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Ierofanga [76]
4 years ago
5

Diamond has an index of refraction of 2.419. What is the critical angle for internal reflection inside a diamond that is in air?

A) 155° B) 105° C) 48.8° D) 24.4° E) 131°
Physics
1 answer:
SSSSS [86.1K]4 years ago
4 0

Answer:

C) 24.4°

Explanation:

let nd = 2.419 be the index of refraction of diamond and na  = 1.0 be the index of refraction of air  and ∅c be the critical angle.

according to Snell's Law:

sin(∅c) = na/nd

sin(∅c) = (1.0)/(2.419)

      ∅c = 24.4°

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If y = 0.02 sin (30x – 200t) (SI units), the frequency of the wave is
leva [86]

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31.831 Hz.

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<u>Given:</u>

  • \rm y = 0.02\sin(30x-200 t).

The vertical displacement of a wave is given in generalized form as

\rm y = A\sin(kx -\omega t).

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  • A = amplitude of the displacement of the wave.
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  • x = horizontal displacement of the wave.
  • \omega = angular frequency of the wave = \rm 2\pi f.
  • f = frequency of the wave.
  • t = time at which the displacement is calculated.

On comparing the generalized equation with the given equation of the displacement of the wave, we get,

\rm A=0.02.\\k=30.\\\omega =200.\\

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