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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.

2 Answers

3 votes

Answer:

1.65

Step-by-step explanation:

User Aardbol
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3 votes

Answer: 1.65m

Step-by-step 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

User Jayesh
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