Lecture Notes
Chapter 12: Diffraction
From course: Physics Grade XII

Diffraction of light

When light waves pass through a small aperture or an obstacle, they spread out into the geometric shadow of the aperture or obstacle. This phenomenon of spreading of light waves into the geometric shadow is called diffraction of light. Depending upon the nature of incident wavefront, diffraction is divided into two types:

Fresnel diffraction

When spherical or cylindrical wavefront is incident on an obstacle or aperture, the diffraction pattern obtained is called Fresnel diffraction. In this case, the source or the screen or both are at finite distances from the obstacle or aperture. The diffracted wavefronts are also spherical or cylindrical and no modification is needed by lenses and mirrors.

Fraunhoffer’s diffraction

When plane wavefront is incident on obstacle or aperture, the diffraction pattern obtained is called Fraunhoffer’s diffraction. In this case, the source and the screen are at infinite distance from the obstacle or aperture. Lenses and mirrors are used to obtain plane wavefront (incident or diffracted).

Diffraction of light at a single slit (Fraunhoffer’s Diffraction)

Let us consider a narrow-slit AB of width ‘d’ is illuminated by monochromatic light source (S). The spherical wavefront XY originating from ‘S’ is made ‘a plane wavefront’ by the lens L1. The plane wavefront passes through the slit AB and is incident on the lens L2. The waves in the direction of OP come to focus at P, which is the position of central maximum.

AR is the direction of secondary waves inclined at an angle [latex]\theta[/latex] to the direction OP. All the wavelets travelling with an angle [latex]\theta[/latex] focused at P’ which will have some path difference (AL). In [latex]\Delta ABL,[/latex]

[latex]sin\theta = \frac{AL}{AB} = \frac{AL}{d}[/latex]

[latex]\therefore\ AL = dsin\theta[/latex]

The whole wavefront can be considered to be two halves OA and OB and if the path difference between the secondary wavelets from A and B is [latex]\lambda[/latex], then, path difference between secondary waves from A and O will be [latex]\frac{\lambda}{2}[/latex]. Similarly, it is [latex]\frac{\lambda}{2}[/latex] for OB as well.

Thus, destructive interference takes place and the point will be of minimum intensity.

  1. For nth secondary minima: we have,
    Path difference = [latex]n\lambda[/latex], where, n = 1, 2, 3, ………
    [latex]\therefore\ dsin\theta_n = n\lambda[/latex]

    [latex]Or,\ sin\theta_n = \frac{n\lambda}{d}[/latex]  …………………. (i)
    where, [latex]\theta_n[/latex] = nth minimum.
    Since, [latex]\theta[/latex] is very small, [latex]sin\theta_n \approx \theta_n[/latex]
[latex]\therefore\ \theta_n = \frac{n\lambda}{d}[/latex]

    (This is condition for nth minima).

For first minima, n = 1 gives, [latex]\theta_1 = \frac{\lambda}{d}[/latex]

For 2nd minima, n = 2 gives [latex]\theta_2 = \frac{2\lambda}{d}[/latex]

  1. For nth secondary maxima: Secondary maxima will be formed if the path difference AL is an odd integral multiple of [latex]\frac{\lambda}{2}.[/latex] i.e.,

Path difference = [latex]\frac{(2n+1)\lambda}{2}[/latex]

[latex]\therefore\ dsin\theta_n = \frac{(2n+1)\lambda}{2}[/latex]

[latex]Or,\ sin\theta_n = \frac{(2n+1)\lambda}{2d}[/latex]  …………… (ii) Where, n = 1, 2, 3, …….

Since,  [latex]\theta[/latex] is very small, [latex]sin\theta_n \approx \theta_n[/latex]

[latex]\therefore\ \theta_n = \frac{(2n+1)\lambda}{2d}[/latex]

 (This is condition of secondary maxima)

For 1st maxima, n = 1, [latex]\theta_1 = \frac{3\lambda}{2d}[/latex]

For 2nd maxima, n = 2, [latex]\theta_2 = \frac{5\lambda}{2d}[/latex]

The diffraction pattern of single slit is shown alongside.

Width of central maximum (2x)

It is defined as the distance between first minima on either side of central maxima. Now,

For [latex]1^{st}[/latex] minima, [latex]\theta = \frac{\lambda}{d}[/latex]

Now, if D be the distance between screen and the slit, then, in [latex]\Delta OPP’,[/latex]

[latex]Tan\theta \approx \theta = \frac{x}{D}[/latex]

[latex]\therefore \frac{x}{D} = \frac{\lambda}{d}[/latex]

[latex]\therefore\ x = \frac{\lambda D}{d}[/latex]

x is the distance of secondary minimum from P.

[latex]\therefore[/latex] width of the central maximum = 2x = [latex]\frac{2\lambda D}{d}[/latex]

Diffraction grating

An arrangement consisting a large number of equidistant, parallel, rectangular slits of equal width separated by opaque space is called diffraction grating. It is made by ruling equidistant parallel lines on an optically transparent plate. The ruled lines act as opaque to light. If a and b are the width of transparent and opaque part of the grating, then, (a + b) = d, is called ‘grating element’. If ‘N’ be the number of ruling lines per unit length, then (a + b) = 1/N.

Or, [latex]d = \frac{1}{N}[/latex]

Let us consider, a parallel beam of monochromatic light having wavelength [latex]\lambda[/latex] falls on the diffraction grating. Let, d = (a + b) be the grating element. The central maximum is formed at the centre of the screen O. Since the rays from the corresponding points of any two consecutive slits will have the path difference [latex]\lambda[/latex], all the waves superimpose each other and produce maximum at point P. Let, [latex]\theta_1[/latex] is the angle of diffraction for first secondary maximum. We have,

[latex](a+b)sin\theta_1 = \lambda[/latex]

Similarly, for second order maximum,

[latex](a+b)sin\theta_2 = 2\lambda[/latex]

In general, for nth order maximum is:

[latex](a+b)sin\theta_n = n\lambda[/latex], where, n = 1, 2, 3,………….

[latex]\therefore\ dsin\theta_n = n\lambda\ [\because (a+b) = d][/latex]

Resolving Power

The resolving power of an optical instrument is the ability of an instrument to produce distinctly separate images of two close objects.

When the central maximum of one image falls on the first minimum of another image, the images are said to be just resolved.

Resolving power of a microscope

Let us consider, a point is illuminated by light of wavelength [latex]\lambda[/latex] and it is observed through microscope. If [latex]\theta[/latex] is the semi vertical angle, the least distance between two objects which can be resolved is

[latex]d = \frac{\lambda}{2\mu sin\theta}[/latex]

Where, [latex]\mu[/latex] = refractive index of the medium between objective lens and object.

The resolving power of microscope is the reciprocal of limiting distance. i.e.

Resolving power = [latex]\frac{1}{d} = \frac{2\mu sin\theta}{\lambda}[/latex]

Resolving power of telescope

It is defined as the reciprocal of smallest angular separation between two distinct objects whose images are separated in the telescope. The angular separation is

[latex]d\theta = 1.22\frac{\lambda}{D}[/latex], where, [latex]d\theta[/latex] = angle made at objective lens, [latex]\lambda[/latex] = wavelength of light and D = diameter of telescope.

The resolving power of telescope = [latex]\frac{1}{d\theta} = \frac{1}{1.22}\frac{D}{\lambda}[/latex]

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