Light and optics

[Temperature101]

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The index of refraction of the water in a certain murky pond is approximately 1.4. An alligator sit 1m below the smooth surface of the pond.The alligators view of objects above the surface is restricted to a certain window at the surface with a radius of:

A. 0.5 m
B. 1 m
C. 2 m
D. 2.6 m

Answer: B

This is from EK 1001 physics. People keep saying EK 1001 is easy...I dont see that for the 1001 physics. May be I am missing something. I have been scoring 9-10 in AAMC PS and think EK 1001 physics is not that easy.

 

[BB8730]

Break the problem down. A restriction in viewing due to refraction always deals with the critical angle, which occurs when the refracted ray is 90 degrees.

n1sinθ1=n2sinθ2
^water air^

(1.4)sinθ1=(1)sin90
(1.4)sinθ1=1
sinθ1=1/1.4 or about 2/3
You should know what the sin of 30, 45, and 60 degrees are. The closest to 0.67 is sin(45) which is ~0.7.

Now you know that the alligator can't see anything past 45 degrees. Draw a triangle like this:

The alligator is at the bottom point and the surface of the water is the horizontal side. You know the length of the right side is 1m and the angle at the bottom is 45 degrees. Since it's a 45-45-90 triangle, the horizontal side must be 1m.

 

[Temperature101]

Break the problem down. A restriction in viewing due to refraction always deals with the critical angle, which occurs when the refracted ray is 90 degrees.

n1sinθ1=n2sinθ2
^water air^

(1.4)sinθ1=(1)sin90
(1.4)sinθ1=1
sinθ1=1/1.4 or about 2/3
You should know what the sin of 30, 45, and 60 degrees are. The closest to 0.67 is sin(45) which is ~0.7.

Now you know that the alligator can't see anything past 45 degrees. Draw a triangle like this:

The alligator is at the bottom point and the surface of the water is the horizontal side. You know the length of the right side is 1m and the angle at the bottom is 45 degrees. Since it's a 45-45-90 triangle, the horizontal side must be 1m.

Thanks...I did not even know where to start since I did not know that a restriction in viewing has something to do with critical angle.

 

[Temperature101]

Break the problem down. A restriction in viewing due to refraction always deals with the critical angle, which occurs when the refracted ray is 90 degrees.

n1sinθ1=n2sinθ2
^water air^

(1.4)sinθ1=(1)sin90
(1.4)sinθ1=1
sinθ1=1/1.4 or about 2/3
You should know what the sin of 30, 45, and 60 degrees are. The closest to 0.67 is sin(45) which is ~0.7.

Now you know that the alligator can't see anything past 45 degrees. Draw a triangle like this:

The alligator is at the bottom point and the surface of the water is the horizontal side. You know the length of the right side is 1m and the angle at the bottom is 45 degrees. Since it's a 45-45-90 triangle, the horizontal side must be 1m.

..................

 

[milski]

Ok, let's start slowly. If the aligator looks straight up (ray going straight up), there will be no bending at all and the aligator will certainly see what's there. Now, if the aligator looks slightly away from the vertical, at angle &#952;1, due to refraction the light will bend a bit, to an angle &#952;2. Since the water has a higher index of refraction &#952;1<&#952;2.

As you keep increasing &#952;1 &#952;2 will increase as well and at some point it will become 90. At that point the ray (aligator's look) will not exit the water and this will be the furthers away from the vertical that the aligator can see. &#952;1 in that case is also known as critical angle. You can imagine the aligator sitting at the tip of an inverted cone and not being able to see outside of it. The bottom of the cone is actually on the surface and the question is asking you what is its radius.

From n1&#952;1=n2&#952;2 you can calculate that the critical angle in this case is 45 degrees and from there the radius of the bottom of the cone is 1m, same as the depth.