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PHYSICS-I

PHYSICS-I. COURSE CODE: 10B11PH111. S 1. y n. d/2. O. d/2. S 2. D. THE INTERFERENCE FRINGES. P. d. At point P for maxima we must have S 2 P – S 1 P = n , n = 0,1,2,3…. L. If d  <D then S 2 PS 1 PD. Thus S 2 P+S 1 P=2D and S 2 P – S 1 P = n .

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PHYSICS-I

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  1. PHYSICS-I COURSE CODE: 10B11PH111

  2. S1 yn d/2 O d/2 S2 D THE INTERFERENCE FRINGES P d At point P for maxima we must have S2P – S1P = n, n = 0,1,2,3… L

  3. If d<D then S2PS1PD. Thus S2P+S1P=2D and S2P – S1P = n

  4. Thus position of nth bright fringe on the screen So we get, position of nth darkfringe on the screen Distance between any two consecutive bright or dark fringes (Fringe width) Fringe width…….. Proportional to the wavelength of source. Proportional to the distance of the screen from plane of S1 and S2. Inversely proportional to the distance between S1 and S2.

  5. Separation between dark and bright fringes • O is equidistant from S1 and S2 so light waves superposed at O are in phase so light intensity at O will be maximum. • At O, we observe the central bright fringe. For this fringe n=0 y = 0. So central bright fringe will be referred as zeroth order bright fringe. S1 O S2

  6. Young’s Double-Slit Experiment Q.1. Two straight narrow slits at 3 mm apart are illuminated by a mono chromatic light (590 nm) and the interference fringes are viewed on screen at 60 cm away from slits. Find fringe width. Comment on the shape of the fringes. Ans: 0.118mm Q.2. Sodium light (589 nm) falls on double slit exp, having d=2mm, and D= 4 cm. Locate 10th bright fringe on screen. Ans: 0.1178mm Q.3. A light source emits two wavelength (430 & 510 nm). Find the separation b/w 3rd bright fringe for both wavelength? Ans: 1.44cm

  7. Young’s Double-Slit Experiment Q4. Green light (510 nm) falling on double slit exp. If overall separation of 10 fringes on screen (200 cm away from slits) is 2 cm, what is the separation of slits? Ans: 0.5mm Q.5. Red light (664 nm) is used in Young’s experiment with slits separated by 0.000120 m. The screen is located a distance 2.75 m from the slits. Find the distance on the screen between the central bright fringe and the third-order bright fringe. Ans: 0.0456m

  8. White Light and Young’s Experiment: Quest: Why does Young’s experiment separate white light into its constituent colors? Quest: In any group of colored fringes, such as the two singled out, why is red farther out from the central fringe than green is? Quest: Why is the central fringe white?

  9. Numerical • In a certain region of interference 45th order maximum for the wavelength  = 5893 Å are obtained. What will be the order of interference at the same place for (a)  = 4820 Å, (b)  = 7576 Å. Ans: (a) 55th (b)35th

  10. c a S b 179º   Fresnel’s Biprism It consists of two thin acute angled prisms joined at the bases. It is constructed as a single prism of obtuse angle of 179º. The acute angle  on both side is about 30´ called angle of prism. The prism is placed with the refracting edge in a way such that Sa is normal to the face bc of prism.

  11. Screen c A E Fringes of equal width S O d a F B Z2 Z1 b D

  12. Z1 = distance of biprism from source Z2 = distance of screen from biprism D= Z1+Z2 Screen c A E Fringes of equal width S O d a Eyepiece F B Z2 Z1 b D • S is source • A and B are coherent sources (Virtual) obtained through ‘ab’ and ‘ac’ surface of biprism Observer measures fringe width. Ques: Can we determine Wavelength of source through this setup ? Ans: Yes d = distance between the sources A and B. Can be determined by two methods (Displacement Method and deviation method)

  13. Displacement Method: By two positions of convex lens. from magnification formula, for L1 we have Multiplying (1) and (2) we get Similarly, for L2 Substituting the values of , d and D we can calculate the value of wavelength () of given monochromatic light.

  14. Deviation Method For small angles µ δ is angle of deviation (d1+d2) From right angle triangle and equation (1). α1 z1 z1 D Hence α2 If base angles are different, then

  15. Fringes with white light When white light is used the center fringe at O is white since all waves will constructively interfere here while the fringes on the both side of O are colored because the fringe width () depends on wavelength of light. For green light, For red light,

  16. Numerical • The inclined faces of a glass prism of refractive index 1.5 make an angle of 1o with the base of the prism. The slit is 10 cm from the biprism and is illuminated by light of  = 5900 Å. Find the fringe width observed at a distance of 1 meter from the biprism. Ans: Fringe width= 0.338 mm (Note: Use angle α in radian)

  17. In a biprism experiment with sodium light, bands of width 0.1963 mm are observed at 100 cm from the slit. On introducing a convex lens 30 cm away from the slit, two images of the slit are seen 0.7 cm apart, at 100 cm distance from the slit. Calculate the wavelength of sodium light. • Ans: 5889 Å • More numerical on • Wave representation; Amplitude, phase, frequency • Intensity distribution; Imax, Imin, ….. • Biprism, double slit interference

  18. Summary • Lecture-1: • Introduction • Lecture-2: • Superposition of two waves, Imax, Imin. • Double slit experiment (Division of wave front), Expression for fringe width. • Lecture-3: • Fresnel’s Biprism- Determine λ of sodium light. • Fringe width • Determination of ‘d’( using convex Lens). • Dependence of ‘d’ on Biprism. • Using white light source. • Displacement of fringes.

  19. P t, µ S1 y d D S2 In Double slit experiment: Insert a thin transparent glass sheet of thickness t and refractive index  in the path of one beam. Fringe pattern will remain same WHAT WILL HAPPEN? or Fringe pattern will shift Because S1P ray will travel more path. hence effective path difference becomes Prove it.

  20. P t, µ S1 y d D S2 Time required for the light to reach from S1 to the point P is where C0 is the velocity of light in air and C its velocity in the medium Time required for the light to reach from S2 to the point P is

  21. So the path difference between the beams reaching P, from S1 and S2 (Δ) As we know So the path difference will be If P is the centre of the nth bright fringe, then At n = 0 the shift y0 of central bright fringe is

  22. It means that the introduction of the plate in the path of one of the interfering beams displaces the entire fringe system through a distance Note: This displacement is towards the beam in the path of which the plate is introduced. Knowing the distance through which the central fringe is shifted, D, d and  the thickness of the material t can be calculated. We have to use white light to determine the thickness of the material. For monochromatic light central fringe will similar to other bright bright fringe. For white light central fringe is white.

  23. Displacement of fringes µ1,t1 • Determine condition of net path difference. C’ S1 C S2 Screen µ2,t2 Case 1: If µ1= µ2=µ and t1>t2 ,Δ is positive ( upward shift) ort2>t1,Δ is negative ( downward shift) Case 2: If t1=t2= t .

  24. THE LLOYD’S MIRROR ARRANGEMENT Light directly coming from the slit S1 interferes with the light reflected from the mirror forming an Interference pattern in the region BC of the screen. Real source Mirror Virtual source L L’ Note: The central fringe will be dark.

  25. Numerical-1 • In Lloyd’s single mirror interference experiment, the slit source is at a distance 2 mm from the plane of mirror. The screen is kept at a distance of 1.5 m from the source. Calculate the fringe width for the wavelength 6000 Å of light used. Ans: 0.225mm. β=λD/d S Mirror Screen

  26. Summary • Lecture-1: • Introduction • Lecture-2: • Superposition of two waves, Imax, Imin. • Double slit experiment (Division of wave front), Expression for fringe width. • Lecture-3: • Fresnel’s Biprism- Determine λ of sodium light. • Displacement of fringes due to one transparent glass sheet. • Displacement of fringes due to two transparent glass sheets.

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