Solutions to Week 9 Assignment

1. For constructive interference, the path difference is a multiple of the wavelength

                      dsinq = ml, m = 0, 1, 2, 3... .

    We find the location on the screen from

                     y = Ltanq.

    For small angle, we have

                    sinq  = tanq, which gives

                    y = L(ml/d) = mLl /d.

    For the second order of the two wavelengths, we have

                   y_a = mL_al = (1.6m)(480 x 10^-9m)(2)/(0.54 x 10^-3m) = 2.84 x 10^-3m = 2.84mm.

                   y_b = mL_bl = (1.6m)(620 x 10^-9m)(2)/(0.54 x 10^-3m) = 3.67 x 10^-3m = 3.67mm.

   Thus the two fringes are separated by 3.67mm - 2.84mm = 0.8mm.

2.  We find the angles of refraction in the gflass from

                  n₁sinq₁ = n₂sinq₂;

                 (1.00)sin60.00° = (1.4820)sinq_2,450, which gives q_2,450 = 35.76°;

                 (1.00)sin60.00° = (1.4820)sinq_2,700, which gives q_2,700 = 35.98°.

      Thus the angle between the refracted beams is

                 q_2,700 - q_2,450 = 35.98° - 35.76° = 0.22°.

3. Because the angles are small, we have

                tanq_1min = 1/2(Dy₁)/L = sinq_1min.

    The condition for the first minimum is 

               Dsinq_1min = 1/2DDy₁/L = l.

   If we form the ratio of the expressions for the two wavelengths, we get

              Dy_1b/Dy_1a  = l_b/l_a;

             Dy_1b/(3.0cm) = (400nm)/(550nm), which gives Dy_1b = 2.2cm.

4. We find the slit separation from

             dsinq = ml;

             dsin15.5° = (1)(589 x 10^-9m), which gives d = 2.20 x 10^-6m = 2.20mm.

    We find the angle for the fourth order from

            dsinq = ml;

           (2.20 x 10^-6m)sinq₃ = (3)(589 x 10^-9m), which gives sinq₃ = 0.8032, which gives q₃ = 53.4°.

5. We find the angles for the first order from

          dsinq = ml = l;

          [1/(7500lines/cm)](10^-2m/cm)sinq₄₀₀ = (400 x 10^-9m), which gives sinq₄₀₀ = 0.300, so q₄₀₀ = 17.5°;

          [1/(7500lines/cm)](10^-2m/cm)sinq₇₀₀ = (700 x 10^-9m), which gives sinq₇₀₀ = 0.563, so q₇₀₀ = 34.2°;

   The distances from the central white line on the screen are

          y₄₀₀ = Ltanq₄₀₀ = (2.30m)tan17.5° = 0.732m;

          y₇₀₀ = Ltanq₇₀₀ = (2.30m)tan34.2° = 1.56m;

   Thus the width of the spectrum is

         y₇₀₀  -  y₄₀₀ = 1.56m - 0.723m = 0.84m.