Figure The Rayleigh criterion. Two images are just resolvable when the center of the diffraction peak of one is directly over the first minimum in the diffraction pattern of the other. The two point objects O and O’ subtend an angle θ at the lens; only one ray (it passes through the center of the lens) is drawn for each object, to indicate the center of the diffraction pattern of its image.

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**Limits of Resolution; Circular Apertures**

Hubble Space Telescope. The Hubble Space Telescope (HST) is a reflecting telescope that was placed in orbit above the Earth’s atmosphere, so its resolution would not be limited by turbulence in the atmosphere. Its objective diameter is 2.4 m. For visible light, say λ = 550 nm, estimate the improvement in resolution the Hubble offers over Earth-bound telescopes, which are limited in resolution by movement of the Earth’s atmosphere to about half an arc second. (Each degree is divided into 60 minutes each containing 60 seconds, so 1° = 3600 arc seconds.) Solution: The Hubble resolution is diffraction-limited; for λ = 550 nm, θ = 2.8 x 10-7 rad. Converting ½ arc second to radians gives 2.4 x 10-6 rad, so the Hubble’s resolution is approximately 10 times better.

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**Limits of Resolution; Circular Apertures**

Eye resolution. You are in an airplane at an altitude of 10,000 m. If you look down at the ground, estimate the minimum separation s between objects that you could distinguish. Could you count cars in a parking lot? Consider only diffraction, and assume your pupil is about 3.0 mm in diameter and λ = 550 nm. Solution: The separation will be hθ (h is the height of the airplane) = 2.2 m. Cars are slightly larger than this, on average, so you could count them. Barely.

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**Resolution of Telescopes**

Telescope resolution (radio wave vs. visible light). What is the theoretical minimum angular separation of two stars that can just be resolved by (a) the 200-inch telescope on Palomar Mountain; and (b) the Arecibo radio telescope, whose diameter is 300 m and whose radius of curvature is also 300 m. Assume λ = 550 nm for the visible-light telescope in part (a), and λ = 4 cm (the shortest wavelength at which the radio telescope has been operated) in part (b). Solution: a. θ = 1.3 x 10-7 rad (although the Palomar telescope is limited by atmospheric effects and not by diffraction) b. θ = 1.6 x 10-4 rad

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Diffraction Grating A diffraction grating consists of a large number of equally spaced narrow slits or lines. A transmission grating has slits, while a reflection grating has lines that reflect light. The more lines or slits there are, the narrower the peaks. Figure Intensity as a function of viewing angle θ (or position on the screen) for (a) two slits, (b) six slits. For a diffraction grating, the number of slits is very large (≈104) and the peaks are narrower still.

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Diffraction Grating Figure Spectra produced by a grating: (a) two wavelengths, 400 nm and 700 nm; (b) white light. The second order will normally be dimmer than the first order. (Higher orders are not shown.) If grating spacing is small enough, the second and higher orders will be missing.

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Diffraction Grating The maxima of the diffraction pattern are defined by d is the grating width between slits. m denotes the principal maxima.

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**Diffraction Grating Diffraction grating: lines.**

Determine the angular positions of the first- and second-order maxima for light of wavelength 400 nm and 700 nm incident on a grating containing 10,000 lines/cm. Solution: The distance between the lines is 1.0 μm. For m = 1, the angles are 23.6° and 44.4°. For m = 2, the angle for 400 nm is 53.1°; the equation for sin θ gives a result greater than 1 for 700 nm, so the second order will not appear.

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**Diffraction Grating Spectra overlap.**

White light containing wavelengths from 400 nm to 750 nm strikes a grating containing 4000 lines/cm. Show that the blue at λ = 450 nm of the third-order spectrum overlaps the red at 700 nm of the second order. Solution: The third-order blue maximum is at sin θ = 0.540; the second-order red maximum is at sin θ = (a greater angle), so the spectra will overlap.

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**Diffraction Grating Compact disk.**

When you look at the surface of a music CD, you see the colors of a rainbow. (a) Estimate the distance between the curved lines (to be read by the laser). (b) Estimate the distance between lines, noting that a CD contains at most 80 min of music, that it rotates at speeds from 200 to 500 rev/min, and that 2/3 of its 6-cm radius contains the lines. Solution: a. The CD acts as a reflection diffraction grating. In order to see rainbow colors, the distance between the lines must be one or a few optical wavelengths, or 0.5 – 1.5 μm. b. If the CD rotates at an average rate of 350 rev/min, and plays for 80 min, it must contain about 28,000 lines. If these are contained within 4 cm, the spacing between them must be about 1.4 μm, in agreement with the estimate in part (a).

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**The Spectrometer and Spectroscopy**

A spectrometer makes accurate measurements of wavelengths using a diffraction grating or prism. Figure Spectrometer or spectroscope.

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**The Spectrometer and Spectroscopy**

The wavelength can be determined to high accuracy by measuring the angle at which the light is diffracted:

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**The Spectrometer and Spectroscopy**

Atoms and molecules can be identified when they are in a thin gas through their characteristic emission lines. Figure Line spectra for the gases indicated, and the spectrum from the Sun showing absorption lines

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**The Spectrometer and Spectroscopy**

Hydrogen spectrum. Light emitted by hot hydrogen gas is observed with a spectroscope using a diffraction grating having 1.00 x 104 lines/cm. The spectral lines nearest to the center (0°) are a violet line at 24.2°, a blue line at 25.7°, a blue-green line at 29.1°, and a red line at 41.0° from the center. What are the wavelengths of these spectral lines of hydrogen? Since these are the lines nearest the center, m = 1. Then the wavelengths are 410 nm, 434 nm, 486 nm, and 656 nm.

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**Peak Widths and Resolving Power**

These two sets of diagrams show the phasor relationships at the central maximum and at the first minimum for gratings of two and six slits. Figure Phasor diagram for two slits (a) at the central maximum, (b) at the nearest minimum. Figure Phasor diagram for six slits (a) at the central maximum, (b) at the nearest minimum.

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**Peak Widths and Resolving Power**

As the number of slits becomes large, the width of the central maximum becomes very narrow: The resolving power of a diffraction grating is the minimum difference between wavelengths that can be distinguished:

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**Peak Widths and Resolving Power**

Resolving two close lines. Yellow sodium light, which consists of two wavelengths, λ1 = nm and λ2 = nm, falls on a diffraction grating. Determine (a) the maximum order m that will be present for sodium light, and (b) the width of grating necessary to resolve the two sodium lines. Solution: a. Using equation and requiring sin θ ≤ 1 shows that the maximum order is m = 2. b. The resolving power needed is Δλ/λ = 1000.

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**X-Rays and X-Ray Diffraction**

The wavelengths of X-rays are very short. Diffraction experiments are impossible to do with conventional diffraction gratings. Crystals have spacing between their layers that is ideal for diffracting X-rays. Figure X-ray diffraction by a crystal.

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**X-Rays and X-Ray Diffraction**

X-ray diffraction is now used to study the internal structure of crystals; this is how the helical structure of DNA was determined. Figure X-rays can be diffracted from many possible planes within a crystal.

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Polarization Light is polarized when its electric fields oscillate in a single plane, rather than in any direction perpendicular to the direction of propagation. Figure Transverse waves on a rope polarized (a) in a vertical plane and (b) in a horizontal plane.

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Polarization Polarized light will not be transmitted through a polarized film whose axis is perpendicular to the polarization direction. Figure (a) Vertically polarized wave passes through a vertical slit, but a horizontally polarized wave will not (b).

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Polarization When light passes through a polarizer, only the component parallel to the polarization axis is transmitted. If the incoming light is plane-polarized, the outgoing intensity is: Figure Vertical Polaroid transmits only the vertical component of a wave (electric field) incident upon it.

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Polarization This means that if initially unpolarized light passes through crossed polarizers, no light will get through the second one. Figure Crossed Polaroids completely eliminate light. I0 I=?

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**Polarization Two Polaroids at 60°.**

Unpolarized light passes through two Polaroids; the axis of one is vertical and that of the other is at 60° to the vertical. Describe the orientation and intensity of the transmitted light. Solution: The first Polaroid reduces the intensity of the unpolarized light by a factor of two. The second Polaroid reduces the intensity by another factor of cos2 θ, giving an overall final intensity of 1/8 of the original intensity. The light will have the polarization of the second Polaroid, 60° to the vertical.

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**Polarization Three Polaroids.**

When unpolarized light falls on two crossed Polaroids (axes at 90°), no light passes through. What happens if a third Polaroid, with axis at 45° to each of the other two, is placed between them? Solution: The first polarizer reduces the initial intensity by a factor of 2. The second reduces it by a factor of (cos 45°)2, or another factor of 2. Finally, the third polarizer reduces the intensity by yet another factor of 2, for an overall reduction of a factor of 8. However, without the central polarizer, the transmitted intensity would be zero.

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Polarization

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Polarization Light is also partially polarized after reflecting from a nonmetallic surface. At a special angle, called the polarizing angle or Brewster’s angle, the polarization is 100%: . Figure Light reflected from a nonmetallic surface, such as the smooth surface of water in a lake, is partially polarized parallel to the surface.

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Polarization

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**Polarization Polarizing angle.**

(a) At what incident angle is sunlight reflected from a lake plane-polarized? (b) What is the refraction angle? Solution: a. Using n = 1.33 for water and n = 1 for air, Brewster’s angle is 53.1°. b. From Snell’s law, the angle of refraction is 36.9°. The sum of these two angles is 90°, as expected.

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**Scattering of Light by the Atmosphere**

Skylight is partially polarized due to scattering from molecules in the air. The amount of polarization depends on the angle that your line of sight makes with the Sun. Figure Unpolarized sunlight scattered by molecules of the air. An observer at right angles sees plane-polarized light, since the component of oscillation along the line of sight emits no light along that line.

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Summary Light bends around obstacles and openings in its path, yielding diffraction patterns. Light passing through a narrow slit will produce a central bright maximum of width Minima occur at

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**Summary Diffraction limits the resolution of images.**

Diffraction grating has many parallel slits or lines; peaks of constructive interference are given by Polarized light has its electric field vectors in a single plane.

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Summary The intensity of plane-polarized light is reduced after it passes through another polarizer: Light can also be polarized by reflection; it is completely polarized when the reflection angle is the polarization angle:

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Physics 1502: Lecture 34 Today’s Agenda Announcements: –Midterm 2: graded soon … –Homework 09: Friday December 4 Optics –Interference –Diffraction »Introduction.

Physics 1502: Lecture 34 Today’s Agenda Announcements: –Midterm 2: graded soon … –Homework 09: Friday December 4 Optics –Interference –Diffraction »Introduction.

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