PI is called the image distance (v)
  PC is called radius of curvature (R)
  According to sign convention mentioned above, we have PO = -u ; PI = v ;PC = R
Substituting these values in equation (4) we get,
n2/v- n1/u =(n2-n1)/R       ................. (5)
n2/v-n1/u=(n2-n1)/ R

    This formula can also be used for plane surfaces. In the case of a plane surface, radius of curvature (R) approaches infinity. Hence 1/R becomes zero. Substituting this in equation 5, we get formula for the plane surfaces

        n2/v- n1/u=0=> n2/v= n1/u

NOTE: The distances u and v are measured from the plane interface. Let us consider the following examples.

Example 1
    A bird is flying down vertically towards the surface of water in a pond with constant speed. There is a fish inside the water. If that fish is exactly vertically below the bird, then the bird will appear to the fish to be:
    a. farther away than its actual distance.

    b. closer than its actual distance.

    c. moving faster than its actual speed.

    d. moving slower than its actual speed.

    Which of the four options are true? How can you prove it?
Solution: For refraction at a plane surface,
      we use n2/v=n1/u     ...............(1)

    Let x be the height of the bird above the water surface at an instant and n be the refractive index of water.
    n1 = refractive index of air,
    Then n1 =1, n2 = n, u= -x and let v = -y, (see fig.- E1) Substituting these values in equation (1)
n/(-y)= 1/(-x) ⇒ y = nx


page no:62


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