The drawing shows a coin resting on the bottom of a beaker...

Name: The drawing shows a coin resting on the bottom of a beaker filled with an unknown liquid. A ray of light from the coin travels to the surface of the liquid and is refracted as it enters into the air. A person sees the ray as it skims just above the surface of the liquid. How fast is the light traveling in the liquid? (Let x = 5.12 cm and y = 5.98 cm.)
Uploaded: 2022-06-07T10:52:34-08:00
Duration: 2 min 53 s
Channel: Christian Dell
Description: The drawing shows a coin resting on the bottom of a beaker filled with an unknown liquid. A ray of light from the coin travels to the surface of the liquid and is refracted as it enters into the air. A person sees the ray as it skims just above the surface of the liquid. How fast is the light traveling in the liquid? (Let x = 5.12 cm and y = 5.98 cm.)

The drawing shows a coin resting on the bottom of a beaker filled with an unknown liquid. A ray of light from the coin travels to the surface of the liquid and is refracted as it enters into the air. A person sees the ray as it skims just above the surface of the liquid. How fast is the light traveling in the liquid? (Let x = 5.12 cm and y = 5.98 cm.)

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Transcript

00:01 Okay, and this question we have light from a coin in a beaker with an unknown liquid is moving up and then it refracts along the boundary between the liquid and the air.

00:11 Now we're given these dimensions here, 5 .12 and 5 .98 for the y and x values respectively so that we can work out this angle theta here.

00:21 Once we work out this angle theta, that will give us actually the angle of instance because these two angles will be equal to each other.

00:26 So let's do that first.

00:28 Let's work out the angle of instance.

00:30 Tan theta is equal to 5 .12 over 5 .98.

00:37 Let's clear up that 8, 5 .98.

00:40 So theta is equal to tan to the minus 1 of 5 .12 over 5 .98.

00:50 That's going to get you to 40 .6 degrees.

00:54 So this angle here is 40 .6 degrees.

00:57 That makes the angle of instance also 40 .6 degrees.

01:03 Okay.

01:04 Now the angle of a fraction, we can see, is actually a right angle.

01:09 So this becomes a critical angle, by which i mean is the angle, at which the angle of refraction is along the boundary.

01:17 So we can use the equation n1 sine r, or sorry, sign i.

01:23 N1, sign i equals n2 sign r.

01:28 This is snell's law and we're here to try and work out the speed, but in order to work out the speed of light and liquid, we need to know the refractive index of the liquid...

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