solutions to Week 7 Assignment

1. Because the rays entering your eye are diverting from the image position behind the mirror, the diameter of  the area on the mirror and the diameter of your pupil must subtend the same angle from the image.

                      D_mirror/d_i = D_pupil/(d_o + d_i);

                      D_mirror/(70cm) = (5.5mm)(70cm + 70cm), which gives D_mirror = 2.75mm

     Thus the area on the mirror is

                     A_mirror= 1/4pD_mirror² = 1/4p(2.75 x 10^-3m)² = 5.9 x 10^-6m².

2. We find the image distance from the magnification:

                     m = h_i/h_o = - d_i/d_o;

                    +4.5 = - d_i/(2.2cm), which gives d_i = -9.90cm.

    We find the focal length from

                   (1/d_o) + (1/d_i) = 1/f;

                   [(1/(2.20cm)] + [(1/(-9.90cm)] = 1/f, which gives f = 2.83cm.

    Because the focal length is positive, the mirror is concave with a radius of

                  r = 2f = 2(2.83cm) = 5.7cm.

3.

Using the mirror equation we have

                1/d_o + 1/d_i = 1/f;

                1(12cm) + 1/d_i = 1/(-5cm), which gives d_i = - 3.53cm.

We find the lateral magnification from

               m = - h_i/h_o = - d_i/d_o = -(-3.53cm)/(12cm) = 0.29.

The image is virtual, reduced, and 3.53cm behind the mirror.

4. We find the index of refraction from

              v = c/n;

              0.85v_water = 0.85c/1.33 = c/n, which gives n = 1.56.

5. For the refraction at the first surface, we have              nairsinq1 = nsinq2;              (1.00)sin45.0° = (1.52)sinq2, which gives q2 = 27.7°.     We find the angle of incidence at the second surface from              (90° - q2) + (90° - q3) + A = 180°, which gives               q3 = A - q2 = 60° - 27.7° = 32.3°.    For the refraction at the second surface, we have             nsinq3 = nairsinq4;             (1.52)sin32.3° = (1.00)sinq4, which gives            q4 = 54.3° from the normal.