In the above equation, we know that n is greater than 1. Hence y is greater than x. Thus the bird appears to the fish to be farther away than its actual distance. We have assumed that bird is flying vertically down with constant speed. For the observer on the ground, bird appears that it has covered ‘x’ distance for certain time. But for fish, it appears that bird has covered a distance ‘y’ in the same time. As y is greater than x , we can conclude that the speed of the bird, observed by the fish, is greater than its actual speed.

    So, options (a) and (c) are correct.

Example 2

    A transparent sphere of radius R and refractive index n is kept in air. At what distance from the surface of the sphere should a point object be placed on the principal axis so as to form a real image at the same distance from the second surface of the sphere?

U = -x, v= ∞ (refracted ray is parallel to the optical axis after refraction at first surface)

    n1and n2 = n, (where n1 is refractive index of air)

    Using n2/v- n1/u=(n2-n1)/R

    n/∞ - 1/-x=(n-1)/R ---> 1/x =(n-1)/R

    ---> x= R/(n-1)

        Object distance from the first surface of the sphere is x= R/(n-1)

Example 3

    A transparent (glass) sphere has a small, opaque dot at its centre. Does the apparent position of the dot appear to be the same as its actual position when observed from outside?

Solution: Let refractive index of glass n1    = nrefractive index of air n2 = 1

Then u = –R (radius of sphere) ; Radius of curvature R = –R


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