QUIZ 24(TEST 4) AND (TEST 3)

4, 20, 28, 30, 42, 44, 46(EC) 54, 56, 58
4. 
Let d*sin theta = m*lambda, where m = , 0, 1, 2, 3, 4, .... for theta greater than or equal to zero. d= slit-to-slit distance. This is the constructive interference  criteria. 

Now, when theta is very small, 
sin theta = tan theta = Ym/L, where m = 0, 1, 2, 3, 4, .....  . L is given. Thus Ym = L*tan theta = L*m*lambda / d. That is,
Ym = L*m*lambda / d,  where m = 0, 1, 2, 3, 4, .....  . 

To get the vertical distance between adjacent bright spots, just subtract consecutive terms in the series: Y(m+1) - Ym = L*lambda / d . 
20.  See my class notes for the condition for minima:  a*sin theta = m*lambda, where a = slit width and  m = +/-1, +/-2, +/-3, +/-4, ... for positive or negative theta.  See equation 24-3b, derived  during  lecture sessions one recent Friday afternoon or Monday night. 

To get the  central maximum's width, you  make a similar small angle approximation:  sin theta = tan theta = Ym/L, where m = +/-1, +/-2, +/-3, +/-4, .....  . L is given.  Thus Ym = L*tan theta = L*m*lambda / a. 

That is,
Ym = L*m*lambda / a,  where m =  +/-1, +/-2, +/-3, +/-4, .....  . To get the vertical distance between  adjacent dark  spots---one above and the other below the central maximum---subtract these two consecutive terms in the series: 
Y1 - Y-1 = 2*L*lambda / a .  
28. d*sin theta = m*lambda, where m = 0, +/-1, +/-2, +/-3, +/-4, .....  .   Set m = 3.  The angle is given. Find lambda.  Note: d = 1 cm/3500. Convert to meters (m). 
30. d*sin theta = m*lambda, where m = 0, +/-1, +/-2, +/-3, +/-4, .....  .   d =  1 cm/8300. Convert to meters (m). For a given wavelength, the maximum value of m produces an angle closest to 90 degrees.  White light has wavelengths ranging from 400 nm to 700 nm.  When theta = 90 degrees, d = m*lambda or  m = d / lambda  = maximum m-value. 

For  lambda = 400 nm,  m = 10-2 m/[8300*400x10-9 m]  = 3.01
For  lambda = 700 nm,  m = 10-2 m/[8300*700x10-9 m]  = 1.72

This shows that at the minimum wavelength, three orders (3 bright spots on either side of the central maximum) appear; for the maximum wavelength, only one order (1 bright spot on either side of the central maximum) appears. Thus, for the  highest wavelengths of visible light, only one order appears.     

This means that for some long wavelengths only one order will be seen; for some short wavelengths,  three will be seen. 
Bottom line: Only one complete order can be seen for all wavelengths. 
These problems incorporate  elements of the  air-glass-air  or glass-air-glass scenarios covered in class. In these cases, our normal definitions for constructive and destructive interference were reversed. We had 2*t = m* lambda for   destructive interference and 2*t = (2m' + 1)*lambda for   constructive interference.  Please note that lambda is computed in the  thin film layer, meaning lambda = lambdao/n. where lambdao = wavelength computed in vacuum and n is the film's index of refraction. Note:  t = thin film thickness.  Below we will have variations of these themes. 
42.  Here we have an air-coating-glass scenario where n for air = 1, n for coating = 1.25 and n for glass = 1.52. At the first interface, between air and coating, there is an 180 degree phase shift since the light is traveling from low n to high n. At the coating-glass interface  another  180 degree shift occurs since again the light travels from lower n to higher n. The two shifts together produce 360 degrees which translates to zero trigonometrically. 

Bottom line, we have the "normal" definitions: 
We have 2*t = m* lambda for   constructive interference and 2*t = (2m' + 1)*lambda for   destructive interference.  Please note that lambda is computed in the  coating  layer, meaning lambda = lambdao/n,  where for example if  lambdao = 570 nm and  n = 1.25, then lambda = 570 nm/1.25.  Be careful. 

Let's find the minimum non-zero value of t for bright green light:  2*t = m* lambda. Set m = 1. Solve for t = 
570 nm/[1.25*2] = 228 nm. Now if you set a coating this thick, let's estimate how much of the visible spectrum, ranging from 400 nm to 700 nm,  will experience constructive interference.   

m = 2*t/lambda = 2*228 nm/lambda = n*2*228 nm/lambdao    ,  where n = 1.25.  Note: When lambdao = 570 nm, m = 1.  
Now if  lambdao > 570 nm, m <1.  So wavelengths between 570 nm and 700 nm will experience no constructive interference. So what fraction of the white light will experience no constructive interference?  Hint: Compute 
(700 - 570)/(700 - 400).   

44. This problem incorporates   the  glass-air-glass scenario covered in class. In this case, our normal definitions for constructive and destructive interference are  reversed. We have 2*t = m* lambda for   destructive interference and 2*t = (2m' + 1)*lambda for   constructive interference.  Please note that lambda is computed in the  thin film layer, meaning lambda = lambdao/n. where lambdao = wavelength computed in vacuum and n is the film's index of refraction. In this case n = 1, since the film is air.  Note:  t = thin film thickness.  

Now, it can be shown that when t = 0, say at the left end of fig.  24-33, a dark line occurs indicating destructive interference. Thus,  2*t = m* lambda for   destructive interference works even when m = 0, which corresponds to t = 0. What we want to do now  is count the number of values of m between the left and the right end, inclusive, that exist for destructive interference.  We are given that 28 dark lines exist, with  m = 0 at the left  and m = 27 at the right end. From this data we can get t: At the right end, 2*t = 27*lambda, where lambda is given.  Solve for t. 
46. This problem is EC because it belongs in a different section of the problems; it should be in section 24-10. The problem 54  hint  below will help.  See page 687, fig. 24-48. Use equation 24-6a. My lecture notes should help.  In this case n2 is unknown and n1 = 1.43.  Plug in 59 degrees for the angle and solve for n2. Is it diamond?   
54. See page 687, fig. 24-48. Use equation 24-6a. My lecture notes should help.  
56. Initially both polarizers' axes are parallel. Let Io be the intensity of the light before going through the first polarizer. If this light is polarized along a direction making an angle theta1 with the first polarizer axis, the intensity after passing through polarizer 1 is I1 = Io*cos2  theta1.  We want the intensity to be reduced by one half  after passing through polarizer 2. We want to rotate polarizer 2 so its axis  makes an angle theta 2 with polarizer 1's axis. Thus we want I2 = I1/2 = I1*cos2 theta2 . That is: I1/2 = I1*cos2 theta2 . Cancel I1 and solve for theta2.  
58.The  polarizers' axes make a  40 degree angle  with each other.  Let Io be the intensity of the light before going through the  first polarizer. Assume this light is polarized along a direction making an angle theta1 with the first polarizer axis. The intensity after passing through polarizer 1 is I1 = Io*cos2  theta1.  The intensity is reduced to 0.15  after transmitting through polarizer 1 and 2. Since polarizer 2's  axis  makes a 40 degree angle with polarizer 1's axis,  we have I2 =0.15 Io = I1*cos2 40.  That is: 0.15Io = Io*cos2  theta1*cos2 40 . Cancel Io and solve for theta1.