Solutions To Week 10 Assignment

1. The phase difference for the reflected waves from the path-length difference and the reflection at the bottom surface is

                       f = (2t/l)2p + p.

    For the dark rings, this phase difference must be an odd multiple for p, so we have

                      f = (2t/l)2p + p = (2m + 1)p, m = 0, 1, 2, . . ., or

                      t = 1/2ml, m =0, 1, 2, . . . .

   Because m = 0 corresponds to the dark center, m corresponds the number of the ring.

   Thus the thickness of the lens is the thickness of the air at the edge of the lens.

                     t = 1/2(31)(550nm) = 8.5 x 103nm = 8.5mm.

2. With respect to the incident wave, the wave that reflects from the air at the tip surface of the air layer has a phase change of

                     f1 = 0. 

   With respect to the incident wave, the wave reflects from the glass at the bottom surface of the air layer has a phase change due to the   additional path-length and a change on reflection:

                     f2 = (2t/l)2p + p.

   For constructive interference, the net phase change is

                    f  = (2t/l)2p + p - 0 = m2p, m = 1, 2, 3, . . ., or t = (1/2)l(m -1/2), m = 1, 2, 3, . . . .

  The minimum thickness is

                    tmin = 1/2(450nm)(1-1/2) =113nm.

  For destructive interference, the net phase change is

                   f  = (2t/l)2p + p - 0 = (2m + 1)p, m = 0, 1, 2, . . ., or t = (1/2)ml,  m = 0, 1, 2, . . . .

  The minimum non-zero thickness is

                   tmin = 1/2(450nm)(1) = 225nm.                   

  

3. If the initial intensity is I0, through the two sheets we have

                  I1 = I0cos2q 1;

                  I2 = I1cos2q 2 = I0 cos2q 1cos2q 2;

                 0.15I0 = I0 cos2q 1 cos240°, which gives q 1 = 60°.

4. We find the effective f-number for the pinhole:

                 f-stop2 = f/D = (70mm)/(1.0mm) = f/70.

    The exposure is proportional to the area and the time:

               Exposure µ At µ D2t µ t/( f-stop2)2.

   Because we want the exposure to be the same, we have

              t1/( f-stop1)2 = t2/( f-stop2)2;

             [(1/250)s]/(11)2 = t2/(70)2, which gives t2 = (1/6)s.

 5. (a) We find the image distance from

             (/1do) + (1/di) = 1/f;

             (1/5.35cm) +(1/di)  = (1/6.00cm), which gives di = -49.4cm.

    (b) From the diagram we see  that the angular magnification is

             M = q '/q  = (ho/do)/(ho/N) = N/do

                 = (25cm)/(5.35cm) = 4.67x.