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9.7 9.7

DiffractionDiffraction

• • Water wavesWater waves

• • Light wavesLight waves

• • Fraunhofer diffractionFraunhofer diffraction


9.7 Diffraction (SB p. 57)

Diffraction of water wavesDiffraction of water waves

Parallel ripples in a ripple tank passing through a narrow gap whose width is less than the wavelength of the ripples.

The ripples that pass through the gap are nearly circular and seem to originate from a point source situated in the gap.



Diffraction of water wavesDiffraction of water waves

Ripples of the same wavelength passing through a wider gap.

The ripples that pass through are almost straight except for some bending at the edges. The wider the gap, the less the bending becomes.



Diffraction of water waves around an Diffraction of water waves around an obstacleobstacle

A smaller “shadow” region is obtained when the size of the obstacle is comparable to thewavelength. That is, its diffraction is greater than that of larger obstacle.

small obstacle large obstacle



Diffraction of lightDiffraction of light

Because the wavelength of light is very short, the diffraction of light waves is noticeable only if the aperture is very small. For this reason shadows are formed.

Sound waves, however, are able to go round objects because of their long wavelength.



Light waves – Light waves – diffraction by a narrow slitdiffraction by a narrow slit

The central maximum is of a high intensity and very broad compared to the other maxima.



Light waves – Light waves – diffraction by a knife edgediffraction by a knife edge

If a knife edge is held up against monochromatic light, the diffraction of light waves round the knife edge produced the pattern as shown.



Fraunhofer diffractionFraunhofer diffraction

Fraunhofer diffraction is the diffraction of light produced by a narrow slit when plane light waves are incident normally on the slit and light waves emerging from the slit are plane waves.



Fraunhofer diffraction - explanationFraunhofer diffraction - explanation

For the point O on the screen and lying on the axis of the slit QO, there is no path difference for waves coming from pairs of points such as A and B or X and Y. Hence, constructive interference occurs at O and a maximum is obtained.



Fraunhofer diffraction - explanationFraunhofer diffraction - explanationIf is the angle between QP and the axis of the slit QO, then angle AQM is also . In triangle AMQ,

If the angle is small,

sin , the first minimum is obtained when

Similarly, all the other minima occur at angle n when

n = 1, 2, 3…

aaAQAM

2

2 sin

a

an

n



Minima (or dark fringes) are also formed on the other side of the axis. In between the dark fringes (minima) are bright fringes (maxima). The central maximum is of highest intensity and the other maxima are of much lower intensities as shown.

Fraunhofer diffraction - explanationFraunhofer diffraction - explanation



Changes in slit width - decreaseChanges in slit width - decrease

If the width of slit (a) decreases, from the equation

the angle θ increases as wavelength λ is constant.

This means that a broader central maximum is obtained but the intensity of all the bright fringes decreases as less light passes through the slit.




The figure shows the diffraction pattern of a narrow slit of width

It shows that the intensity at the point O drops to Io when the slit width is halved.

On the other hand, the width of the central maximum is doubled. In usual situation, less bright fringes are observed because the intensity of the secondary bright fringes drops below that detectable by the eye.



Changes in slit width - increaseChanges in slit width - increase

If the width of the slit increases, the width of the central maximum decreases but its intensity increases.

When the slit becomes very wide compared to the wavelength λ, the angle θ is almost zero.

That is, only the part of the screen directly in front of the slit is illuminated. This explains the rectilinear propagation of light.




The figure shows the diffraction obtained when the slit width is doubled.

Go to

Example 7Example 7

Go to

Example 8Example 8


End


QQ: : In the diffraction pattern of a single slit, the separation between the first minimum on one side and the first minimum on the other side is 5.2 mm. The distance of the screen from the slit is 80.0 cm and the wavelength of light used is 546 nm. What is the width of the slit?

Solution



Solution:Solution:Distance between first minimum and the central maximum x = (5.2) mm = 2.6 × 103 m

Using the equation, = (since is small)

and = =

Slit width, a =

= = 1.68 × 104 m

21

a

Dx

a

Dx

xD

3-

9-

10 6210 546

.

)(0.80)(

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QQ: : What is the angular separation between the first minimum and the central maximum for a diffraction pattern of a single slit if the slit width is equal to(a) 1 wavelength,(b) 5 wavelengths and(c) 10 wavelengths?

Solution



Solution:Solution:Since the slit width a is comparable to λ, the equation sinθ =

is used rather than θ = . The equation θ = is used when the slit

width a is large compared to λ.(a) When a = λ, sinθ= = = 1 Angular separation, θ = 90

(b) When a = 5λ, sinθ= = = 0.2 Angular separation, θ = 11.54

(c) When a = 10λ, sinθ= = = 0.1 Angular separation, θ = 5.74

a

a

a

a

a

5

a

10

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