Intensity of Interference Patterns

Transcription

1 Intensity of Interference Patterns Wish to find I on a screen far away Let consider the E fields coming from the double slits: S 2 S 1 r r 2 1 E field from S 2 has a phase lag due to the extra path difference, r 2 - r 1. 2 E () t E cos( t ) E () t E cos( t ) 1

2 Intensity of Interference Patterns Phasor Representation of an E Field: phasor E E(t) E o t t E(t) Et () Ecos( t) o - E field as a vector (phasor) E rotating in the x-y plane with an angular frequency. - The time variation of this E field, E(t) is given as the horizontal projection (light red) of the phasor E (dark red).

3 Phasor in Action time E(t) Et () Ecos( t) - E field as a vector (phasor) rotating in the x-y plane with an angular frequency. - The time variation of this E field, E(t) is given as the horizontal projection (light red) of the phasor (dark red). o Copyright George Watson, Univ. of Delaware, 1997

4 Intensity of Interference Patterns Recall that there are two coherent E fields with a slight phase difference coming from the double slits: r r 2 1 E field from S 2 has a phase lag due to the extra path difference, r 2 - r 1. S 2 S 1 E () t E cos( t ) 2 E () t E cos( t ) 1

5 Phase Difference relates to Path Difference Here, we have the lighter cyan wave slightly ahead of the blue wave. r 2 r 1 (path difference measured in length) (phase difference measured in radians) (a complete cycle measured in wavelength) (a complete cycle measured in phase) 2 This gives the relation, 2 1 r r 2 r r kr r where k = 2 is called the wave number.

6 Phase Difference depends on Path Difference Recall from our geometry, we have the following picture for the path difference: r2 r1 dsin Substituting this into our previous equation, we have: d r r is the angular position of the observation point P sin NOTE: We expressed one full cycle as 2 so that has to be in radian!

7 Intensity of Interference Patterns At a given point on the screen far away from the two slits, the total E - field at P, E P, is given by the vector-sum of the two phasors. E and E 1 2 To find the magnitude of the resultant phasor E P, E P, we use the law of cosines. E E E 2E cos( ) P E 1 E 2 E 1 P E 2 E 1 E 2 E

8 Intensity of Interference Patterns Using the symmetry of the cosine function, cos( ) cos we have, E E E 2E cos( ) P E 2E 2E cos E P 2 E (1cos ) 2 2 P 2 Using another trig identity, 1cos 2cos we have, EP 4E cos. EP 2E cos 2 2 This gives,.

9 Intensity of Interference Patterns The intensity of an electromagnetic wave is given by the average magnitude of the Poynting vector, S av. In general, the Poynting vector is proportional to the square of the magnitude of the electric field so for the intensity at P, We can write the expression as, 2 I I cos 0 2 where I 0 is the maximum intensity when = 0. Note: when the two waves are in phase ( = 0, straight ahead), the resultant intensity is at maximum (I=I 0 ) and when the two waves are exactly half-cycle out of phase ( = ), the resultant intensity is identically zero.

10 Intensity in Two-Slit Interference Putting this expression for the phase difference into our previous intensity equation for a two-slit interference pattern, we then have, d I I0cos I0cos sin 2 2 I 2 d I0 cos sin

11 Intensity in Two-Slit Interference I 2 d I0 cos sin From the intensity equation, we can re-derive the conditions for the bright (maximum) and dark (minimum) fringes: d Maximum occurs when: sin m dsin m ( m0, 1, ) Minimum occurs when: d 1 1 sin ( m ) dsin ( m ) ( m0, 1, ) 2 2

12 Interference in Thin Films Color fringes observed from an oil slick on water or on a soap bubble are the white-light interference patterns produced by the reflected light off a thin film of oil or soap.

13 Phase Shifts During Reflection From Maxwell s Equations, one can show that the reflected wave will suffer a 180 o or /2 phase shift if it is reflected off from a medium with a larger n. na nb Er Ei (for normal incidence) n n a b

14 Interference from a Thin Air Gap 12 D air gap t glass air glass Assumptions: Thickness of air gap t is small Thickness of glass is large Incident light is nearly normal at the upper plate. Interference due to this small gap wave #1: reflected from top interface of air gap: n air n glass no phase shift wave #2: reflected from bottom interface of air gap: n n 180 o (or ) phase shift glass air

15 Interference from a Thin Air Gap Now, consider the conditions for interference: Constructive: wave #1 and wave #2 upon reflection must have a net phase difference of multiples of 2, i.e., m(2 ), m 0,1, 2, wave #1: suffers no phase shift during reflection wave #2: acquires a (180 o ) phase shift during reflection and it also gains an extra phase shift due to path difference = 2t in the air gap. So, the net phase diff accumulated between wave #1 and #2 = 2 t (2 ) For const interf, 2 t (2 ) m (2 ), m 1,2,3, 0, 2, 4, 2 t (2 ),,3,

16 Interference from a Thin Air Gap Solving for only positive t s, we have Rewriting, 1 2 t m, m 0,1, 2, 2 (condition for Con. Int. from thin film where one of the waves suffers a phase shift) Destructive: wave #1 and wave #2 upon reflection must have a net phase difference of m1 2 (2 ), m0,1, 2, t,,, Again, the net phase diff accumulated between wave #1 and #2 = 2 t (2 ) 2t 1 (2 ) m (2 ), m0,1,2, 2,3,5,

17 Interference from a Thin Air Gap Rearranging the equation, we have, 2 t (2 ),3,5 2 t (2 ) 0,2,4 Finally, we have 2 t m, m0,1,2, (condition for Des. Int. from thin film where one of the waves suffers a phase shift)

18 Thin and Thick Films Interference effects can be observed Interference effects are difficult to observe

19 Another Thin Film Example (nonrefractive coating on lens) 12 n n n air coating glass t air coating glass wave #1: reflected from top interface of the coating: n n 180 o (or ) phase shift coating wave #2: reflected from bottom interface of the coating: n n 180 o (or ) phase shift glass air coating

20 Interference from a Thin Film Since both wave #1 and #2 suffers the same phase shift upon reflection, the net phase difference will be from the path difference (2t) only. So, we have the standard condition (net phase diff. due to path diff. only), There is one more consideration: the path difference is accumulated in a medium with n coating so that the relevant wavelength should be n = /n coating. Constructive: 2 t m, m 0,1, 2, 2 n t m, m 0,1, 2, coating n 1 Destructive: 2 t m n, m 0,1, 2, ncoati ngt m, m 0,1, 2, 2 Note: In addition to wavelength modification, the RHS dependence are switched with respect to the air gap case.

21 Example 35.4: Thin Film Question: a. Will there be a bright or dark fringe at the point of contact? b. What is the distance x to the next bright fringe?

22 Example 35.4 Since n n and n n plastic silicone silicone glass Both wave #1 and #2 suffer a phase shift. So, at the point of contact (t=0), the reflected wave #1 and #2 will arrive at the eyes in phase (bright fringe). To find the location of the next constructive interference, we use, 2 n t ( m1) silicone 0

23 Example 35.4 From the two similar triangles, we have h t hx t l x l Substitute t into the previous eq, 2n t 2n x silicone silicone mm50010 mm l 2n h mm 0 silicone hx l 0.833mm

24 Newton s Rings When viewed in monochromic light, the interference pattern is a set of concentric rings called the Newton s rings. Since each fringe corresponds to a path difference ~, the lack of symmetry of these rings can be used to check for precision in lens making extremely accurately.

25 Michelson Interferometer C and D are cut from the same piece of glass so that Ray 1 & 2 will go thru the same thickness of glass. Distances comparable to can be measured with ease using this device by counting fringes.