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katen-ka-za [31]
3 years ago
5

A ray of light is projected into a glass tube that is surrounded by air. The glass has an index of refraction of 1.50 and air ha

s an index of refraction of 1.00. At what minimum angle will light in the glass tube be totally reflected at the glass/air interface?
Physics
1 answer:
Tamiku [17]3 years ago
4 0

Answer:

θ = 41.8º

Explanation:

This is an internal total reflection exercise, the equation that describes this process is

         sin θ = n₂ / n₁

where n₂ is the index of the incident medium and n₁ the other medium must be met n₁> n₂

        θ = sin⁻¹ n₂ / n₁

let's calculate

       θ = sin⁻¹ (1.00 / 1.50)

       θ = 41.8º

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At the interface of two transparent media, light ray experiences both refraction and reflection. Does the angle of reflection de
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Answer:

No, the angle of reflection is independent on the angle of refraction. The total internal reflection is when the angle of incidence is more than the critical angle.

Explanation:

No, the angle of reflection is independent on the angle of refraction. The angle of reflection is only affected by the angle of incidence.

The total internal reflection is when the angle of incidence (I) is more than the critical angle (C). When the light ray pass through the rectangular block, with time, a critical angle will be formed when the refracted ray is exactly 90 degrees to the normal. As soon as the angle of incidence is more than the critical angle, a total internal reflection is formed (as shown in the attached figure).

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6. In baseball, a fastball takes about 1/2 second to reach the plate. Assuming that such a pitch is thrown horizontally, compare
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Answer:

<em>The ball falls 0.3 m in the first 1/4 seconds, and it falls 0.92 m in the second 1/4 second</em>

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<u>Horizontal Launch </u>

When an object is thrown horizontally with a speed v from a height h, it describes a curved path ruled exclusively by gravity until it eventually hits the ground.

To calculate the vertical distance (y) traveled by the object in terms of the time (t) we use:

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\displaystyle y_1=\frac{9.8*0.25^2}{2}

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The distance the ball falls in t=1/2 seconds = 0.5 seconds is:

\displaystyle y_2=\frac{9.8*0.5^2}{2}

\displaystyle y_2=1.225\ m

The distance it falls in the second 1/4 second is:

y_2-y_1=1.225\ m-0.30625\ m

y_2-y_1=0.91875\ m

The ball falls 0.3 m in the first 1/4 seconds, and it falls 0.92 m in the second 1/4 second

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4 years ago
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