Solutions to Week 8 Assignment

1. We find the critical angle for light leaving the water:                    nsin q1  = sin q2;                   (1.33)sin qC = sin90°, which gives qC = 48.8°.     If the light is incident at a greater angle than this it will totally reflect.     We see from the diagram that                    R > Htan qC = (62.0cm)tan48.8° = 70.7cm.

2. (a) We find the image distance from

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

                   (1/10.0 x 10³mm) + (1/d_i) = 1/80mm, which gives d_i = 81mm.

    (b) For an object distance of 3.0m, we have

                  (1/3.0 x 10³mm) + (1/d_i) = 1/80mm, which gives d_i = 82mm.

    (c)  For an object distance of 1.0m, we have

                  (1/1.0 x 10³mm) + (1/d_i) = 1/80mm, which gives d_i = 87mm.

   (d) We find the smallest object distance from

                  (1/d_omin) + (1/120mm) = 1/80mm, which gives d_i = 240mm = 24cm.

3. We can relate the image and object distance from the magnification:

                 m = -d_i/d_o,  or d_o = - d_i/m.

    We use this in the lens equation:

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

                - (m/d_i) + (1/d>_i) = 1/f, which gives d_i = (1 -m)f.

    (a) If the image is real, d_i > 0. With f < 0, we see that m > 1; thus m = +2.00. The image distance is

                 d_i = [1 - (+2.00)](-50.0mm) = 50.0mm.

         The object distance is

                d_o = -d_i/m = -(50.0mm)/(+2.00) = -25.0mm.

         The negative sign means the object is beyond the lens, so it would have to be an object formed by a preceding optical device.

  (b) If the image is virtual, d_i < 0. With f < 0, we see that m < 1; thus m = -2.00. The image distance is

                d_i = [1 - (-2.00)](-50.0mm) = -150.0mm.

         The object distance is

               d_o = -d_i/m = -(-150.0mm)/(-2.00) = -75.0mm.

         The negative sign means the object is beyond the lens, so it would have to be an object formed by a preceding optical device.

4. We find the image formed by the refraction of the first lens:

               (1/d_o1) + (1/d_i1) = 1/f₁;

              (1/35.0cm) + (1/d_i1) = 1/27.0cm, which gives d_i1 = +21.3cm.

   This image is the object for the second lens. because it is beyond the second lens, it has a negative object distance: d_o2 = 16.5cm - 118.1cm = -101.5cm.

   We find the image formed by the refraction of the second lens:

              (1/d_o2) + (1/d_i2) = 1/f₂;

              [1/(-101.5cm)] + (1/d_i2) = 1/27.0cm, which gives d_i2 = +21.3cm.

   Thus the final images is real, 21.3cm beyond second lens.

   The total magnification is the product of the magnifications fro the two lenses:

              m = m₁m₂ = (-d_i1/d_o1)(-d_i2/d_o2) = d_i1d_i2/d_o1d_o2

                  = (+118.1cm)(+21.3cm)/(+35.0cm)(-101.5cm) = -0.708 (inverted).

5. We find the focal length of the lens from

              1/f = (n - 1)[(1/R₁) + (1/R₂)]

                   = (1.56 - 1){[(1/(-21.0cm)] + [1/(+18.5cm)]}, which gives f = 277.5cm.

   We find the image distance from

             (1/d_o) + (1/d_i) = 1/f;

             (1/100cm) + (1/d_i) = 1/277.5cm, which gives d_i = -156cm (in front of the lens).

  The magnification is

             m = -d_i/d_o = -(-156cm)/(100cm) = +1.56(upright).