water has an index of refraction of 1.33. What is the critical angle for light leaving a pool of water into air? 0 37 90 49
1 answer:
When light travels from a medium with higher refractive index into a medium with lower refractive index, there is a maximum angle (called critical angle) for which all the light is reflected, so there is no refraction.
The value of the critical angle is given by:
when n1 is the refractive index of the first medium, while n2 is the refractive index of the second medium. In our case, n1=1.33 (the water) and n1=1.00 (the air). Putting numbers in, we get
The value of the critical angle is given by:
when n1 is the refractive index of the first medium, while n2 is the refractive index of the second medium. In our case, n1=1.33 (the water) and n1=1.00 (the air). Putting numbers in, we get
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Answer:
hmax= 28.2 m
Explanation:
Horizontal movement:
- For the horizontal movement, as once released, nothing exerts any influence on the ball in the horizontal direction, it will keep moving at a constant speed, so, the equation for the horizontal displacement is as follows:
x = v₀*t = 56.1523 m
Vertical movement:
- For the vertical movement, once thrown upward, the ball is under the sole influence of gravity, g, which causes the ball to accelerate downward, and which magnitude is 9.8 m/s².
- At any time, applying the definition of acceleration, we can find the value of the velocity, as follows:
- When the ball reaches to the maximum height, the ball will come momentarily at rest, so vf =0.
- At this point, we can find the time at which the ball reached to its highest point, as follows:
- In the same way, we can find the maximum height reached by the ball, just replacing this value of time in the equation for the displacement, in the vertical direction, as follows:
- Now, returning to the horizontal movement, as the time must be the same for both movements, we can replace the value for time we have just found, in the equation for the horizontal displacement:
- But as we know that vox = voy, we can rewrite the equation above as follows:
- We can solve for v₀, as follows:
- Now, we can replace this value in the expression for hmax, as follows: