Summarized notes - 12th Science JEE - Interference and Diffraction
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This document discusses the principles of interference and diffraction of light waves. It defines constructive and destructive interference based on the path difference between light waves. The conditions for constructive and destructive interference are outlined. The document also discusses Young's experiment to demonstrate light interference and derive an expression for the bandwidth of interference fringes. It explains how the biprism experiment can be used to measure the wavelength of light using interference principles.
![9011041155 /09011031155
Condition
for
Destructive
Constructive
Interference
Interference
(Or
Conditions
and
for
Brightness & Darkness of a point)
a. Condition
for
Constructive
(brightness):1. Overlapping
2. Path difference = n λ = (2n) λ /2
[n = 0, 1, 2, ...........]
3. Phase difference
4. 4. Intensity
4
Interference](https://image.slidesharecdn.com/summarizednotes-12thsciencejee-interferenceanddiffraction-131120044147-phpapp02/85/Summarized-notes-12th-Science-JEE-Interference-and-Diffraction-4-320.jpg)
![9011041155 /09011031155
b. Condition
for
Destructive
Interference
(darkness):1. Overlapping
2. Path difference =(n – 1/2)λ
= (2n – 1) λ/2
[n = 1, 2, 3, 4..........]
3. Phase difference
4.
Intensity
5](https://image.slidesharecdn.com/summarizednotes-12thsciencejee-interferenceanddiffraction-131120044147-phpapp02/85/Summarized-notes-12th-Science-JEE-Interference-and-Diffraction-5-320.jpg)
![9011041155 /09011031155
Let xn, xn -1 = distances of nth and n - 1th dark
bands from O. Using these in (9).
Xn = (2n – 1) λ D/2d
Xn – 1 = [2(n + 1) – 1] λD/2d = (2n + 1) λD/2d
X = Band width of dark band
= Xn-1 – Xn
= (2nλD/2d) + (λ D/2d) λ – (2n D/2d) + (λ D/2d)
X = λ D/d
...................... (10)
11](https://image.slidesharecdn.com/summarizednotes-12thsciencejee-interferenceanddiffraction-131120044147-phpapp02/85/Summarized-notes-12th-Science-JEE-Interference-and-Diffraction-11-320.jpg)
![9011041155 /09011031155
Assuming the Expression for Path Difference
Obtain an Expression for Band Width of An
Interference Band
a. From above diagram, bright and dark bands are
placed alternately. Let, n = 0, 1, 2, 3....................
λ = Wavelength of the light used.
D = Distance between the sources and the screen.
d = Distance between the two sources.
For a bright band.
xn = n λ D/d and xn+1 = (n + 1) λ D/d
X = λ D/d
b. For a dark band.
Xn = (2n -1) λ D/2d
Xn-1 = [2 (n + 1) - 1] λ D/d = (2n + 1) λ D/2d
X = λ D/d
12](https://image.slidesharecdn.com/summarizednotes-12thsciencejee-interferenceanddiffraction-131120044147-phpapp02/85/Summarized-notes-12th-Science-JEE-Interference-and-Diffraction-12-320.jpg)
Summarized notes - 12th Science JEE - Interference and Diffraction
- 1. 9011041155 /09011031155 Interference and Diffraction Principle of Superposition of Waves a. Statement : When two or more waves arrive at a point in the medium simultaneously then each wave produces its own displacement independent of the other waves. The resultant displacement at that point is given by the vector sum of individual displacements produced by each wave. 1
- 2. 9011041155 /09011031155 Interference of Light Redistribution of light intensity in the region of medium is due to the physical process called as “Interference of light” Constructive Interference 2
- 3. 9011041155 /09011031155 Destructive Interference Path Length and Path Difference 3
- 4. 9011041155 /09011031155 Condition for Destructive Constructive Interference Interference (Or Conditions and for Brightness & Darkness of a point) a. Condition for Constructive (brightness):1. Overlapping 2. Path difference = n λ = (2n) λ /2 [n = 0, 1, 2, ...........] 3. Phase difference 4. 4. Intensity 4 Interference
- 5. 9011041155 /09011031155 b. Condition for Destructive Interference (darkness):1. Overlapping 2. Path difference =(n – 1/2)λ = (2n – 1) λ/2 [n = 1, 2, 3, 4..........] 3. Phase difference 4. Intensity 5
- 6. 9011041155 /09011031155 M.C.Q Q.1 In interference of light (a.1) Light energy is created (c.1) Light energy is redistributed Q.2 (b.1) Light energy is destroyed (d.1) Light energy is doubled Select the correct statement (a.2) Only longitudinal waves produce interference. (b.2) Only transverse waves produce interference. (c.2) Only standing waves produce interference. (d.2) Both longitudinal and transverse waves produce interference. Q.3 Two sources of light are said to be coherent if they emit light waves of the same (a.3) Frequency and speed (b.3) Wavelength and constant phase difference (c.3) Frequency and amplitude Q.4 (d.3) Intensity and frequency Which one of the following quantities is conserved in the interference of light waves? (a.4) Phase difference (b.4) Amplitude 6 (c.4) Intensity (d.4) Path difference
- 7. 9011041155 /09011031155 Young’s Experiments to Demonstrate the Phenomenon of Interference 7
- 8. 9011041155 /09011031155 Theory of Interference Band (Expression for bandwidth or fringe width of an interference band.) Expression for path difference S2P – S1P :In ∆PNS2 and ∆PMS. By Pythagoras theorem. S2P2 = S2N2 + PN2 = D2 + (x + d/2)2 ..................... (1) S1P2 = S1M2 + PM2 = D2 + (x - d/2)2 ...................... (2) S2P2 – S1P2 = (x + d/2)2 – (x – d/2)2 = x2 + 2x d/2 + d2/4 – x2 + 2x d/2 – d2/4 = 2xd ..................... (3) 8
- 9. 9011041155 /09011031155 If x, d < < D, then S1P ≈ S2P ≈ D Equation (3) becomes. 2D (S2P – S1P) = 2xd S2P – S1P = xd/D ..................... (4) For a point P to be bright. S2P – S1P = (2n) λ/2 .................... (5) From (3) and (4), xd/D = n λ x = n λ D/d Let xn, xn – 1 = distances of nth and n + 1th bright bands from O. 9
- 10. 9011041155 /09011031155 Using these in (6), xn = nλ D/d and xn - 1 = (n + 1)λ D/d X = Band width of bright band = xn+1 – xn = (nλ D/d) + (λ D/d) – (n λ D/d) X = λD/d ..................... (7) e. For a point P to be dark S2P – S1P = (2n – 1) λ /2 ..................... (8) From (3) and (8), xd/D = (2n – 1) λ/2 x = (2n – 1) λD/2d ...................... (9) 10
- 11. 9011041155 /09011031155 Let xn, xn -1 = distances of nth and n - 1th dark bands from O. Using these in (9). Xn = (2n – 1) λ D/2d Xn – 1 = [2(n + 1) – 1] λD/2d = (2n + 1) λD/2d X = Band width of dark band = Xn-1 – Xn = (2nλD/2d) + (λ D/2d) λ – (2n D/2d) + (λ D/2d) X = λ D/d ...................... (10) 11
- 12. 9011041155 /09011031155 Assuming the Expression for Path Difference Obtain an Expression for Band Width of An Interference Band a. From above diagram, bright and dark bands are placed alternately. Let, n = 0, 1, 2, 3.................... λ = Wavelength of the light used. D = Distance between the sources and the screen. d = Distance between the two sources. For a bright band. xn = n λ D/d and xn+1 = (n + 1) λ D/d X = λ D/d b. For a dark band. Xn = (2n -1) λ D/2d Xn-1 = [2 (n + 1) - 1] λ D/d = (2n + 1) λ D/2d X = λ D/d 12
- 13. 9011041155 /09011031155 Conditions for Obtaining Well Defined Steady Interference Pattern a. Statement :Two sources must be i. equally bright ii. monochromatic iii. coherent iv. as narrow as possible v. as closed as possible b. Explanation :i. This will not give a well-defined interference pattern. ii. This will not give a well defined interference pattern. iii. This produces unstable interference pattern. iv. These will not a well defined interference pattern. 13
- 14. 9011041155 /09011031155 v. Since X = λ D/d. For λ, D = constant X α 1/d. Therefore smaller the distance between the sources higher the bond width. This gives a well defined interference pattern. 14
- 15. 9011041155 /09011031155 Biprism Two thin prisms of extremely small refracting angle are connected bases to base. Such a prism formed is called as “Biprism”. Biprism is used for producing two coherent sources from a single source. Biprism Experiment This experiment is performed to find wavelength of monochromatic source of light. 15 unknown
- 16. 9011041155 /09011031155 a. Measurement of D :- Distance between slit and the eyepiece is measured directly from the scale on the optical bench. d. Measurement of X :- For this purpose vertical wire of the cross wire is made coincide with one edge of the white band and corresponding reading (xn) on the micrometer scale is recorded using slow motion screw. The cross wire is moved through known number of bands (n). Vertical wire is again made coincide with one edge of the bright band and corresponding heading (xn) on the micrometer is recorded. Mean band width X is obtained by the formula. 16
- 17. 9011041155 /09011031155 X = Xn – X0/n e. Measurement of d (method of conjugate foci) :S1, S2 are virtual coherent sources. Distance between these sources (d) cannot be measured directly. 1. A convex lens is interposed in between biprism is so adjusted that real, bright and enlarged images S1’ and S2’ of S1, S2 are observed in the focal plane of the eyepiece. Using slow motion screw distance (d1) between S1’ and S2’ is measured. u = object distance v = Image distance object size / Image size = object distance / Image distance 17
- 18. 9011041155 /09011031155 d/d1 = u/v ......................... (1) 2. Convex lens is displaced towards eyepiece and its position is so adjusted that real, diminished and reduced images S1’’ and S2’’ of S1, S2 are observed in the focal plane of the eyepiece. Using slow motion screw distance (d2) between S1” and S2” is measured. d/d2 = v/u .........................(1) Multiplying (1) by (2), d/d2 = u/v × v/u d2/d1d2 = 1 d2 = d1d2 d= d1d 2 18
- 19. 9011041155 /09011031155 f. Unknown wavelength l is calculated by :λ = X d/D = X d1d 2 / D • Ask Your Doubts • For inquiry and registration, call 9011041155 / 9011031155. 19