3}

Chapter 17

Interference and Diffraction

17.1 ^fe use the method of Example 17.3.

A = 600.0 nm = 600 x 10~9 m= 6.000 x 10~7 m

Two narrow slits separated by

d = 1.00 mm= 1.00 x 10~3 m

Screen is 2.00 m from the slits:

L = 2.00 m

Solution: The slits are far away from the pointof observation =» we can use for maxima

dsind = mA

OT Asin B = m—

aFor L ^> Ax we can use

sin d fa tan 6

so

tan* ~ ^~<?a

AxOT mA~~L~ ~ ~d~

XL _ (6.000 x 1Q-7 m)(2.00 m)d ' 1.00 x 10-3 m

~3

2nd maximum: m = 2

2XLd

= 2(1.2 x 10-3 m)= 2.40 x 10~3 m= 2.40 mm

3rd maximum m = 3

Ax3 = 3(1.2 x 10~3 m)

= 3.60 x 10~3 m = 3.60 mm

In our domain of d <C L ie B <C 1

(m+ 1}XL mXAxm+i - Axm_j = -i —^ -a a

= 1.2 x 10~3XLd

= 1.2 x 10~3 m = 1.20 mm 71

=>• Separation of maxima is

1.20 x 10~3 m = 1.20 mm

Since we used the approximation sin 5 w t,for 9 <C 1 the separation is not constant ex<near the center of the screen.

17.2 /2 = 435 Hz. The musician can increase thequency of the string by increasing the tensi

^17.3jFbr destructive interference, the reflected ra}LU« exterior or tne coating must exhibit a ptdifference of an odd number of TT from thereflected from the glass surface. Model norincidence. The thickness of the coating is #

L7. Interference and Diffraction 327

air \V/ n = 1.

MeFl Vglass

00 \= 1.38 <5t

or

n ~ 1.5

The phase difference through the coating is

There is a phase change of TT at each reflection,so there is no net phase difference between thebeams due to reflections. For destructive inter-ference, we want

Ac/> = (2m + I)TT

pick the lowest order m. (:then

* Aodd -

thinnest coating)

7T =

6t = -P-

XX_

4n550 nm4 x 1.38

= 100 nm (2 sig. figs.)

L7.4 140 nm

L7.5

= 488 nm = 488 x 10~9 m

= 4.88 x 10~7 m

-7A2 = 514.5 nm = 5.145 x 10"' m

slit d = 3.00 iJ.ro. = 3.00 x 10~b m

We consider Fraunhofer Diffraction.

The blue-green (Ai .= 488 nm) light:

The green (A2 = 514.5 nm) light:

sin #2 =

There are two diffraction patterns superim-posed on the screen. Since •£"• is not a sim-ple fraction, there will be no order mi or 7712where the minima occur at 0\ = $2- So thereis a bright blue-green band at the center of thepattern, then alternating blue then blue-greenbands. Since sin 9 oc A and for 9 <C 1, I =distance to screen, then

Ax~T

mX

a.

Ax ex A

=> the green is wider. Consider the approximatewidth of the central maximum of each:

AX2 =

2V

2 (A2l)a

2(5.145 x 1Q-7 m-4.88 x IP"7 m)(1.00 m](3.00 x 10~6 m)

= 2 (8.83 x 10~3 m) = 17.7 mm

17.6 6>min = 4 x 10~6 rad

17.7

A = 600.0 nm = 6 x 10~7 ma = 1.0 x 10~6 m

d = 5.00 x 10~6 m

a) Interference maxima occur when

dsmff — m\

or

mX 6 x 10~7

sin 0 = —— = -—: -—r- x md 5.0 x 10-6

= 0.120,0.240,0.360,0.480,0.600,... 0.960

or

9 = 6.89°, 13.9°, 21.1°, 28.7°,36.9°,..., 73.7°

b) Diffraction minima occur when

mX (6 x ID"7 m)sin 9 = = 7717- —-£ ^

a (1 x 10~6 m)

for m = 1:sin 0 = 0.600

No other minima are present.

=> 0 = 36.9°

So:

t = (2m -f 1) ~4nfor some integer m. Then:

t=(2m + l)|̂ L = (2m + l)86.47nm (1]

Transmission: Each reflection occurs at a boundary with a medium (air) with a lowerrefractive index than the soap solution: there is no phase change due to reflection. But onebeam travels an extra distance It and so has an extra phase

Thus the total phase difference is A<£2 and for constructive interference:

= Im'w.

= 2m V/\

so, A .690 run

= m = m =

So we need to find a value of t that satisfies both equation 1 for some integer m ANDequation 2 for some integer m'. From equation 1, we get t = 86.5 nm, t — 3 (86.47) =259 nm, and so on. So t = 259 nm is a solution to equation 1 with m = 1 and equation 2witb-in/= 1. Thus the thickness of the film is 260 nm. (2 sig fig.)

\l732fThe beam that is refelected at the top surface of the oil film has a phase changeof TTA.0J = IT. The beam reflected at the bottom surface has no phase change dueto reflection since noii > nair. There is a phase difference due to the extra path lengthtravelled:

t

AThus the total phase difference between the beams is:

= 4irn— — TTA

47rn— — ir =

and for constructive interference:

Thus4 = (2m + 1) = (2m + 1) - = (2m + l)92nm

4n 4(1.50)Values of TO up to about 5 will give a possible thickness less than the coherence length.

17.24 Bright fringes are at <isin0 = mA. For small 9, we may write sin0 ~ y/D , andthen we have:

A _ (488 nm) (3.00 m) ,1()-9

y = m-D = m^ - — ii - 1 = ml. 464 x lO3-^ m = 1. 464 x 10~3m ma l.UU mm 10""

Thus the bright fringes are at y = 1.46 mm, 1.464 mmx2 = 2. 93 mm, 1.464 mmx3 =

332 Lea & Burke Physics: The Nature of Thing!

The first 3 rings are at

r = J(589 nm)(1.00 m) (m- i j , m = l,2,3

r = 0.54 mm, 0.94 mm, 1.2 mm

b) Aside from intensity effects (if ns > n0;ithere is no phase change on reflection at thetop of the oil layer) the optical path length dif-ference in the wedge will increase, i.e,

2 - 2 7 TA^2 = —r—ny + TT, A<fo = 0

A

For constructive interference

Ad>2 — A(j>i — 2mir

SO

then

ie, the rings become more closely spaced.

17.30 R — 0.85 m, radii are 0.50 mm, 0.87 mm, and1.1 mm.

1.7.31 Tilt one of the mirrors in the interferometer byangle a

tilt introduces extra path length 26 = lay since

tan a K a = —y

for

a = 1 arc second =1

206265o

extra path length is

rad

206265If LI is measured from the lower end of MI then

2(L2 -Li+6) = mX

Now set LI and L\ so that at y — 0, LI — L\ =0then 26 = mX — X for m = I which is the nextbright fringe

0-,= A

206265A

for A = 650 nm = 650 x 10~9 m (red)

650 xIP'9 my = x 20626o

Zi

= 67 mm

for A = 430 nm (blue)

430 x10-9 my = x 206265

= 44 mm

17.32 n = 1.00013

17.33 1 = 632.8 nm, I = 1.005 cm. Full of air, the'apparatus is set up so that

2L2 - 2 [(Li - 1) + nl] = mi A

2[L2 -Li -(n- 1)1} = mi A

Evacuate the cylinder, then n —> 1 and

Subtract the above two equations:

-1+n =21

9A

= 1 +9(632.8 x 10~9 m)

2(1.005 cm) x 10~2 m/cmn = 1.00028

17.34 Light has a typical wavelength of 550 nm whilesound wavelengths are of order 0.35 m. Thus,diffraction of sound by household size objects isvery much more pronounced than for light.

17.35 Assume there is no direct line of sight to the ra-dio stations, and that the hillsides absorb 100%of the radio waves. The tunnel looks somethinglike this:

y = x 206265

100 MHz

Longer wavelengths are diffracted more aroundobstacles. Thus the 600 kHz signals, with theirlonger wavelengths are bent further into thetunnel were they are picked up, than are the100 MHz signals.

334 Lea &: Burke Physics: The Nature of Things

17,44 A = 500 nm, D = 31 m; A = 300 mn, D -19 m; .-. Can't resolve Algol with any visible

_. light telescope.

<17.45)-D = 10 m, A = 600 nm = 600 x 10~9 m.^ ^Using the Rayleigh criterion:

(1.22)(600 x 1Q~9 m)10 m

= 7.32 x 10~8 rad x 206265

= 0.02"

arcsecrad

In reality, atmospheric distortions limit the an-gle to ~ 0.5" on really good nights,

17.46 D = 15 cm . Astronomers use larger telescopesbecause most objects of interest are faint and alarger telescope gathers more light. The lightgathering capacity is oc D2. 9min = 0.063", how-ever it gathers 256 times as much light as the15 cm telescope.

17.47 A = 633 nm = 6.33 x 10~7 m. Two slits,separated by

d = 0.5 mm = 0.5 x 10~3 m

Every third interference maximum is missing.For two slit interference, maxima are given by

— ,o

= 0,1,2...

For slit diffraction, minima are given by

sin0 = m'-, m' = 1,2,3...a

Every third maximum is missing; this would bem = 3, m' = 1; m = 6, m' = 2, etc

d

for m = 3, which corresponds to the first diffrac-tion rninimurn m' = 1, ie

-a

Equate sin d:

Aa1a

a

a

= ^4d3dd

— — —3= 0.17

0.5 mm— — .

3mm

17.48 d=14jum.

17.49 Want a two slit system, each slit 2.0 /xm widewith every other fringe missing

A = 488 nm

Interference maxima are at

Asin a = m— . . .

d

diffraction minima are at

sin 9 = mi —a

If every other interference maximum is missingthen

A7

d =

A

See appendix for plot of intensity vs. angle foitwo 2.0 fim slits.

17.50 d = 4.8 /zm. We see 13 -fringes on the screen.

17.51 a)

8P\ = (3 Pa)cos \kx — ut + —I 8J

SP-z = (5 Pa)cos \kx — ut + —

where

rad

fc = 10 —m

u> = 3.5 x 103

Placing 6PI along the y-axis, SP^ is at | :22.5° to 5PI. Measuring on the figure,

9 = 14° = 0.087T

Result:

SP = (8 Pa) cos (kx - wt + 0.

= (8 Pa) cos (kx -wt + 0.203vr)

use nigger telescopes oecause tney can gather more light Many interestingobjects are very feint. The area of the telescope determines how much light it can gather,and A <x D2. The Shane telescope, with D = 3.0 m, gathers (300/15)2 = 400 times asmuch light as a 15 cm diameter telescope.

The Hubble telescope (D = 2.4 m) has a resolution of:

g. ,^.600x10-^ 7 3 x Mr* rad2.4m ~ 4. 85 x 10-6 (rad/") -

at 600 nm. It can make use of its higher resolution because it is above the earth's at-mosphere, in orbit.

17.48 The maxima in a double slit interference pattern are given by

Am— (3)

The minima in the diffraction pattern for a rectangular slit are given by:

Asin S = n— (4)

aIf the ninth interference fringe is missing, then m = 9 in equation 3 gives the same angleas n = 1 in equation 4:

9-: = -=>d = 9a = 9 (1.5 /un) = 13. 5 u.m

So the-slit separation is 14 fan.

( 17.50/The maxima in a double slit interference pattern are given by

sin# = m- (5)a

The minima in the diffraction pattern for a rectangular slit are given by:

sin# = n— (6)

If the fourth interference fringe is missing, then m = 4 in equation 5 gives the same angleas n = 1 in equation 6: «

4- = -=»<f = 4a = 4 (1.2 Mm) = 4. 8 um

The maximum value of sintf is sinf = 1, so the maximum value of m is:

d _ 4 (1.2 /xm) 4(l.2xlQ-6m) =m ~ A ~ 589 nm 589 x 10~9 m

Thus we find fringes at m = 0,±1,±2,±3, missing, ±5, ±6, ±7, missing The totalnumber of fringes on the screen is 12 plus the central maximum for a total of 13.

17.52 Assuming mat we are looking at a first order maximum, then

The spacing between rulings is 1.1 Mm.

17.54 (a) If we put yi along the y-axis, then yz points along the negative i-axis and y-<

is at 45° , as shown in the diagram.

4cm