Diffraction The phenomenon of bending of light round

$Diffraction The phenomenon of bending of light round the corner of an obstacle and$

Diffraction The phenomenon of bending of light round the corner of an obstacle and their spreading into the region of geometrical shadow is called diffraction and the distribution of light intensity resulting in dark and bright fringes is called diffraction pattern.

$Diffraction The diffraction phenomenon is of two types: 1) Fresnel diffraction: The source of$

Diffraction The diffraction phenomenon is of two types: 1) Fresnel diffraction: The source of light and the screen are at finite distance from the diffracting aperture. Source Aperture 2) Fraunhofer Screen diffraction: The source and the screen are at infinite distance from the aperture (Using convex lens) Source Note: Wavelength of light must be comparable to the size of an obstacle

$Fraunhofer Diffraction Fresnel Diffraction Wave fronts Planar wave fronts Cylindrical wave fronts Observation distance$

Fraunhofer Diffraction Fresnel Diffraction Wave fronts Planar wave fronts Cylindrical wave fronts Observation distance is infinite. In practice, often at focal point of lens. Source of screen at finite distance from the obstacle. Fixed in position Move in a way that directly corresponds with any shift in the object. Movement of diffraction pattern Surface of calculation Diffraction patterns Fraunhofer diffraction Fresnel diffraction patterns on spherical surfaces. flat surfaces. Shape and intensity of a Change as we propagate Fraunhofer diffraction pattern them further ‘downstream’ of stay constant. the source of scattering.

Resultant of ‘N’ simple harmonic waves of equal Amplitude, Periods and Phases increasing in arithmetic progressions

Let us suppose that the number of vibrations are infinitely large but the amplitude ‘a’ and the common phase ‘ ’ are infinitely small, so that ‘Na’ and ‘N ’ are finite. Also suppose N =2 Then (because /n is very small) From eq. (6) (since for large N, )

$Single slit diffraction Fraunhofer diffraction b$

Single slit diffraction Fraunhofer diffraction b

Position of maxima and minima 1. PRINCIPLE MAXIMA The resultant amplitude is given by So maximum value of resultant amplitude R is A So Imax = I 0 = A 2

2. Minima Re-consider the equation The resultant amplitude R will be minimum when sin = 0 but 0 Therefore the condition for minima is Where n = 1, 2, 3…. gives the condition of 1 st, 2 nd, 3 rd, minima. Here n 0, because for n = 0, = 0 which corresponds central maxima.

Maximum Possible orders Let us consider the condition for minima The first minima would occur (n = 1) at And the second minima (n = 2) would occur at Further, Since, Therefore,

3. Secondary maxima Apply the method of finding maxima and minima In order to determine the position of secondary maxima differentiate the above equation with respect to Thus either or

3. Secondary maxima Consider From the graph, roots which satisfy the equation are (the more exact values are = 1. 430 , 2. 46 , 3. 47 , 4. 471 …) exclude β = 0 because central maximum

3. Secondary maxima So the direction of secondary maxima are approximately:

$INTENSITY CALCULATIONS Intensity of diffraction pattern is given as$

INTENSITY CALCULATIONS Intensity of diffraction pattern is given as

$In a single slit Fraunhoffer diffraction, show that more than 90% of transmitted intensity$

In a single slit Fraunhoffer diffraction, show that more than 90% of transmitted intensity is in the central maxima.

$INTENSITY CALCULATIONS Intensity of diffraction pattern is given as Intensity of principle maxima: for$

INTENSITY CALCULATIONS Intensity of diffraction pattern is given as Intensity of principle maxima: for 0 Imax = I 0 = A 2 The intensity of secondary maxima: Substituting = ± 3 /2 1 st secondary maxima (n =1), Thus I 1 is 4. 6% of I 0. Similarly, substituting = ± 5 /2 2 nd secondary maxima (n = 2) Thus I 2 is 1. 61% of I 0. Relative intensities of successive maxima 1: 4 /9π2 : 4 /25π2: 4 /49π2

$3. Spread of central diffraction maximum Direction of first minima is given by$

3. Spread of central diffraction maximum Direction of first minima is given by

$Single Slit Fraunhofer diffraction: Effect of slit width • Principle maxima, = 0, I$

Single Slit Fraunhofer diffraction: Effect of slit width • Principle maxima, = 0, I =I 0 =A 2 • Minima, sin = 0, = n • , sin = n /b First minima at sin = /b As the opening size gets smaller, the wave front experiences more and more curvature DISTINCTION BETWEEN SINGLE SLIT DIFFRACTION PATTERN AND DOUBLE SLIT INTERFERENCE PATTERN

$Example: Calculate the wavelength of light whose first diffraction maximum in the diffraction pattern$

Example: Calculate the wavelength of light whose first diffraction maximum in the diffraction pattern due to single slit falls at 30 deg and coincides with the first minimum for red light of wavelength 650 nm. Ans: 433. 33 nm

$Minima Secondary maxima Principal maxima Secondary maxima Minima Schematic illustration of diffraction pattern from$

Minima Secondary maxima Principal maxima Secondary maxima Minima Schematic illustration of diffraction pattern from a single slit.

DISTINCTION BETWEEN SINGLE SLIT DIFFRACTION PATTERN INTERFERENCE PATTERN AND DOUBLE SLIT

$Single-Slit Diffraction: Slit Width b • http: //webphysics. ph. msstate. edu/javamirror/ipmj/java/slitdiffr/index. html • http:$

Single-Slit Diffraction: Slit Width b • http: //webphysics. ph. msstate. edu/javamirror/ipmj/java/slitdiffr/index. html • http: //www. lightlink. com/sergey/java/slitdiffr/index. html Larger a Slits farther Maxima closer Smaller a Slits closer Maxima farther

Practice Problem - 1 Coherent laser light of wavelength 633 nm is incident on a single slit of width 0. 25 mm. The observation screen is 2. 0 m from the slit. (a) What is the width of the central bright fringe? (b) What is the width of the bright fringe between the 5 th and 6 th minima? Answer: (a). 01 m, (b) 0. 5 cm Practice Problem - 2 Monochromatic light is incident on a single slit of width 0. 30 mm. On a screen located 2. 0 m away, the width of the central bright fringe is measured and found to be near 7. 8 mm. What is the wavelength of the incident light? Answer: 585 nm Practice Problem - 3 The Fraunhofer diffraction pattern produced by a narrow slit of width 0. 3 mm is observed on a screen placed at the focal plane of a lens of focal length 25 cm. Calculate the linear width of the first secondary maxima and the linear separation of the first two secondary maxima. Assume λ = 600 0 Å and the lens is placed very close to the slit. Answer: (a) 0. 05 cm, (b) 0. 05 cm

$In other words, the diffraction pattern acts like an envelope containing the interference pattern.$

In other words, the diffraction pattern acts like an envelope containing the interference pattern. Considering only interference: The image below is taken from the central maximum area of a display.

$The resultant intensity pattern due to a combination of both the single-slit diffraction pattern$

The resultant intensity pattern due to a combination of both the single-slit diffraction pattern and the double slit interference pattern. The amplitude of the diffraction pattern modulates over the interference pattern.

FRAUNHOFER DOUBLE SLIT DIFFRACTION

$Missing orders in double slit diffraction$

Missing orders in double slit diffraction

Missing orders of int. pattern (m) Because of min of diff pattern (n)

Consider the case when b = 8. 8 10– 3 cm, d = 7. 0 10– 2 cm, and = 6. 328 10– 5 cm. How many interference minima will occur between the two diffraction minima on either side of the central maximum? In the experimental arrangement the screen was placed at a distance of 15 ft. Calculate the fringe width.

$Numerical: Double slit diffraction If λ=0. 5µm, slit width ‘b’=2 µm and separation between$

Numerical: Double slit diffraction If λ=0. 5µm, slit width ‘b’=2 µm and separation between slits ‘a’=2 µm. Calculate Ans: 3. 58 o, 10. 8 o Ans: 7. 18 o , 14. 47 o Ans: 3 (m=0, +1, -1) Ans: m=2 order is missing

DIFFRACTION FROM N NUMBER OF SLITS (DIFFRACTION GRATING)

$Intensity of diffraction pattern is given as MAXIMA (Principal Maxima) Where, m = 0$

Intensity of diffraction pattern is given as MAXIMA (Principal Maxima) Where, m = 0 , 1, 2, 3, … Intensity of the principal maxima MINIMA Where n can have the all integer values except 0, N, 2 N, 3 N…

SECONDARY MAXIMA Intensity of the secondary maxima

WIDTH OF A PRINCIPAL MAXIMA Angular width of a principal maxima is equal to the angular separation between first two minima on either side of the principal maxima.

WIDTH OF A PRINCIPAL MAXIMA (m. N+1)th order minima mth order principal maxima (m. N-1)th order minima Slit Screen

For very small d m and Thus, But, Therefore,

Here 2 d m is the width of mth order principal maxima. From the equation it is obvious that larger the number slits sharper will be principal maxima.

Calculation of missing ORDERS IN N slit DIFFRACTION PATTERN 1) If d=2 b, m = 2 n= 2, 4, 6…. Since n=1, 2, 3… Absent orders are m = 2, 4, 6, …. . of principal maxima Similarly 2) If d=3 b, m = 3 n = 3, 6, 9…. Since n = 1, 2, 3… 3) If d=4 b, m = 4 n= 4, 8, 12…. Since n = 1, 2, 3…

HIGHEST POSSIBLE ORDER WITH A DIFFRACTION GRATING ØThe maximum value for ØSubstituting this value in equation Here n is the highest order for principal maximum visible. If n is not an integer, then the highest possible order is given by the integer lower than n.

N slit as DIFFRACTION GRATING N - SLITS d numbers of lines per inch If there are 15000 lines per inch then grating element ‘a+b’ b a d = a+b

DIFFRACTION GRATING Typically there are 15000 lines in 2. 54 cm in the laboratory grating. Thus only two orders are possible in grating.

LASER DIFFRACTION

ØIf Then not possible Hence 0 th order will be present and all the higher order spectra will be absent. Ø If ØSimilarly 1 st, 2 nd, 3 rd orders will be visible respectively. ØIn general if only (n-1)th orders will be obtained i. e. nth order and all other higher orders will be absent.

$Dispersive power of grating Rate of change of angle of diffraction with respect to$

Dispersive power of grating Rate of change of angle of diffraction with respect to change in the wavelength of light. For principal maxima of grating we know Where, d = a+b Note: Sometimes ‘dθ’ is called angular dispersion. What about Linear dispersive power of grating?

Linear dispersive power If dx is the linear separation of two spectral lines differing in wavelength dλ in the focal plane of a lens of focal length f, then λ 1 dx f dθ f λ 2 λ 1 f dθ f dx λ 2

For two wavelengths

Overlapping orders: many wavelengths Colour & Wavelength Orders d sinθ = nλ Red (7000Ǻ) 3 3 x 7000Ǻ = 21000Ǻ Green (5250Ǻ) 4 4 x 5250Ǻ 5 th order of principal maxima of 4000Ǻwill Violet 4200Ǻ coincide with 4 th Order 5 principal 5 x maxima of 5000Ǻwavelength. (4200Ǻ)

Irradiance is the power of electromagnetic radiation per unit area (radiative flux) incident on a surface.

Quest: Two spectral lines have wavelengths and +d respectively. Show that if d their angular seperation d in a grating spectrometer is given by: Where (a+b) is the grating element and n is the order at which the lines are observed. Sol:

$Intensity of diffraction pattern is given as MAXIMA (Principal Maxima) Where m = 0$

Intensity of diffraction pattern is given as MAXIMA (Principal Maxima) Where m = 0 , 1, 2, 3, … Intensity of the principal maxima (N-1) Secondary Minima Between Two Successive Principal Maxima Where n can have the all integer values except 0, N, 2 N, 3 N… (N-2 )Secondary Maxima Between Two Successive Principal Maxima Intensity of the secondary maxima