Question

B. For a fish in water, outside world would appear to be contained within a circle of area 12.56 m² at the
surface. If the refractive index of water is 4/3, find the depth at which the fish is present. [1.76 m)​

Answer 1

pls refer the attachment first

Area of the circle = 12 .56 m ²

we know that,

Area of the circle= π R ²

Now equating both the results,

π R ² = 12.56

R² = 12.56 / 3.14

R ² = 4

R = 2 m

The refractive index of water is 4 / 3

Now ,we need to find the critical angle

sin c = 1 / n

sin c = 3 / 4

Now,

sin c = r  / √ r²+ h²

equating both the results,

3 / 4 = r / √ r²+ h²

9 / 16 = r² / r² + h ²

9 r²+9 h² = 16 r²

9 h²= 16 R ² - 9 R ²

9 h² = 7 R ²

3 h= √ 7 R

3 h = √ 7 × 2

3 h = 2 .64× 2

3 h = 5.28

h = 5.28 /3

h = 1.76 m