PHYS 1112
Chapter 24 Problems

Selected Solutions $$

1. In order to get the light to be strongly reflected by this layer, we need for the wave to come back (traverse the fluid twice) to the interface in phase with the light that is reflected from the interface. Since the index of refraction for glass is less than n = 1.756, we know that the light that is reflected from the interface from the glass side will experience a 180^o phase shift. If the light is reflected from the other side, it will experience no phase shift. This means that the light that traverses the liquid and returns must travel a distance equal to 1/2 wavelength of the light in the liquid in order to be in of phase with the light. We can write this as
   
   2t = l'/2
   
   where l' = l/n = 600 nm/1.756 = 342 nm. If we plug this into the equation, we get
   
   t = l'/4 = 342 nm/4 = 85.4 nm
   
   Of course, other possible thicknesses of this liquid layer that would also give a strong reflection would be odd integer multiples of this.
   
   
2. In oder to find out how many dark bands that there will be, we need to find out how many wavelengths will "fit" in the air gap at the paper end of the system, i.e. we need to find out m where m is given by
   
   2t = ml
   
   m = 2t/l = 2(4.00 x 10^-5 m)/(5.461 x 10^-7 m) = 146.5
   
   Therefore, if we include the first dark band, there will be a total of 147 dark bands.
   
   
3. Since we are at the polarizing angle (Brewster's angle), we know that the sum of the angle of reflection and the angle of refraction must equal a right angle, i.e.
   
   q_reflected + q_refracted = 90^o
   
   These angles must also be consistent with Snell's Law. This means that
   
   sin q_incident = sin q_reflected = n sin q_refracted = (1.65) sin q_refracted
   
   sin (90^o - q_refracted ) = cos q_refracted = (1.65) sin q_refracted
   
   cot q_refracted = 1.65
   
   q_refracted = cot^-1(1.65) = 31.2^o
   
   
4. This problem is really asking "How many bright fringes appear in the first 1 cm away from the primary maximum that occurs at q = 0^o ?" Mathematically, it is asking for the value of m that solves the equation
   
   d sin q₁= ml
   
   where q₁ is the angle that subtends the first 1 cm on the screen. Since 1 cm << 2 m, we can approximate sin q₁ by
   
   sin q₁ = (.01 m)/(2 m) = .005
   
   m = d (.005)/l = (.001 m)(.005)/(6 x 10^-7 m) = 8.3

If you find an error in these, or just have a comment, send an e-mail message to jpratte@kennesaw.edu