1. A 1.00-cm-high object is placed 10.0cm from a concave mirror whose radius of curvature is 30.0 cm. (a) Draw a ray diagram to locate (approximately) the position of the image. (b) Determine the position of the image and the transverse magnification analytically.
    (a)

    (b) Use mirror equation. 1/p + 1/q = 1/f. f = 1/2R = 1/2(30.0 cm) = 15.0 cm.
          1/(10.0 cm) + 1/q = 1/(15.0 cm), solve q = - 30.0 cm.
         The image is virtual, upright, its position is 30.0 cm from the mirror and behind the mirror.
          m = - q/p = -(-30.0 cm)/(10.0 cm) = 3. The image is larger than the object (magnified).
  2. A triangular glass prism with apex angle 60.0º has an index of refraction n = 1.50. (a) What is the smallest angle of incidence θ1 for which a light ray can emerge from the other side? (b) For what angle of incidence θ1 does the light ray leave at the same angle θ1?

    Solution.

    (a)At the first refraction, sin θ1 = nsin θ2. Total internal reflection occurs for

                      1.50 sin θ2’ =1.00 sin 90º

                         θ2’ = sin-1(1/1.5) = 41.8 º, θ2 = 60º - θ2’ = 18.26º

           Since sin θ1 = nsin θ2, θ1 = sin-1(nsin θ2) =27.9º

    (b) When light passes symmetrically through the prism, θ1 = θ3  requires θ2 = θ2‘.

          With θ2 + θ2‘ = 60º, θ2 = 30º.
          And since 1.00 sin θ1 =  1.50 sin 30º, θ1 = 48.6º.
  3. A swimmer had dropped her goggles in the shallow end of a pool, marked as 1.0 m deep. But the goggles don't look that deep. How deep do the goggles appear to be when you look straight down into the water?

    Use Snell' law:
    sin θ1= n2sin θ2;
    When θ1 is small,  sin θ1 »θ1» tanθ1 = x/d' and sin θ2 » tan θ2 = x/d.
    θ1= n2θ2;
    x/d' = n2x/d
    d' = d/n2 = (1.0 m)/(1.33) = 0.75 m.