A man in a boat is looking straight down at a fish in the water directly beneath him. The fish is looking straight up at the man. They are equidistant from the air–water interface. To the man, the fish appears to be 2.0 m beneath his eyes. To the fish, how far above its eyes does the man appear to be?
the equation given is n(water)d' = n(air)d
at first i just plugged the numbers in but then i realized it's the apparent depth if i do that. how do i actually solve this?
angle/depth question ?
2020-09-30 1:07 pm
回答 (1)
2020-09-30 1:32 pm
Let D m be the equidistance from the air-water to the man or to the fish.
n(water) = 1.33 and n(air) = 1.00
When the man looks at the fish:
Real depth, d = D m
Apparent depth, d' = (2 - D) m
n(water) d' = n(air) d
1.33 (2 - D) = 1 × D
2.66 - 1.33 D = D
2.33D = 2.66
D = 1.14 (m)
When the fish looks at the man:
n(water) d = n(air) d'
1.33 × 1.14 = 1 × d'
d' = 1.52 m
The distance that the man appears above the fish's eye
= (1.14 + 1.52) m
= 2.66 m
n(water) = 1.33 and n(air) = 1.00
When the man looks at the fish:
Real depth, d = D m
Apparent depth, d' = (2 - D) m
n(water) d' = n(air) d
1.33 (2 - D) = 1 × D
2.66 - 1.33 D = D
2.33D = 2.66
D = 1.14 (m)
When the fish looks at the man:
n(water) d = n(air) d'
1.33 × 1.14 = 1 × d'
d' = 1.52 m
The distance that the man appears above the fish's eye
= (1.14 + 1.52) m
= 2.66 m
收錄日期: 2021-04-24 08:10:44
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