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3 years ago

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A Ray of light in air is incident on an air to glass boundary at an angle of 30. degrees with the normal. If the index of refrac

tion of the glass is 1.52, what is the angle of the refracted ray within the glass with respect to the normal? A. 19 degrees B. 22 degrees C. 30 degrees D. 1.52

1 answer:

elena55 [62]3 years ago

5 0

Answer:

A

Explanation:

Snell's law states:

n₁ sin θ₁ = n₂ sin θ₂

where n is the index of refraction and θ is the angle of incidence (relative to the normal).

The index of refraction of air is approximately 1. So:

1 sin 30° = 1.52 sin θ

θ ≈ 19°

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3 years ago

Read 2 more answers

In a game of pool, the cue ball strikes another ball of the same mass and initially at rest. After the collision, the cue ball m

ikadub [295]

(a) $-39.4^{\circ}$

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where

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$v_2 = 3.40 m/s$ is the velocity of the second ball after the collision

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$\phi_2$ is the angle of the second ball

Solving for $\phi_2$ , we find the angle between the direction of motion of the second ball and the original direction of motion:

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First, we calculate the total momentum along the x-direction after the collision, which is:

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The initial total momentum along the x-direction as

$p_x = m u$

where

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$u$ is the initial velocity of the cue ball

The momentum along this direction must be conserved, so we can equate the two expressions and find the value of u:

$mu = 6.69 m\\u = 6.69 m/s$

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3 years ago

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Answer:

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b)

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2 years ago

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hope this helps please give brainliest!

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3 years ago