lecture33 - thin films and two slit

lecture33 - thin films and two slit - What you will be...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: What you will be able to do a/er today … Explain … Calculate … •  the path difference for light interference in a 2 ­slit experiment •  what are dark and bright ‘fringes’ •  the y ­posiAon of interference fringes Apply … •  knowledge of interference from sound to light •  the concept of path difference Surveys •  UBC teaching evaluaAon (email sent out) •  CLASS survey on VISTA worth 0.5% BONUS marks (by DEC 10) Surveys: 101 survey I depend on feedback from you. •  worksheets •  clickers •  pre ­reading •  general improvement as an incenAve: STARBUCKS gi/ cards $10 EACH! Lo\ery on the last day of class $20 in the pot now  for every 10 students who fill in the survey another $5 goes into the pot $10 Interference from a Soap Film •  Ray 1 undergoes a phase change of 180° with respect to the incident ray •  Ray 2, which is reflected from the lower surface, undergoes no phase change with respect to the incident wave 1 2 Air  initial phase difference  path difference Soap Film n = 1.35 t Air 3 4 The initial phase difference between the two reflected rays, 1 and 2, is: incident A.  zero radians B.  π/2 radians C.  π radians D.  3π/2 radians E.  2π radians 1 2 n=1 n=2 n = 1.5 Worksheet A thin film with n = 2 is located on material with a refractive index of n = 1.5. When viewed straight on (perpendicular), the film appears blue (λ=512nm) and has no red (λ= 640nm) color. What is the minimum thickness of the film? Ans. 320 nm incident 1 2 n=1 n=2 n = 1.5 ‘thin film’ applied to an air gap air gap increases in width from center to edge  changing path difference At the center, where the path difference is 0, there is only a difference in phase At a given radius, there will be either destructive or constructive interference Young s Double Slit Experiment, Setup         The light passing through two slits form a visible pattern on a screen The pattern consists of a series of bright and dark bands called fringes Constructive interference occurs where a bright fringe occurs Destructive interference results in a dark fringe S1 S2 d L equally spaced (path difference change of λ) Interference Patterns •  Constructive interference occurs at the center point •  The two waves travel the same distance therefore, they arrive in phase Interference Patterns •  At R, The upper wave travels one-half of a wavelength farther than the lower wave. The trough of the bottom wave overlaps the crest of the upper wave. This is destructive interference A dark fringe occurs   At Q, The upper wave travels one wavelength farther than the lower wave. Therefore, the waves arrive in phase. A bright fringe occurs Young’s double slit interference !x = sin ! ! ! d ! ! some nm (10"9 ) d ! some mm (10"3 ) L ! some meters (1) y = tan ! ! ! L can use small angle approximaAon Δx Δx = d sinθ Δx = r2 − r1 = d sin(θ ) This assumes that the paths are parallel   Not exactly, but a very good approximation (L >> d)   The positions of a point on the screen can be given by: (1) the angle that the point makes with respect to the central axis (2) or the vertical distance of the point from the central axis y = L tan(θ ) ! ! some nm (10"9 ) d ! some mm (10"3 ) L ! some meters (1) can use small angle approximaAon !x = sin ! ! ! d y = tan ! ! ! L bright fringes start with m = 0 then count dark fringes start with m = 0 third bright fringe m = 2 second bright fringe m = 1 first bright fringe central bright fringe (central bright fringe) m = 0 m = 0 m = 1 m = 2 ybright = m λL d m = 0, ± 1, ± 2 K 1 ȹ λ L ȹ ydark = ȹ m + ȹ m = 0, ± 1, ± 2 K 2 Ⱥ d ȹ Interference Equations: final •  For bright fringes sources in phase: constructive Δx = mλ ybright = m λL d m = 0, ± 1, ± 2 K •  For dark fringes ydark 1 ȹ λ L ȹ = ȹ m + ȹ m = 0, ± 1, ± 2 K 2 Ⱥ d ȹ sources in phase: destructive Δx = (m+ ½)λ Examples •  Red light (λ=664 nm) is used in Young s experiment according to the drawing. Find the distance y on the screen between the central bright and the third-order bright fringe. Ans y = 0.046 m In the previous example, what is the total phase difference between the waves from the two sources at the location of the third (m = 3) bright fringe? A.  zero radians B.  π radians C.  2π radians D.  4π radians E.  6π radians one wavelength is 2π, so m = 3 has three wavelengths, or 6π, separation from the central fringe worksheet A gas mixture is used as a light source in a double slit experiment. The source emits light at 550 nm and 400 nm. (a) What is the lowest order 550 nm bright fringe that will fall on a 400 nm dark fringe? Ans. m = 4 (b) What is the order for the 400 nm dark fringe? Is it Ans. m = 5, m values are NOT the same the same? worksheet !"#$%&'() Young’s double slit experiment is*( performed with a Nd-YAG +#,-./0( measured carefully on a laser (λ= 1064 nm). Fringes are1#,$%&(0%23(&45&"2'&-3(20(5&"6#"'&1(,-1&"(783&"(7239(%2.93(6"#' @ABC from the double slit. You insert screen some distance L away (-'DE(F"2-.&0(8"&('&80,"&1(G8"&6,%%H(#-(8(0G"&&-(@E)A('(878H(6"#' J"1(18"K(6"2-.&(20(6#,-1(3#($&(J(''(6"#'(39&(G&-3&"(#6(39&(G&-3"8%($"2.93 a small piece of glass (n=1.5) into the lower slit. What 0'8%%(52&G&(#6(.%800(7 is the 9#7(',G9(1#&0(39&(G&-3"8%($"2.93(6"2-.&('#M&(26(8( minimum thickness of the glass;B(needed2-(39&(%#7&"(0%23(>80(09#7-(2-(39&(62.,"&DN(( @EB(4(@A '(20(2-0&"3&1( to turn the central +#,(G8-(800,'&(8-(2-1&4(#6("&6"8G32#-(#6(-?OEBB(6#"(39&(52&G&(#6(.%800(8bright fringe to a dark fringe? (Remember: we’re dealing a phase change?) 6#"(783&"E( >!2GK(39&(G%#0&03(8-07&"D( with TRANSMITED light; (((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((( is there 19 worksheet Light traveling in different medium will changes its wavelength (and its speed, but NOT its frequency). So a different number of wavelengths will fit in the glass as compared to the air. We want the fit some (m+1/2)λ in the film to cause destructive interference = 1064 nm QUESTION In a two ­slit interference experiment, the slits are 0.200 mm apart, and the screen is at a distance of 1.00 m. The third bright fringe (not counAng the central bright fringe straight ahead from the slits) is found to be displaced 9.49 mm from the central fringe. Find the wavelength of the light used. (Pick closest answer) (A)  375nm (B)  486nm (C)  512nm (D)  632nm (E)  750nm In a two ­slit interference experiment, the slits are 0.200 mm apart, and the screen is at a distance of 1.00 m. The third bright fringe (not counAng the central bright fringe straight ahead from the slits) is found to be displaced 9.49 mm from the central fringe. Find the wavelength of the light used. Path difference is: , and . Thus, (A)  375nm (B)  486nm (C)  512nm (D)  632nm (E)  750nm Y= ...
View Full Document

Ask a homework question - tutors are online