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OPTICS Wave optics. Interference. Diffraction. Polarization 1. Young's Double Slit Experiment Constructive interference: 8 = x; n, - x; n; Screen 8 = k) = 2k Destructive interference: 8 = (2k +1) Coordinates of bright and dark fringes on the screen: (*.+) Ymax =± Ymin =± Ay = Here d is path difference (m), 1 is wavelength of light (m), I is distance from the slits to the screen (m), k = 0, 1, 2, 3 ... is minimum or maximum order number, d is distance between the slits (m), Ymax are coordinates of bright fringes on the screen (m), ymin are coordinates of dark fringes on the screen (m), Ay is width of the fridges (m). 2. Interference in thin films: Air Constructive interference: Film 21 /n² – sin² a : (2k + 1) B Destructive interference: Air 21 Vn² – sin² a = k) Here n is refractive index of the film, a is angle of light incidence, k = 0, 1, 2, 3 ... is minimum or maximum order number, 1 is wavelength of light (m), l is thickness of the film (m). 3. Single-Slit Diffraction Constructive: sin- 2/a sine/a a sin© = ±(2k + 1) : 2 sine0 Destructive: sine/ * sine --2 a sin© = ±kl a is the slit width, k = 0, 1, 2, 3 ... is minimum or maximum order number, 1 is wavelength of light (m), 0 is the angle on corresponding minimum or maximum. 4. Diffraction grating: Viewing sereen d sin o ±kA d = /N is period of diffraction grating and N is number of lines per unit length, ø is angle on corresponding minimum or maximum, k = 0, 1, 2... is minimum or maximum order number, 1 is wavelength of light (m). 5. Resolution of the optical instruments Limiting angle of resolution 0min: 1.222 Өmin Omin = D Minimum separation distance d: d = L0min 1 is wavelength of light (m), D is aperture diameter, for human eye pupil diameter (m). 9. For the human eye, with a pupil diameter of about 2 mm and using a wavelength of 500 nm, calculate the minimum angle separating two just resolvable points. Then find the actual minimum distance between such just resolvable points as well as the distance between their images on the retina.

OPTICS Wave optics. Interference. Diffraction. Polarization 1. Young's Double Slit Experiment Constructive interference: 8 = x; n, - x; n; Screen 8 = k) = 2k Destructive interference: 8 = (2k +1) Coordinates of bright and dark fringes on the screen: (*.+) Ymax =± Ymin =± Ay = Here d is path difference (m), 1 is wavelength of light (m), I is distance from the slits to the screen (m), k = 0, 1, 2, 3 ... is minimum or maximum order number, d is distance between the slits (m), Ymax are coordinates of bright fringes on the screen (m), ymin are coordinates of dark fringes on the screen (m), Ay is width of the fridges (m). 2. Interference in thin films: Air Constructive interference: Film 21 /n² – sin² a : (2k + 1) B Destructive interference: Air 21 Vn² – sin² a = k) Here n is refractive index of the film, a is angle of light incidence, k = 0, 1, 2, 3 ... is minimum or maximum order number, 1 is wavelength of light (m), l is thickness of the film (m). 3. Single-Slit Diffraction Constructive: sin- 2/a sine/a a sin© = ±(2k + 1) : 2 sine0 Destructive: sine/ * sine --2 a sin© = ±kl a is the slit width, k = 0, 1, 2, 3 ... is minimum or maximum order number, 1 is wavelength of light (m), 0 is the angle on corresponding minimum or maximum. 4. Diffraction grating: Viewing sereen d sin o ±kA d = /N is period of diffraction grating and N is number of lines per unit length, ø is angle on corresponding minimum or maximum, k = 0, 1, 2... is minimum or maximum order number, 1 is wavelength of light (m). 5. Resolution of the optical instruments Limiting angle of resolution 0min: 1.222 Өmin Omin = D Minimum separation distance d: d = L0min 1 is wavelength of light (m), D is aperture diameter, for human eye pupil diameter (m). 9. For the human eye, with a pupil diameter of about 2 mm and using a wavelength of 500 nm, calculate the minimum angle separating two just resolvable points. Then find the actual minimum distance between such just resolvable points as well as the distance between their images on the retina.

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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OPTICS Wave optics. Interference. Diffraction. Polarization 1. Young's Double Slit Experiment Constructive interference: 8 = x; n, - x; n; Screen 8 = k) = 2k Destructive interference: 8 = (2k +1) Coordinates of bright and dark fringes on the screen: (*.+) Ymax =± Ymin =± Ay = Here d is path difference (m), 1 is wavelength of light (m), I is distance from the slits to the screen (m), k = 0, 1, 2, 3 ... is minimum or maximum order number, d is distance between the slits (m), Ymax are coordinates of bright fringes on the screen (m), ymin are coordinates of dark fringes on the screen (m), Ay is width of the fridges (m). 2. Interference in thin films: Air Constructive interference: Film 21 /n² – sin² a : (2k + 1) B Destructive interference: Air 21 Vn² – sin² a = k) Here n is refractive index of the film, a is angle of light incidence, k = 0, 1, 2, 3 ... is minimum or maximum order number, 1 is wavelength of light (m), l is thickness of the film (m). 3. Single-Slit Diffraction Constructive: sin- 2/a sine/a a sin© = ±(2k + 1) : 2 sine0 Destructive: sine/ * sine --2 a sin© = ±kl a is the slit width, k = 0, 1, 2, 3 ... is minimum or maximum order number, 1 is wavelength of light (m), 0 is the angle on corresponding minimum or maximum. 4. Diffraction grating: Viewing sereen d sin o ±kA d = /N is period of diffraction grating and N is number of lines per unit length, ø is angle on corresponding minimum or maximum, k = 0, 1, 2... is minimum or maximum order number, 1 is wavelength of light (m). 5. Resolution of the optical instruments Limiting angle of resolution 0min: 1.222 Өmin Omin = D Minimum separation distance d: d = L0min 1 is wavelength of light (m), D is aperture diameter, for human eye pupil diameter (m).
Transcribed Image Text:OPTICS Wave optics. Interference. Diffraction. Polarization 1. Young's Double Slit Experiment Constructive interference: 8 = x; n, - x; n; Screen 8 = k) = 2k Destructive interference: 8 = (2k +1) Coordinates of bright and dark fringes on the screen: (*.+) Ymax =± Ymin =± Ay = Here d is path difference (m), 1 is wavelength of light (m), I is distance from the slits to the screen (m), k = 0, 1, 2, 3 ... is minimum or maximum order number, d is distance between the slits (m), Ymax are coordinates of bright fringes on the screen (m), ymin are coordinates of dark fringes on the screen (m), Ay is width of the fridges (m). 2. Interference in thin films: Air Constructive interference: Film 21 /n² – sin² a : (2k + 1) B Destructive interference: Air 21 Vn² – sin² a = k) Here n is refractive index of the film, a is angle of light incidence, k = 0, 1, 2, 3 ... is minimum or maximum order number, 1 is wavelength of light (m), l is thickness of the film (m). 3. Single-Slit Diffraction Constructive: sin- 2/a sine/a a sin© = ±(2k + 1) : 2 sine0 Destructive: sine/ * sine --2 a sin© = ±kl a is the slit width, k = 0, 1, 2, 3 ... is minimum or maximum order number, 1 is wavelength of light (m), 0 is the angle on corresponding minimum or maximum. 4. Diffraction grating: Viewing sereen d sin o ±kA d = /N is period of diffraction grating and N is number of lines per unit length, ø is angle on corresponding minimum or maximum, k = 0, 1, 2... is minimum or maximum order number, 1 is wavelength of light (m). 5. Resolution of the optical instruments Limiting angle of resolution 0min: 1.222 Өmin Omin = D Minimum separation distance d: d = L0min 1 is wavelength of light (m), D is aperture diameter, for human eye pupil diameter (m).
9. For the human eye, with a pupil diameter of about 2 mm and using a wavelength of 500 nm, calculate the minimum angle separating two just resolvable points. Then find the actual minimum distance between such just resolvable points as well as the distance between their images on the retina.
Transcribed Image Text:9. For the human eye, with a pupil diameter of about 2 mm and using a wavelength of 500 nm, calculate the minimum angle separating two just resolvable points. Then find the actual minimum distance between such just resolvable points as well as the distance between their images on the retina.
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