Single Slit Diffraction

Diffraction

Diffraction may be thought of as the 'spreading out' of waves as they pass through or by an aperture or edge. We now investigate this phenomenon more closely for a single slit of. Under the Fraunhofer conditions, a single slit will exhibit a light curve following the single slit diffraction intensity expression.The narrower the slit, the broader the peaks of light.The shape or 'envelope' of this light curve will serve to set limiting intensities for multiple slit arrangements, assuming that all.

• Diffraction is the phenomenon by virtue of whichlight bendswhile passing through aslit or an opening.
• The extent of bending depends upon the diameter of the slit.
• Both Interference and Diffraction are closely related to each other.
• Young replaced the two slits by a single slit in his single slit experiment.Therefore this experiment is also referred as Young’s single slit experiment.
• When the light passed through one slit a different type of pattern was observed on the screen.
• The pattern which was observed had a central maximum band and which was very wide as compared to interference pattern.
• There were alternate dark and bright bands and their intensity was decreasing on both the sides.
• The central maxima,was very wide whereas corresponding secondary maxima and minima were reduced in the intensity.
• The change in the pattern was formed due to diffraction instead of interference.

Conclusion:-

2. Alternate dark and bright bands on either side.
3. Intensity was decreasing on both sides.

Diffraction Fringe Pattern

• In diffraction there is an incoming wave which passes through a single slit and as a result diffraction pattern was obtained on the screen.

Diffraction pattern Maxima

• The incident wavefront is parallel to the plane of the slit. This shows they are in phase with each other.
• From the given figure:-
• Path difference =NP – LP
• =>NQ = a sinθ where a=width of the slit.
• The slit was divided into smaller parts M1, M2, N and L, and when contribution from each part is added up.
• At the central point θ =0. This implies path difference =0.
• This means all the parts of the slit will contribute completely. Therefore the intensity is the maximum.
• Central Maximum occurs at θ =0.

Problem:-The light of wavelength 600nm is incident normally on a slit of width 3mm.Calcluate the linear width of central maximum on a screen kept 3m away from the slit?

Wavelength λ =600nm =600 x10-9m.

Width of the slit a =3mm=3x10-3m.

Distance of the screen D=3m.

Condition for minima:-a sinθ =nλ

=>a sin θ =λ

=>sin θ = (λ/a) = (600 x10-9)/ (3x10-3)

=>θ =(600 x10-9)/(3x10-3)

=>(x/D) = (600 x10-9)/ (3x10-3)

=> x= (600 x 10-9 x3)/ (3 x10-3)

x=600 x 10-6m.

Therefore width =2x = 1200 x 10-6m

Width = 1.2 mm

Diffraction pattern Secondary Maxima

Scientist Fresnel found that secondary maxima occurred when the value of θ is:

• θ =(n+(1/2)) (λ/a)
• For n=1 =>θ =(3λ)/(2a) = (1.5λ/a) (equation 1)
• Where = (3λ)/ (2a) it lies midway between 2 dark fringes.
• Suppose if the slit is divided into 3 parts ,
• Consider the first 2/3rd part of the slit ,
• Path difference = (2/3) a x θ
• =(2/3) a x(3λ/2a) using (equation 1)
• Path difference =λ.
• The λ is getting divide into 2 halves with path difference =λ/2 and λ/2.
• Each of λ/2 gets cancel with another λ/2.
• Contribution from (2/3rd) part gets cancel out with each other.
• Therefore only (1/3rd) contributes to I(Intensity).
• Intensity ≠ 0.
• Intensity is reduced.
• For n=2 => θ =(5 λ)/(a)
• Only (1/5th) part will contribute.
• For n=3 =>θ =(7 λ)/(2a)
• Only (1/7th) part contributes.

Slit divided into 3 parts

Diffraction pattern for Minima

• Condition to get minima on the diffraction pattern:
• θ = (n λ/a); n= (+-) 1, (+-2)…
• For n=1, =>θ = (λ/a)
• Suppose the slit is divided into small parts. For every M1in the portion LM there exists another M2 in MN.
• From the figure it is clear that the contribution from all elements inLM (M1) will cancel out the contribution from all elements MN (M2).
• Therefore net intensity I=0 at θ = (λ/a).
• Minima occur at θ = (nλ/a). => aθ =nλ
• => a sinθ =nλ.
• Path difference = (nλ/a).
• Minima will occur ata sin θ =n λ. Where a =width of the slit,λ = wavelength of the light used.

Problem:- A beam of light consisting of two wavelengths, 650 nm and 520 nm, is used to obtaininterference fringes in a Young’s double-slit experiment.

(a) Find the distance of the third bright fringe on the screen from the centralmaximum for wavelength 650 nm.

(b) What is the least distance from the central maximum where the bright fringes dueto both the wavelengths coincide?

Wavelength of the light beam, λ1 =650nm

Wavelength of another light beam,λ2 =520nm

Distance of the slits from the screen = D

Distance between the two slits = d

(a) Distance of the nth bright fringe on the screen from the central maximum is given bythe relation,

x=nλ1(D/d)

For third bright fringe, n=3

Therefore x=3x650(D/d) =1950(D/d) nm

(b) Let the nth bright fringe due to wavelength and (n − 1)th bright fringe due towavelength λ1coincide on the screen. We can equate the conditions for bright fringesas:

2=(n-1)λ1

520n=650n-650

650=130n

Therefore, n=5.

Hence, the least distance from the central maximum can be obtained by the relation:

Note: The value of d and D are not given in the question.

Problem:- In a double-slit experiment the angular width of a fringe is found to be 0.2° on a screenplaced 1 m away. The wavelength of light used is 600 nm. What will be the angularwidthof the fringe if the entire experimental apparatus is immersed in water? Take refractiveindex of water to be 4/3.

Distance of the screen from the slits, D = 1 m

Wavelength of light used, λ1 =600nm

Angular width of the fringe in air θ1 =0.2o

Angular width of the fringe in water = θ2

Refractive index of water, μ= (4/3)

Refractive index is related to angular width as:

μ = (θ12)

θ2 = (3/4)θ1

Single And Double Slit Diffraction

= (3/4)x0.2 =0.15

Therefore, the angular width of the fringe in water will reduce to 0.15°.

Problem:- A parallel beam of light of wavelength 500 nm falls on a narrow slitand the resulting diffraction pattern is observed on a screen 1 maway. It is observed that the first minimum is at a distance of 2.5mm from the centre of the screen. Find the width of the slit.

Wavelength of light beam, λ = 500 nm = 500 × 10−9 m

Distance of the screen from the slit, D = 1 m

For first minima, n = 1

Distance between the slits = d

Distance of the first minimum from the centre of the screen can be obtained as:

x = 2.5 mm = 2.5 × 10−3 m

It is related to the order of minima as:

n λ =x(d/D)

d= (n λD/x)

d= (1x500x10-9x1)/ (2.5 x10-3) = 2x 10-4 m

d=0.2mm

Single Slit Diffraction Of Light

Therefore, the width of the slits is 0.2 mm.

In classical physics, we can classify optical phenomena into one of two categories: ray optics and wave optics.

Light is a transverse electromagnetic wave. In many situations, the wavelengths of the light being studied are very small compared to the dimensions of the equipment used to study the light. Under these conditions we can make an approximation called geometrical optics or ray optics. Consider the wavelength scale of light waves. Wavelengths in the middle of the visible band are on the order of 500 nm. So a laser beam with a diameter of 1 mm has a diameter of 2000 wavelengths. Individual atoms in a solid are separated by distance on the order of 0.1 nm. So, as far as visible light is concerned, matter is quasi-continuous.
If the wavelengths of the light become comparable to the dimensions of the equipment, then we study optical phenomena using the classical theory of radiation, or wave optics. Wave optics contains all of ray optics, but the mathematical treatment is much more involved.

Diffraction and interference are phenomena observed with all waves. Diffraction can only be observed with waves traveling in two or three dimensions. (We already encountered interference when studying mechanical waves on a string and sound waves in physics 221.)

Diffraction is the tendency of a wave emitted from a finite source or passing through a finite aperture to spread out as it propagates. Diffraction results from the interference of an infinite number of waves emitted by a continuous distribution of source points in two or three dimensions. Huygens' principle lets us treat wave propagation by considering every point on a wave front to be a secondary source of spherical wavelets. These waveletspropagate outward with the characteristic speed of the wave. The wavelets emitted by all points on the wave front interfere with each other to produce the traveling wave. Huygens' principle also holds for electromagnetic waves. When studying the propagation of light, we can replace any wave front by a collection of sources distributed uniformly over the wave front, radiating in phase.

When light passes through a small opening, comparable in size to the wavelength λ of the light, in an otherwise opaque obstacle, the wave front on the other side of the opening resembles the wave front shown on the right.

The light spreads around the edges of the obstacle. This is the phenomenon of diffraction. Diffraction is a wave phenomenon and is also observed with water waves in a ripple tank.

Water waves in a ripple tank

 A single large slit A single small slit

The single slit

When light passes through a single slit whose width w is on the order of the wavelength of the light, then we can observe a single slit diffraction pattern on a screen that is a distance L >> w away from the slit. The intensity is a function of angle. Huygens' principle tells us that each part of the slit can be thought of as an emitter of waves. All these waves interfere to produce the diffraction pattern. Where crest meets crest we have constructive interference and where crest meets trough we have destructive interference.

Single Slit Diffraction Minima

Very far from a point source the wave fronts are essentially plane waves. This is called the Fraunhofer regime, and the diffraction pattern is called Fraunhofer diffraction. The positions of all maxima (constructive interference) and minima (destructive interference) in the Fraunhofer diffraction pattern can be calculated fairly easily.
The Fraunhofer approximation, however, is only valid when the source, aperture, and detector are all very far apart or when lenses are used to convert spherical waves into plane waves. Being very far apart means that the distances between source, aperture, and detector must be much greater than the width of the aperture.

The Fresnel regime is the near-field regime. In this regime the wave fronts are curved, and their mathematical description is more involved.

The positions of all maxima and minima in the Fraunhofer diffraction pattern from a single slit can be found from the following simple arguments.

• Consider a slit of width w, as shown in the diagram on the right. A plane wave is incident from the bottom and all points oscillate in phase inside the slit.
• For light leaving the slit in a particular direction defined by the angle θ, we may have destructive interference between the ray at the right edge (ray 1) and the middle ray (ray 7). To arrive at a distant screen perpendicular to the direction of propagation of the rays, the rays coming from different points inside the slit have to travel different distances. They have a different optical path length. If ray 7 has to travel an extra distance of one-half wavelength (λ/2) compared to ray 1, then ray 1 and ray 7 destructively interfere. Crest meets trough.
• The optical path length (OPL) of a light ray traveling from point A to point B is defined as c times the time it takes the ray to travel from A to B. In free space the optical path length is just the distance d between the points. In a transparent medium with index of refraction n, it equals n times this distance, OPL = nd, because the light moves with speed c/n.

If the optical path length of two rays differs by λ/2, the two rays interfere destructively. For ray 1 and ray 7 to be half a wavelength out of phase we need

(w/2)sinθ = λ/2 or w sinθ = λ.

But from geometry, if these two rays interfere destructively, so do rays 2 and 8, 3 and 8, and 6 and 10, 5 and 11, and 6 and 12.
In effect, light from one half of the opening interferes destructively and cancels out light from the other half.

Destructive interference produces the dark fringes. Dark fringes in the diffraction pattern of a single slit are found at angles θ for which

w sinθ = mλ,

where m is an integer, m = 1, 2, 3, ... . For the first dark fringe we have w sinθ = λ.

When w is smaller than λ , the equation w sinθ = λ has no solution and no dark fringes are produced.

If the interference pattern is viewed on a screen a distance L from the slits, then the wavelength can be found from the spacing of the fringes. We have λ = w sinθ/m and sinθ = z/(L2 + z2)½), or

λ = zw/(m(L2 + z2)½),

where z is the distance from the center of the interference pattern to the mth dark line in the pattern.

If L >> z then (L2 + z2)½ ~ z/L and we can write

λ = zw/(mL).

Single Slit Experiment

Problem:

When a monochromatic light source shines through a 0.2 mm wide slit onto a screen 3.5 m away, the first dark band in the pattern appears 9.1 mm from the center of the bright band. What is the wavelength of the light?

Solution:

• Reasoning:
Dark fringes in the diffraction pattern of a single slit are found at angles θ for which w sinθ = mλ, where m is an integer, m = 1, 2, 3, ... . For the first dark fringe we have w sinθ = λ. Here we are asked to solve this equation for λ.
• Details of the calculation:
z = 9.1 mm = 9.1*10-3 m.
L = 3.5 m.
w = 0.2 mm = 2*10-4 m.
L >> z, therefore sinθ ~ z/L and λ = zw/(mL).
λ = (9.1*10-3 m)(2*10-4 m)/(3.5 m).
λ = 5.2*10-7 m = 520 nm.

Problem:

Consider a single slit diffraction pattern for a slit width w. It is observed that for light of wavelength 400 nm the angle between the first minimum and the central maximum is 4*10-3 radians. What is the value of w?

Solution:

Single Slit Diffraction Wavelength

• Reasoning:
Dark fringes in the diffraction pattern of a single slit are found at angles θ for which w sinθ = mλ, where m is an integer, m = 1, 2, 3, ... . For the first dark fringe we have w sinθ = λ. Here we are asked to solve this equation for w.
• Details of the calculation:
First minimum: w sinθ = λ, w = (400 nm)/sin(4*10-3 radians) = 1*10-4 m.