In the figure, a fish watcher at point P watches a fish through a glass wall of

Question

In the figure, a fish watcher at point P watches a fish through a glass wall of a fish tank. The watcher is level with the fish; the index of refraction of the glass is 8/5, and that of the water is 4/3. The distances are d1 = 9.4 cm, d2 = 3.6 cm, and d3 = 8.0 cm. (a) To the fish, how far away does the watcher appear to be? (Hint: The watcher is the object. Light from that object passes through the wall's outside surface, which acts as a refracting surface. Find the image produced by that surface.Then treat that image as an object whose light passes through the wall's inside surface, which acts as another refracting surface. Find the image produced by that surface, and there is the answer.) (b) To the watcher, how far away does the fish appear to be?

Explanation / Answer

a) n1/d1 = -n2/d1'

1/9.4 = -(8/5)/d1'

d1' = -(8*9.4)/5 = ?15.04 cm

S2 = 15.04+d2 = 18.64cm

n2/S2 = -n3/S2'

S2' = (4/3)*18.64/(8/5)cm = ?15.5 cm

distance = D = d3 + S2' = 23.5 cm

b) n3/d3 = -n2/S1'

(4/3)/8 = -(8/5)/S1'

S' = -(8*8*3)/(5*4) = ?9.6 cm

S2 = ?9.6+d2 = -13.2 cm

n2/S2 = - n1/S2'

S2' = ?8.26cm

D = d1 + S2' = 9.4 + 8.26 = 17.66 cm

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