Double slit diffraction

5. Double Slit Diffraction

Background

Huygens’s principle

Interference

Fraunhofer and Fresnel diffraction

Coherence

Laser

Aim of the experiment

1. To plot the intensity distribution of the Fraunhofer diffraction pattern due to

two slits of same width and to estimate the width of the slits and separation

between the slits from the intensity pattern.

2. To compare intensity distribution of diffraction patterns due to two double

slits with same widths, but different distances between the slits. The widths of

the slits and separation between the them are determined. Finally intensity

relations of the peaks and ‘missing orders’ are studied.

Apparatus required

Laser, He-Ne, 1 mW

Universal measuring amplifier

Optical profile bench

Slide-mounts

Lens holder

Object holder

Lens, mounted, f + 20 mm.

Lens, mounted, f + 100 mm.

Photoelement

Diaphragm, 4 double slits

Multi-range meter

Connecting cords

Theory

When plane monochromatic light wave front of wavelength, λ, falls normally

on a system of two slits of equal widths, b, separated by a distance, g, the luminous

intensity I of the beam diffracted in the direction φ is given by:

)./sin..(cos)/sin..(

)/sin..(sin)( 2

2

2

0

g

b

bII …………………(1)

The equation (1) shows that the intensity is a product of a single slit diffraction term

and a double slit interference term (see figures1 (a) and 1 (b) below).

Date :

57

Double slit diffraction

Fig. 1(a) Double slit diffraction pattern g/b=5, Fig 1(b) Same pattern resolved as a

product of single slit diffraction pattern and double slit interference pattern with I0 =1.

If one considers only a single slit (1st class interference), this gives minimum intensity

when the numerator of the first factor becomes zero. This always is the case when the

following applies:

mb m sin , (m = 1, 2, 3, ……) ….………………….(2)

Principal maximum is obtained for =0, when the first factor of expression (1)

becomes 1, there are other secondary maxima approximately at =1.43, 2.46,

3.47, 4.48 etc.

On the other hand, the double slit interference pattern has maxima at angles satisfying

the condition,

ng n sin , (n=0, 1, 2, 3, . . . .) …………………….(3)

An interesting situation arises when both equations, (2) and (3) are satisfied for a

particular value of . This happens when,

.m

n

b

g ………………….(4)

In this case thn order interfernce maximum will have zero intensity and will therefore

be missing. In the figure 1, g/b=5, hence we find that the 5th

secondary peak is

missing. Condition (4) is again satisfied for n=10 and m=2, so the 10th

peak is again

absent. Similarly 15th

,20th

etc. peaks will also be missing in this case. These are

known as ‘missing orders’ in the literature. The central envelope has 9 peaks within it,

where as the side lobes will have 4 peaks each for the case shown above.

Procedure

1. Place the photocell at the center of the shifting range at the one end of the

optical bench. The laser is mounted on the opposite end of the bench. Connect

the photocell to the 410 -input of the measurement amplifier (amplification

factor 53 1010 ) and the output of the amplifier to the multimeter.

58

Double slit diffraction

2. Align the laser source on the optical bench to get collimated laser beam. A

broadened and parallel laser beam, obtained with the lenses f= 20 mm and f=100

mm, must impinge centrally the photocell. To achieve this the distance between

the laser and the lens f=20mm is kept 11.5 cm whereas the distance between the

lenses f=20mm and f=100mm is adjusted to 13 cm.

3. Place the double slit diaphragm perpendicular to the beam at a distance of 5 to 6

cm from the lens f=100mm. This makes a separation about 1m between the slit

and the photocell.

4. Choose any one of the double slits available. Make sure that both slits of the

two slits are equally illuminated.

5. Measure the micrometer constant of the scale attached to the photocell detector.

6. Turn the circular scale attached at the base of the photocell in equal intervals(say

0.2 or 0.3mm) and note that the intensity of the fringes from one end (say from

left to right) using a multimeter. Reach the other end of the fringe system this

way noting down many readings of the intensities in between.

7. Repeat 6 for other available double slits. Observe number of peaks in the central

envelope and side lobes and identify missing orders.

8. Measure the distance, D, separation between slit and the detector, using the scale

on the optical bench.

9. Plot the intensity distribution of the diffraction pattern and calculate the slit

width, b, and the separation between the slits, g, from graphs. Match these

estimates with the ones written on the diaphragm. Furthermore, calculate the

relative intensities of the peaks with respect to the central maximum. Match

your results with the theoretical estimates obtained from the expression (1).

Observations

Wavelength of the light () = ……………… cm

Position of the laser =……………………..cm

Position of the lens f/20mm =………………..cm

Position of the lens f/100mm=………………….cm

Position of the double slit diaphragm=………………….cm= a

Position of the photocell=………………………….cm = b

Distance of the slit from the cell, D = b-a = …………….cm

59

Double slit diffraction

Table I: For intensity distribution (x0 corresponds to the central maximum)

Double Slit I b= g=

Position (x cm) = (x-x0)/D V (volt) Position (x cm)

= (x-x0)/D

V (volt)

70 71 72 73 74 75 76 77 78 79 80 81 82 70 60

Double slit diffraction

Double Slit I b= g=

Position (x cm) = (x-x0)/D V (volt) Position (x cm)

= (x-x0)/D

V (volt)

61

Double slit diffraction

Double Slit II b= g=

Position (x cm) = (x-x0)/D V (volt) Position (x cm)

= (x-x0)/D

V (volt)

62

Double slit diffraction

Double Slit II b= g=

Position (x cm) = (x-x0)/D V (volt) Position (x cm)

= (x-x0)/D

V (volt)

63

Double slit diffraction

Double Slit III b= g=

Position (x cm) = (x-x0)/D V (volt) Position (x cm)

= (x-x0)/D

V (volt)

64

Double slit diffraction

Double Slit III b= g=

Position (x cm) = (x-x0)/D V (volt) Position (x cm)

= (x-x0)/D

V (volt)

65

Double slit diffraction

Graph : Double slit intensity pattern

66

Double slit diffraction

Results and Calculations:

1. Plot Intensity distribution of the diffraction as a function of parallel to the plane of the slit.

Table – 2: Estimation of slit width, b, and slit separation, g .

Slit width b Slit separation g No. peaks inside

the central

envelope

No. peaks within

side lobes

Missing orders

observed Given

(mm)

Estimated from

plot (mm)

Given

(mm)

Estimated from

plot (mm)

Double slit I

Double slit II

Double slit III

Table – 3: Relative intensities of the peaks with respect to the central maximum (assume I0= 1).

I1 I2 I3 I4 I5

From the

plot

From

theory

From the

plot

From

theory

From the

plot

From

theory

From the

plot

From

theory

From the

plot

From

theory

Double slit I

Double slit II

Double slit III

67

Double slit diffraction

Error calculation For the minimum of the 1

st class interference pattern, mb m sin

Therefore,

is supplied and, D and x are obtained as a difference of two scale readings.

Similarly evaluate error in the estimation of separation between the slits g.

Precautions

(i) Do not look into the laser light directly.

(ii) Adjustment of lens, slit, laser must be made properly so that fringes are

bright and distinct.

(iii) Since the linear shift d is proportional to D, it should be fairly large. A

value of D of about 1.0 m is preferable.

(iv) Make sure that a strong monochromatic source of light is used.

(v) Take care of the backlash error while photocell measurements.

Questions Same as for single slit diffraction

1. What are missing orders?

2. What do you expect if the width of the slits is gradually increased?

3. What is the effect of increasing the width of the opaque space, keeping the

slit width constant?

References

1. PHYWE LEP 2.3.05 Diffraction intensity at double slit systems

2. Fundamental of Optics by F. Jenkins and H. White 535 JEN/F

3. Optics by E. Hecht 535 HEC/O

x

x

D

D

b

b 22

68