Physics 24100 Electricity & Optics Lecture 27 Chapter 33 sec. 7-8 Fall 2017 Semester Professor Koltick
Clicker Question Bright light of wavelength 585 nm is incident perpendicularly on a soap film (n = 1.33) of thickness 1.21 microns, suspended in air. Is the light reflected by the two surfaces of the film closer to interfering fully destructively or fully constructively? (A) fully constructively (B) fully destructively
Interference Three ways to introduce a phase difference in two beams of coherent light: Incident light Reflected light Different path length Different index of refraction Reflection from a material with a larger n t = 1.21 μm n= 1.33 λ air = 585 nm λ soap = (585 nm)/1.33 = 440 nm Number of wavelengths in the soap film: 2t/λ soap = 5.5 In phase with the reflected light. (constructive interference) (Phase reversal since n>1)
Constructive Interference (1) The geometric optics of the previous chapter cannot be used to explain interference To understand these interference phenomena we must take into account the wave nature of light Interference takes place when light waves of the same wavelength are superimposed If the light waves are in phase, they interfere constructively, as shown to the right + = 14 Thursday, December 10, 15
Constructive Interference (2) The statement that the two waves interfere constructively is the same as saying that the phase difference between the two waves is zero A phase difference of 2π radians, 360 (one wavelength) will also produce two waves that are in phase If the light waves are traveling from some common point, then the phase difference can be related to the path difference between the two waves The criterion for constructive interference is given by a path difference Δx given by 15 Thursday, December 10, 15
Destructive Interference (1) If the two light waves are out of phase, as shown to the right, the amplitudes of the waves will sum to zero everywhere and the two waves will destructively interfere Here the phase difference between the two waves is π radians, 180, or λ/2 + = 16 Thursday, December 10, 15
Destructive Interference (2) If we think of the two light waves as being emitted from the same source, the phase difference can be related to the path difference, and we can see that destructive interference will take place if the path difference is a half wavelength plus an integer times the wavelength 17 Thursday, December 10, 15
Geometrical optics: Diffraction Edge, obstacle or aperture much larger than λ Light travels in straight lines Shadows have distinct, sharp edges Wave optics: Feature sizes are similar to the wavelength, λ Waves spread out around objects and can interfere constructively or destructively Interference patterns of light and dark fringes
Wave properties of light Diffraction through a slit. Antinodal lines can be seen. 8 Thursday, December 10, 15
Single Slit Diffraction (1) Diffraction of light through a single slit of width a that is comparable to the wavelength of light that is passing through the slit Approach: Using a Huygens construction Spherical wavelets emitted from a distribution of points located in the slit Light emitted from these points will superimpose and interfere based on the path length difference for each wavelet at each position 44 Thursday, December 10, 15
Single Slit Diffraction (2) At a distant screen we will observe an intensity pattern characteristic of diffraction Bright and dark fringes For diffraction we will only analyze the dark fringes as destructive interference 45 Thursday, December 10, 15
Single Slit Diffraction: First Dark Fringe (1) To study the diffraction let s expand and simplify our previous figure for single slit diffraction We assume coherent light with wavelength λ incident on a slit with width a that produces an interference pattern on a screen at distance L We analyze pairs of light waves emitted from points in the slit 46 Thursday, December 10, 15
Single Slit Diffraction: First Dark Fringe (2) We start with light emitted from the top edge of the slit and from the center of the slit as shown To analyze the path difference we show an expanded version of our figure to the right Here we assume that the point P on the screen is far enough away that the rays r 1 and r 2 are parallel and make an angle θ with the central axis C 47 Thursday, December 10, 15
Single Slit Diffraction: First Dark Fringe (3) Therefore the path length difference for these two rays is The criterion for the first dark fringe is Although we chose one ray originating from the top edge of the slit and one from the middle of the slit to locate the first dark fringe, we could have used any two rays that originated a/2 apart inside the slit 48 Thursday, December 10, 15
Single Slit Diffraction: Dark Fringes The dark fringes from single slit diffraction can be described by 51 Thursday, December 10, 15
Single Slit Diffraction Angle of first minimum: sin θ θ = λ a When a < λ there is no minimum and the slit emits light uniformly in all directions. The larger the aperture, the less flaring of the transmitted light.
Single Slit Intensity If the screen is placed a sufficiently large distance from the slits, the angle θ will be small and we can make the approximation We can express the position of the dark fringes as The intensity of light passing through a single slit can be calculated but we will not present the derivation here The intensity I relative to I max that we would get if there were no slit is We can see that this expression for the intensity will be zero for sin(α) = 0, which means α = mπ for m = 1, 2, 3, 52 Thursday, December 10, 15
Single Slit Intensity If the screen is placed a sufficiently large distance from the slits, the angle θ will be small and we can make the approximation We can express the position of the dark fringes as The intensity of light passing through a single slit can be calculated but we will not present the derivation here The intensity I relative to I max that we would get if there were no slit is We can see that this expression for the intensity will be zero for sin(α) = 0, which means α = mπ for m = 1, 2, 3, 52 Thursday, December 10, 15
Single Slit Intensity: Dark Fringes We can write If the screen is placed a sufficiently large distance from the slits we can make the small angle approximation and write 53 Thursday, December 10, 15
Single Slit Diffraction Waves from all points in the slit travel the same distance to reach the center and are in phase: constructive interference.
Fresnel Diffraction Two waves that touch the top and bottom parts of the disk travel the same distance to the center and interfere constructively.
Diffraction from and Edge The shadow has a diffraction pattern within a few wavelengths of the edge.
Rayleigh Criteria: Resolvability The minimum angular separation α c of two marginally resolvable points is such that the maximum of the diffraction pattern from one falls on the first minimum of the diffraction pattern of the other. First minima is at sin θ = 1.22 λ d so α c = θ 1.22 λ d
Rayleigh Criteria Resolvable
Resolvability Rayleigh Criteria: α c = 1.22 λ d The minimum resolvable angle can be decreased by Increasing d Decreasing λ (use ultraviolet light or observe it in a medium with larger n) Electrons behave like waves with very short wavelengths λ(e) λ(light)/10 5
Diffraction from a Circular Aperture The central bright spot is called Airy disk. About 85% of the power is in this area. The dark fringes are found at: sin θ 1 = 1.22 λ d sin θ 2 = 2.23 λ d sin θ 3 = 3.24 λ d d is the diameter of the circular aperture.
Diffraction from a Circular Aperture The bright fringes are found at: sin θ 1 = 1.63 λ d sin θ 2 = 2.68 λ d sin θ 3 = 3.70 λ d Image of two nearby binary stars can t be clearly resolved because their diffraction patterns overlap. The Airy disk limits the resolvability of nearby objects.
Diffrac5on and Resolving close sources Brighter is Mizar, Alcor is to the le2 The Pair can be resolved by the human eye Mizar A and Mizar B Can not be resolved by the human eye first-known double, the pairing discovered by G. B. Riccioli in 1650
G. Seurat Poin5llist Art If dots are 2-mm apart how far away most we stand to have the dots blend together? Assume the eye pupil diameter is 2.2 mm.
Diffraction Gratings (3) A portion of a diffraction grating is shown below In this drawing we see coherent light with wavelength incident on a series of narrow slits each separated by a distance d A diffraction pattern is produced on a screen a long distance L away 92 Thursday, December 10, 15
Lots of Slits Constructive interference when δ = d sin θ = mλ Lines become sharper: half-width of central line is θ = λ Nd d When N is very large (10 4 /cm) the lines are very sharp.
Diffraction Gratings Sharp, bright fringes occur when d sin θ = mλ, m = 1,2,3, Resolving power: can we detect two spectral lines separated by Δλ? d Δθ = m Δλ Half width of central line: θ = λ Nd Resolving power: R = λ = mλ = mn Δλ d Δθ
Reflection grating Note: in ray optics this would have produced a single image by tilted mirror. In wave optics we get diffraction pattern 49 Thursday, December 10, 15