Question

12. Ria is walking slowly with her head above the water surface in the swimming pool that is about \( 1.5 m \) deep. She spots a coin at the bottom of the pool directly below. The coin appears raised to a depth that is less than the actual depth of the pool. The refractive index of water in the pool can be taken as 1.3. Ria then gets under the water and begins looking up at the water surface. A friend of hers holds a coin directly above her eyes in the air. To Ria, the coin appears to be at a height more than the actual height. In both the above cases, the virtual image of the coin is formed due to the refraction of light at the water-air interface. The bending of the light towards or away from the normal determines the position of the image formed at a distance that is more or less than its real depth/height. 

(a) State the formula Ria can use to find the apparent depth of the coin placed at the bottom of the pool. 

(b) Find the apparent depth of the coin placed at the bottom of the pool as seen by Ria from the air above the water surface. 

(c) If the water in the pool gets polluted with an unknown liquid, thereby decreasing its refractive index, what will be the effect on the apparent depth of the coin placed in water as seen by Ria above the water surface? 

(d) Represent the ray diagram for the second case when Ria looks at the coin in the air from inside the water. 

(e) Find the apparent height of the coin from the water surface as seen by Ria underwater, when the coin is in the air at a height of \( 1.5 m \) from the water surface. [5 marks] 

Answer 1

(a) Given μ = 1.3

μ = \(\frac{Real\,depth}{Apparent\,depth}\) 

1.3 = \(\frac{1.5}{Apparent\,depth}\)

Apparent depth = \(\frac{1.5}{1.3}\)

 = 1.15 m

(b) n = \(\frac{Real\,depth}{Apparent\, depth}\) 

1 = \(\frac{1.5}{Apparent\,depth}\) (n = 1  for air)

Apparent depth  = 1.5 m

(C)  When the rays of light from the coin travel from denses medium to rarer medium, the refracted ray band away from the normal The point from which the refracted rays appear to come gives the apparent position of the coin. Since the rays appear to come from a point above the coin, it appears to be raised. It appears raised from its position

(d) 

(e) We know that

\((\frac h{h-1.5})=\mu\)

\((\frac h{h-1.5})=1.3\) 

h = 1.3h - 1.95

0.3h = 1.95

h = \(\frac{1.95}{0.3}\)

h = 6.5 m