Selected Solutions $$

• 5. (a) If the slots are small enough, then we know that the beam will only make it through the slot if the time it takes it to go the distance s is equivalent to the time that it takes the disk to rotate an angle q. The time that it takes the light to travel s is given by Dt = s/c. The time that it takes the disk to rotate is given by Dt = q/w. Setting these two equal to one another gives

  s/c = q/w
  c = s w/q

  (b) If s = 2.5 m, q = 1^o/60 = 2p/((360)(60)) = .00029 rad, and w = 5555 rev/s = 34855 rad/s, we get that the speed of light is

  c = (2.5 m)(34855 rad/s)/(.00029 rad) = 3.0 x 10⁸ m/s.

• 6. The first thing to notice is that the beam of light is striking the interface at B at a 60^o angle with respect to the perpendicular. Using Snell's Law, we get that

  n₁ sin(60^o) = n₂ sin q₂
  sin q₂ = (1.7)()/(1.5) = .981
  q₂ = 79.0^o

  Now comes the hard part; we have to do a little geometry. At the interface between the 2 and 3 media, we need to know the angle that the beam makes relative to a horizontal line. Since the beam was diverted by 19.0^o relative to the original beam (which was horizontal), this means that the beam will strike the interface with medium 3 at an angle of 19.0^o. We can again use Snell's Law to determine the angle that it travels through 3, thus giving

  n₂ sin(19.0^o) = n₃ sin q₂
  sin q₂ = (1.5)(.325)/(1.3) = .375
  q₂ = 22.0^o

  In solving for q₂, we have already used the fact that the angle between the mirror and the beam coming in is equal to the angle between the beam and the perpendicular at the 2-3 interface. Since q₂ and q₃ are complementary angles, we know that

  q₃ = 90 - q₂ = 68.0^o

  At the last interface, we note that the angle of incidence is just q₂. We again use Snell's Law to give

  n₃ sin (22.0^o) = 1 sin q₄
  sin q₄ = (1.3)(.375)
  q₄ = 29.1^o

• 7.The first thing that we need to know is the angle of incidence of the beam with the circular surface. To do this, draw a line from the incidence point to the center of the circle and note that the angle of incidence is the same as the angle that the line makes with the horizontal. To find this angle, we note that we have formed a triangle with a hypotenuse of 10 cm and an opposite side of 5 cm. This gives the angle of incidence as sin q_i = (5 cm)/(10 cm) = .5, or q_i = 30^o. Using Snell's Law, we get that the refracted angle is

  n sin q_r = (1)(.5)
  sin q_r = .5/2 = .25
  q_r = 14.5^o

  Once again, we have to use a little geometry to figure out the angle of incidence at the flat surface. Since the angle of refraction was 14.5^o, this means that the beam was diverted by an amount 30^o - 14.5^o = 15.5^o from the horizontal. This is the angle of incidence at the flat surface. Using this, we get

  n sin(15.5^o) = 1 sinq
  sinq = 2(.267) = .534
  q = 32.3^o