A small but bright light is at the bottom of a pool 2.2 m deep. How wide is the circle of light that exits the surface of the water when that light shines in the middle of the night? Remember that nwater=1.33 .

Answer: 1.65m

Explanation:

Refractive index in terms of the depth of liquid is the ratio of the real depth to the apparent depth of the liquid i. e Refractive index = Real depth/apparent depth

Refractive index of water given = 1.33

Real depth is the measure of how deep is the liquid while apparent depth is the depth at the surface of the liquid.

Real depth = 2.2m

Apparent depth = ?

Applying the formula above

Apparent depth = Real depth/refractive index

= 2.2/1.33

= 1.65m

Therefore, the circle of light that exits the surface of the water when that light shines in the middle of the night is 1.65m wide