The absolute refractive index is equal to the speed of light of the wave in air divided by the speed of light in the second medium. This means that it is equal to 3 x10^8 / 1.71 x10^8. This means the answer is 1.75
Answer:
The distance of the goggle from the edge is 5.30 m
Explanation:
Given:
The depth of pool (d) = 3.2 m
let 'i' be the angle of incidence
thus,
i = 
i = 67.75°
Now, Using snell's law, we have,
n₁ × sin(i) = n₂ × 2 × sin(r)
where,
r is the angle of refraction
n₁ is the refractive index of medium 1 = 1 for air
n₂ is the refractive index of medium 1 = 1.33 for water
now,
1 × sin 67.75° = 1.33 × sin(r)
or
r = 44.09°
Now,
the distance of googles = 2.2 + d×tan(r) = 2.2 + (3.2 × tan(44.09°) = 5.30 m
Hence, <u>the distance of the goggle from the edge is 5.30 m</u>
As long as you describe the chemical reactions within the torpedo and explain how torpedoes advanced modern warfare.
see
below
Explanation:
refractive index = speed of light in vacuum / speed of light in medium
light travels at a speed of 3.0 x 10^8 m/s in vacuum
refractive index = 3.0 x 10^8 / 2.0 x 10^8
refractive index = 1.5
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