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love history [14]
4 years ago
11

A beam of light, traveling in air, strikes a plate of transparent material at an angle of incidence of 56.0°. It is observed tha

t the reflected and refracted beams form an angle of 90.0°. What is the index of refraction of this material?
Physics
1 answer:
tankabanditka [31]4 years ago
3 0

Answer:

the index of refraction for the transparent material is 1.48

Explanation:

let n1 = 1.0 be the refractive index of air and n2 be the refractive index of the trasnparent material.

then:

Snell's Law state that:

n1×sin(∅1) = n2×sin(∅2)

where ∅1 = 56.0° is the angle of incidence and  ∅2 = 90 - 56.0 = 34° be the angle refraction.

n1×sin(∅1) = n2×sin(∅2)

          n2 = n1×sin(∅1)/sin(∅2)

               =  (1.0)×sin(56.0°)/sin(34.0°)

               =  1.48

Therefore, the index of refraction for the transparent material is 1.48

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Answer:

(A) 88.92 cm

(B) 22.22 cm

Explanation:

distance (s) = 200 cm = 0.2 m

initial velocity (v) = 0 m/s

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lets first find the time (T) it takes for the first drop to strike the floor

from  s = ut + 0.5at^{2}

         200 = 0 + 0.5 x 9.8 x T^{2}

         200 = 4.9 x T^{2}

         200 / 4.9 = T^{2}

         T = 6.4

(A) When the first drop strikes the floor, how far below the nozzle is the second drop.

we can find how far the second drop was when the first drop hits the ground from the formula s = ut + 0.5at^{2}

where

  • s = distance
  • u = initial velocity = 0
  • t = time, since the drops fall at regular (equal) intervals of time, the first drop striking the floor at the instant the fourth drop begins to fall there wold be 3 time intervals and this can be seen illustrated in the attached diagram. therefore the time of the second drop = 2/3 of the time it takes the first drop to strike the ground (\frac{2}{3}.T)
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substituting all required values we have

s = 0 + (0.5 x 9.8 x (\frac{2}{3}T)^{2})

s = 0 + (0.5 x 9.8 x (\frac{2}{3} x 6.4)^{2})

s = 88.92 cm

(B) When the first drop strikes the floor, how far below the nozzle is the third drop.

we can find how far the third drop was when the first drop hits the ground from the formula s = ut + 0.5at^{2}

where

  • s = distance
  • u = initial velocity = 0
  • t = time, since the drops fall at regular (equal) intervals of time, the first drop striking the floor at the instant the fourth drop begins to fall there wold be 3 time intervals and this can be seen illustrated in the attached diagram. therefore the time of the third drop = 1/3 of the time it takes the first drop to strike the ground (\frac{2}{3}.T)
  • a = acceleration due to gravity = 9.8 m/s^{2}

substituting all required values we have

s = 0 + (0.5 x 9.8 x (\frac{1}{3}T)^{2})

s = 0 + (0.5 x 9.8 x (\frac{1}{3} x 6.4)^{2})

s = 22.22 cm

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Given

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