Question

A bird is flying down vertically towards the surface of water in a pond with constant. There is a fish inside the water.if that fish 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 ?​

Answer 1

Answer:

(a)farther away than its actual distance

Explanation:

According to the image formula for refraction at plane surface. Here river water surface acts as a plane surface.

We have n2/v = n1/u

Let x be the height of the bird at O above the water surface and y be the height of the image at I.

Here n1 = 1 (refractive index of Air)

          n2 = n (refractive index of water)

According to sign convention,

Height measured vertically above from the point of axis is taken as positive and below is taken as negative.

Therefore u = x (object distance vertically above)

                    V = y(image distance vertically above)

n/(y) = 1/(x)

y = nx from above expression which we got, we can say that n is greater than 1 and therefore y is greater than x.

So from this we can say that bird appears to fish to be farther away from the actual distance and it appears to fish that the bird covers more distance i.e y distance than the actual x distance. It appears that the speed of the bird is greater than the actual speed.