n = c/v

l_of = c                  (Equation 1)

l_nf = v                 (Equation 2)

l_o/l_n = c/v = n  or  l_n=l_o/n.

1. Constructive interference occurs when a crest falls on a crest and a trough falls on a trough. Destructive interference occurs when a crest falls on a trough.
   
   
2. For constructive interference, the path difference must equal an integral number of wavelengths. For destructive interference, the path difference must equal half-integral number of wavelengths.
   
   
3. In Fig. 1 below, the wave from source 1 has traveled a distance S₁P to point P and the wave from source 2 has traveled a distance S₂P to point P. The path difference = S₁P - S₂P.
   
   
   
   
4. For constructive interference, S₁P - S₂P = ml, m = 0, 1, 2, etc.
   
   
5. For destructive interference, S₁P - S₂P = (m + 1/2)l.
   
   
   

y₁(r₁,t) = A cos (2pt/T - 2pr₁/l)

y₂(r₂,t) = A cos (2pt/T - 2pr₂/l).

1. Every point on a wave vibrates with simple harmonic motion of the same amplitude. The phase 2pr/l of each point depends on the distance r from the source.
   
   
   1. Points ml (m = 1, 2, 3 . . . ) apart have a phase difference of 2p(ml/l) = 2p, 4p, . . .(360^o, 360^o, . . .), which is equivalent to no phase difference.
      
      
   2. Points 2(m + 1/2)l (m = 0, 1, 2, 3, . . .) apart have a phase difference of 2(m + 1/2)p = p, 3p, . . . . (180^o, 540^o, . . .)
      
      
2. The 2pr/l term is the phase of each wave and the difference
   [2pr₁/l - 2pr₂/l] is the phase difference DF between the waves.
   DF = (2p/l)(r₁ - r₂), where (r₁ - r₂) is the path difference.
   
   
   1. A phase difference of 2pm (2p, 4p, . . .) gives constructive interference. You can also see this since (2p/l)(r₁ - r₂) = 2pm is equivalent to (r₁ - r₂) = ml.
      
      
   2. A phase difference of 2p(m + 1/2) (p, 3p, . . .) gives destructive interference. You can also see this since (2p/l)(r₁ - r₂) =
      2p(m + 1/2) is equivalent to (r₁ - r₂) = (m + 1/2)l.
      
      
   3. If the waves travel through media of different index of refraction n₁ and n₂,  DF = (2p)(r₁/l_i - r₂/l₂), or since l₁ = l/n₁ and l₂ = l/n₂,  DF = (2p/l)(r₁n₁- r₂n₂). There will even be a phase difference in this case if r₁ = r₂ = L. Then DF = (2pL/l)(n₁- n₂).
      
      

1. If point P is one of maximum intensity (constructive interference),

   S₁B = ml.

   If point P is one of minimum intensity (destructive interference),

   S₁B = (m + 1/2)l.

   
   
2. From triangle BS₁S₂, we see that sin Q = S₁B/d  or  d sin Q = S₁B.
   
   
   1. For constructive interference, d sin Q = ml.
      
      
   2. For destructive interference, d sin Q = (m + 1/2)l.
      
      

1. For y_m << L,  sin Q is approximately = y_m/L = tan Q
   
   
2. When y_m << L
   
   
   1. For constructive interference,
      
      
      1. sin Q = ml/d = y_m/L  or  y_m = mlL/d
         
         
      2. y_{m + 1} – y_m = [(m + 1) – m]lL/d = lL/d =
         distance between maxima.
         
         
   2. For destructive interference,
      
      
      1. sin Q = (m + 1/2)l/d = y_m/L  or  y_m = (m + 1/2)lL/d
         
         
      2. y_{m + 1} – y_m = [(m + 1 + 1/2) – (m + 1/2)]lL/d = lL/d = distance between minima.
         
         

1. In Fig. 5a above, the incident wave I is reflected at the first interface of a less dense-more dense medium, where the index of refraction in the incident medium  n₁ < n,  the index of refraction in the film. There is an 180^o or p change in phase at the first interface. There is no change in phase for the transmitted wave T. And there is no change in phase at the second interface of a more dense-less dense medium with  n₃ < n.
   
   The "real" path difference for 5a or 5b above is 2t, where t is the thickness of the medium. A change of phase of p for the incident ray is equivalent to an additional path difference of l_n/2, where l_n is the wavelength in the film. For Fig. 5, the index of refraction of the film is n. Remember that the wavelength in a medium l_n = l/n where l is the wavelength in a vacuum.
   
   The condition for
   
   
   1. maximum:

      2d = (m +1/2)l_n   or
      2d = (m +1/2)l/n   or
      2nd = (m +1/2)l
      

   2. minimum:

      2nd = ml
      

   3. Because of the change in phase of p, the conditions for maxima and minima have been reversed.
      
      
2. In Fig. 5b, the incident wave I is reflected at the first interface of a less dense-more dense medium where the index of refraction in the incident medium n₁ < n,  the index of refraction in the film. There is an 180^o or p change in phase at the first interface. There is no change in phase for the transmitted wave T. There is another change of phase of p at the second interface since n₃ > n. The total phase change is 360^o or 2p.  That is equivalent to no phase change overall.
   
   The condition for
   
   
   1. maximum:

      2nd = ml
      

   2. minimum:

      2nd = (m + 1/2)l
      

1. Since the condition for a maximum is the same for N slits with a separation d as it is for 2 slits with a separation d, you get a maximum when sin Q= ml/d = Nml/a.
   
   
2. Since N is a very large number, sin Q will be greater than 1 unless m = 0. There is only one principal maximum and that occurs at the center of the screen.
   
   

1. The width of the slit is a. Think of light coming from an element of length dy at the top of the slit and an element of length dy half way down on the slit as shown in Fig. 7b above. Obviously there is a path difference between them when the light from the top and middle of the slit arrive at P.
   
   
   1. Destructive interference occurs if this path difference for the element at the top of the slit and the element at the center of the slit is l/2. This will also be true for other elements that you pair as you go down the slit with path differences of l/2.
      
      
   2. In the little triangle on the left, for destructive interference, sin Q = (l/2)/(a/2) = l/a, the condition for the first minimum.
      
      
2. For the second minimum, divide the slit into four parts and match elements that give a path difference of l/2. The second minimum occurs for sin Q = (l/2)/(a/4) = 2l/a.
   
   
3. Continuing in this way, you find for the mth minimum:

   sin Q = ml/a

1. From Fig. 7 above,  tan Q= y/L.   For small Q,  tan Q ≈ sin Q.
   
   
2. From VI, E, 3 above, sin Q = ml/a.
   For sin Q ≈ tan Q,   ml/a = y_m/L,   or
   y_m = mlL/a,  where y_m corresponds to the mth mimimum.
   
   
3. The distance between minima =
   Dy = y_m+1 - y_m = (m + 1)lL/a - mlL/a = lL/a.
   
   
4. Since the width of the central maximum = the distance between the first minimum on the right and the first minimum on the left of the central maximum, the width of the central maximum = 2lL/a.
   
   

1. Assume light from two point sources far from a slit illuminates the slit. Light diffracting at the slit produces two single-slit diffraction patterns, one for each source as shown in Fig. 9 below.
   
   
   
   
2. Because the sources are at different angular positions, the central maxima of the diffraction patterns overlap.
   
   
3. The result in Fig. 9 is one big blob and you can't distinguish two separate objects.
   
   

1. If the angular separation of the two objects is great enough, you see a separation of the two maxima and can distinguish the existence of two objects.
   
   
2. The two peaks are barely distinguishable if the central maximum of one coincides with the first minimum of the other. We found earlier that sin Q = l/a for the first minimum in single-slit diffraction. The angular separation between the diffraction peaks equals the angular separation between the sources. In most optical systems, the wavelength of the light is much smaller than the aperture. In the small angle approximation, sin Q is approximately equal to Q, the Rayleigh condition that the two sources be just resolvable gives for a slit  

   Q_min = l/a

3. The Rayleigh criterion for circular apertures is                                  

   Q_min = 1.22 l/D,

   where D is the aperture diameter.
   
   

1. The direction of the polarization is parallel to the electric field of the electromagnetic wave and perpendicular to the direction of the propagation of the wave.
   
   
2. Light waves are transverse waves that may be polarized.
   
   

1. The electric field vectors can be resolved into two vectors perpendicular to each other.
   
   
2. When you send light through a polaroid, the electric field vector perpendicular to the transmission axis is absorbed, leaving only the component along the transmission axis of the polaroid.
   
   
3. When unpolarized light with intensity of the incident light equal
   to I_i is sent through a polaroid, the intensity of the light is reduced by a factor of 2. The transmitted intensity I_t = I_i/2.
   
   

1. Figure 13a below represents polarized light with incident electric field E_i approaching a polaroid. The direction of the transmission axis is indicated with a dash line. The transmitted electric field E_t is smaller and rotated through an angle Q along the transmission axis.
   
   
   
   
2. Figure 13b above gives an end view. The component of E₁ along the transmission axis is the transmitted electric field E_t = E_i cos Q.
   
   
3. The Poynting Vector = the intensity of the light I = E_max²/2µ_oc.
   The intensity is proportional to the electric field, that is,
   I = kE², where k is a constant.
   
   
   1. The initial intensity I_i = kE_i²
      
      
   2. The transmitted intensity I_t = kE_t² = kE_i² cos² Q
      
      
   3. I_t/ I_i = cos² Q
      
      

1. The incident ray in Fig. 15 below is unpolarized. It has components of the electric field perpendicular to the plane of the paper represented by a dot • and a parallel component in the plane of the paper represented by an arrow .
   
   
   
   
2. When the angle between the reflected ray and the transmitted ray is 90^o, the reflected ray cannot contain any of the parallel component because then the electric field would be in the direction of the propagation and that is not possible. The transmitted ray has parallel components of the electric field and some perpendicular components of the electric field.
   
   
3. When the reflected ray has only components of the electric field perpendicular to the paper, the reflected light is plane polarized. The angle of incidence when you get plane polarized light in the reflected ray is called the Brewster angle Q_B.
   
   
   1. Snell's law states that n_i sin Q_i = n_t sin Q_t.
      
      
   2. For Q_i = Q_B,  Q_r + Q_t = 90^o.
      
      
   3. Since Q_i = Q_r,  and for this case Q_i = Q_B,
      Q_B + Q_t = 90^o  or  Q_t = 90^o - Q_B
      
      
   4. Snell's law becomes n_i sin Q_B = n_t sin (90^o - Q_B) = n_t cos Q_B  or
      tan Q_B = n_t/n_i.
      
      
   5. If the light is incident from air with n_I = 1,
      tan Q_B = n_t.