How Does Light Refract When Emerging from Water into Air?

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SUMMARY

The diameter of the light cone emerging from water into air, when light is emitted from a source 10.0 cm in diameter at the bottom of a 70 cm deep swimming pool, is approximately 1.8 meters. This conclusion is derived using Snell's Law, where the refractive indices of air (1.0) and water (1.33) are applied. The angle of refraction calculated is 48.6 degrees, leading to the final diameter through trigonometric calculations. The correct answer to the posed question is option d) 1.8m.

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  • Understanding of Snell's Law and refractive indices
  • Basic trigonometry, specifically tangent functions
  • Knowledge of total internal reflection concepts
  • Familiarity with light behavior in different mediums
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daveed
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this is the entire question:
at the shallow end of a swimming pool, the water is 70 cm deep. The diameter of the cone emerging from the water into the air above, emitted by a light source 10.0 cm in diameter at the bottom of the pool and measured by an observer on the edge of the pool 2.5 meters away is:

a)1.6m
b)1.7m
c)1.75m
d)1.8m

it would seem to me that you would use snell's law to find an angle of refraction, however, how do you know at what angle the light goes in the water? if it goes straight it wouldn't refract at all... I am confused...
 
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total internal reflection

Consider the light emitted at all angles. What's the angle of the refracted light if the angle of incident light is [itex]\theta[/itex]? As the angle of incidence gets larger, at some angle (the critical angle) you won't get any refracted light. You'll have so-called "total internal reflection". Figure out that angle, then use it to figure out the diameter of the "cone" of light that emerges from the water.
 


The answer to this question can be found by using the formula for the angle of refraction, which is given by Snell's law: n1sinθ1 = n2sinθ2. In this case, n1 is the refractive index of air (which is approximately 1), n2 is the refractive index of water (which is approximately 1.33), θ1 is the angle of incidence (which is equal to 90 degrees since the light is entering the water at a right angle), and θ2 is the angle of refraction.

To find the angle of refraction, we can rearrange the formula to solve for θ2: θ2 = sin^-1(n1/n2 * sinθ1). Plugging in the values, we get θ2 = sin^-1(1/1.33 * sin90) = sin^-1(0.75) = 48.6 degrees.

Next, we can use trigonometry to find the length of the light cone emerging from the water into the air. The diameter of the light source at the bottom of the pool is 10.0 cm, which means the radius is 5.0 cm. The distance from the light source to the edge of the pool (where the observer is located) is 2.5 meters, which is equal to 250 cm.

Using the tangent function, we can find the length of the light cone (which is the hypotenuse) by taking the tangent of half the angle of refraction (which is 48.6/2 = 24.3 degrees) and multiplying it by the distance from the light source to the edge of the pool. So the length of the light cone is tan24.3 * 250 = 99.7 cm = 0.997 m.

Finally, we can find the diameter of the light cone by doubling the length we just found, which is 0.997 * 2 = 1.994 m. This is the approximate diameter of the light cone emerging from the water into the air. Therefore, the answer to the question is d) 1.8m.

To summarize, the light cone emerging from the water into the air has a diameter of approximately 1.8 meters, which is the answer d) in the given options.
 

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