Lecture 16 Diffraction Chp. 37

• Topics– Young’s double slit interference experiment– Diffraction and the wave theory– Single slit diffraction– Intensity of single slit diffraction– Circular aperture and double slit diffraction– Diffraction grating– Dispersion and resolving power– Warm-up problem

• Demos– Diffraction grating and slits– Inverted mirage– Measuring diameter of a strand Debra’s hair

Young’s Double SlitInterference Experiment

m=0

m=1

m=1

m=2

m=2

D

y

ym mDd

d sin

d sin m m = 0,1,2,3... Maximum

d sin (m 1

2) m = 0,1,2,3... Minimum

tan ymD

or ym Dtan Dsin

m ym +/-

0

1

2

3

0

D/d

2D/d

3D/d

Maximam ym +/-

0

1

2

3

D/2d

3D/2d

5D/2d

7D/2d

Minima

ym (m 1/2)D

d

What about the intensity of light along the screen?

I 4I0 cos2 12

2d

sin

13E Suppose that Young’s experiment is performed with blue-green light of 500 nm. The slits are 1.2mm apart, and the viewing screen is 5.4 m from the slits. How far apart the bright fringes?

From the table on the previous slide we see that the separation between bright fringes is

D /d

D /d (5.4m)(500 10 9m) /0.0012m

0.00225m2.25mm

Diffraction of a single slit

path length difference between r1 and r2 a

2sin

2for the first minimum

asin

asin m m =1,2,3.. minima

Find minimum

Find maximumFirst maximum lies on the axis at = 0or ym=0. Other maxima lie in betweenthe minima. To find them we need to find the intensity along the screen.

ym

tan ymD

or ym Dtan Dsin

ym Dtan mDa

Intensity of single slit diffraction

Maxima/Minima conditions

phase difference 2path length difference

2

x sin

I() I0sin

22

2

2asin

Maxima when 2

(m 1

2) m = 0,1,2,3..

sin (m 1

2)a

Minima when 2

m m =1,2,3..

sin ma

Exact solution for maxima

I() I0sin2 2

I0(sin2)( 2)

dI

d0

I02sin cos 2 I02sin2 3 0

cos sin / tan

let 2

Transcendental equation. Solve graphically

To find maxima of a function, take derivative and set equal to 0

sin ma

m

m1

sin ma

m

5m

50.2,0.4,0.6,0.8,1.0

sin ma

m

10m

10

8. A 0.10-mm-wide slit is illuminated by light of wavelength 589nm. Consider a point P on a viewing screen on which the diffraction patters of the slit is viewed; the point is at 30o from the central axis of the slit. What is the phase difference between the Huygens wavelets arriving at point P from the top and midpoint of the slit? (Hint: see Eq. 37-4.)

We note that nm = 10-9 m = 10-6 mm. From Eq. 37-4,

sin

2x

This is equivalent to 266.7 - 84 = 2.8 rad = 160o

omm

mm30sin

2

10.0

10589

26

rad 7.266

6. Sound waves with frequency 3000 Hz and speed 343 m/s diffract through a rectangular opening of a speaker cabinet and into a large auditorium. The opening, which has a horizontal width of 30.0 cm, faces a wall 100 m away. Where along that wall will a listener be at the first diffraction minimum and thus have difficult hearing the sound? (Neglect reflections).

Let the first minimum be a distance y from the central axis which is perpendicular to the speaker. Then

1).m(for sin22

aa

m

yD

y

Therefore,

11 22

svaf

D

a

Dy

m

Hzm

m

sm

2.4113433000300.0

1002

y

Diffraction and Interference by a double slit

I = I (double slit interference) x I(diffraction)

I() Im cos2() sin2 2

d sin

dsin = m2 m2 = 0,1,2.. maxima

asin

asin = m m =1,2,3.. minima

Sample problem 37-4

405nm

d 19.44nnm

a4.050nm

(a) How many bright interference fringes fall within the centralpeak of the diffraction envelope?

GivenThe idea here is to find the angle where the first minimum occurs of the diffraction envelope.

asin d sin m2m2 d /a19.44 /4.050 4.8

We have m=0 and m=1,2,3 and 4 on bothsides of central peak.

Ans is 9

Diffraction by a circular aperature

The first minimum for the diffraction of light from a circular aperature is given by:

sin 1.22 /d where d is the diameter of the aperature.

Our ability to resolve two distant point like objects is determined when the first minimum of one objects diffraction pattern overlaps the central maximum of another. This is called Raleigh’s criterion.

Raleigh's Criterion

R 1.22 /d

Example

Example 15E. The two headlights of an approaching automobile are 1.4 m apart. Assume the pupil diameter is 5.0 mm and the wavelength of light is 550 nm. (a) At what angular distance will the eye resolve them and (b) at what distance?

R s /DDs /RD1.4m /134.2 10 6 rad

D0.010 106m10km

R 1.22 /d

R 1.22(550 10 6mm) /5.0mm

R 134.2 10 6rad

(a)

(b)s

D

R 1.22 /d

R

Diffraction Grating

Double slit -- N slits or rulings.

d sin m m = 0,1,2,3..

d = w/N where w is the entire width of the grating

w

Can be used to measure wavelength of light

tan (ym y0) /D

Measure angles of diffracted lines with a spectroscope using formula below. Then relate to wavelength

= d sin /m

R / Nm

Resolving power of grating. Measure of the narrowness of lines

Dispersion of a grating. Measure of how well lines are separated

D / m

dcos

Highest R

Highest D

Show diffraction gratings with increasing N and single slit diffraction with varying slit

width a.

Babinets Complementarity Principle

In the diffraction region the intensity is the same whether you have an aperature or opaque disk. You can also replace a slit with a wire or hair strand. A compact disk is an example of a diffraction grating in reflection instead of transmission.

Experiment: Measure diameter of a strand of hair from Debra.

asin for the first diffraction minimum

a /sinsin tan y1 /D

aD / y1

a560 nm (D / y1)

Mirage

1.06

1,091.08

1.07 1.071.08

1.09

sky eye

Hot road causes gradient in the index of refraction that increasesas you increase the distance from the road

In the demo before you the gradient is in the opposite direction

Warm-up

HRW6 37.CQ.02. [73994] You are conducting a single-slit diffraction experiment with light of wavelength

(a) What appears, on a distant viewing screen, at a point at which the top and bottom rays through the slit have a path length difference equal to 5

the m = 5 maximum the m = 5 minimum the m = 4 minimum the m = 4 maximum

(b) What appears, on a distant viewing screen, at a point at which the top and bottom rays through the slit have a path length difference equal to 4.5

the minimum between the m = 4 and m = 5 minima the minimum between the m = 4 and m = 5 maxima the maximum between the m = 4 and m = 5 maxima the maximum between the m = 4 and m = 5 minima