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Flauer [41]
3 years ago
11

A light beam is directed parallel to the axis of a hollow cylindrical tube. When the tube contains only air, the light takes 8.7

2 ns to travel the length of the tube, but when the tube is filled with a transparent jelly, the light takes 1.82 ns longer to travel its length. What is the refractive index of this jelly?
Physics
1 answer:
monitta3 years ago
3 0

Answer:

Explanation:

velocity of light in a medium of refractive index V = V₀ / μ

V₀  is velocity of light in air and μ is refractive index of light.

time to travel in tube with air =  length of tube / velocity of light

8.72 ns = L / V₀  L is length of tube .

time to travel in tube with jelly =  length of tube / velocity of light

8.72+ 1.82  = L / V  L is length of tube .

10.54 ns = L / V

dividing the equations

10.54 / 8.72 = V₀  / V

10.54 / 8.72 =   μ

1.21 =  μ

refractive index of jelly = 1.21 .

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The ball would hit the floor approximately 0.55\; \rm s after leaving the table.

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Let \Delta h denote the change to the height of the ball. Let t denote the time (in seconds) it took for the ball to hit the floor after leaving the table. Let v_0(\text{vertical}) denote the initial vertical velocity of this ball.

If the air resistance on this ball is indeed negligible:\displaystyle \Delta h = -\frac{1}{2}\, g\, t^{2} + v_0(\text{vertical}) \cdot t.

The ball was initially travelling horizontally. In other words, before leaving the table, the vertical velocity of the ball was v_0(\text{vertical}) = 0 \; \rm m \cdot s^{-1}.

The height of the table was 1.5\; \rm m. Therefore, after hitting the floor, the ball would be 1.5\; \rm m \! below where it was before leaving the table. Hence, \Delta h = -1.5\;\rm m.

The equation becomes:

\displaystyle -1.5 = -\frac{9.81}{2} \, t^{2}.

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In other words, it would take approximately 0.55\; \rm s for the ball to hit the floor after leaving the table.

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The ball was in the air for approximately t = 0.55\; \rm s and would have travelled approximately v(\text{horizontal})\cdot t \approx 5.5\;\rm m horizontally during the flight.

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