A giant ocean tank at an aquarium has acrylic plastic walls 18...

Question

A giant ocean tank at an aquarium has acrylic plastic walls 18 cm thick. The index of refraction of acrylic plastic is 1.49. A fish is 220 cm from the inside wall. To a viewer on the outside, how far does the fish appear to be from the outside wall? Hint: The of the first refraction is the object for the second refraction.

Accepted Answer

Hello, fellow physicists today, we're gonna solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. A greenhouse uses glass panes with a thickness of 3.0 millimeters. The index of refraction of glass is about 1.52. A scientist is observing a plant that is 100.0 centimeters away from the inside of the glass to an observer outside the greenhouse. How far does the plant appear to be from the outside of the glass pane? So that's our end goal is we're considering what the how far. So we need to figure out how far does the plant appear to be from the outside of the glass pane from an observer located outside of the greenhouse? That's our end goal. What we're ultimately trying to solve for is what is the distance that the observer outside of a greenhouse is when it's observing a plant from the observing a plant that's within the greenhouse from the outside of the glass pane? OK. So with that in mind, let's look at our multiple choice answers. And let's also note that they're all in units of meters. Let's read them off to see what our final answer might be. A is 0.66 B is 1.0 C is 1.7 and D is 2.3. OK. So first off, let us make the following assumption, let us assume that all of the light will pass through the glass without being absorbed or reflected. With that said, we can now start solving for this problem by determining what the apparent distance between the plant to the inner surface of the glasses. And this is our first step. So we need to note that in this case, the medium is air and the refractive index of air is, and let's write this down in blue. So the refractive index of air, I'm gonna denote it as N A is equal to 1.00. Also, let us note that we are told that the actual distance from the plant to the inside of the glass is 100 centimeters. Now, we can write the equation to solve for the apparent distance. And let's call the apparent distance, let's call it D prime is equal to negative N A divided by NG where NG is the refractive index of glass multiplied by D and D. In this case is the actual distance and D the actual distance is equal to 100 centimeters as we mentioned before. And once again D prime is the apparent distance. So now at this stage, we can plug in all of our known variables and solve for D prime. So let's do that. So D prime is equal to negative 1.00 divided by 1.52 multiplied by 100 centimeters. But as we can tell all of our multiple choice answers are in meters. So let's convert centimeters to meters. So we need to multiply 100 by 10 to the power of negative two. So when we plug that into a calculator, we will get negative 0.66 m when we rounded two decimal places. So the light from the plant appears to be 66 centimeters from the inner surface of the glass or 0.66 m. So now our second step is that we need to solve for the distance value between the inner surface of the glass to the observer outside of the greenhouse, which is our final answer that we are looking for. So we need to note that the image distance found in step one serves as the exact object distance on the same side of the interference as the plant. And please note that the object distance must be a positive value. We must also consider that the 3.0 millimeter thickness of the glass. So we must write our total object distance value as D prime plus the thickness which is 3.0 multiplied by 10 to the power of negative 3 m because we need to convert millimeters to meters. And for this particular condition, our medium will be glass instead of air. So we must therefore write our equation for the apparent distance of the plant to the observer outside of the greenhouse. And we need to call this value D so or D prime prime. So D prime prime, which I'll say once again, since this is a lot of information all at once bear with me. So the apparent distance of the plant to the observer found outside of the greenhouse. And this is denoted by D prime prime. And D prime prime is equal to negative negative NG divided by N A multiplied by D prime plus 3.0 multiplied by 10 to the power of negative 3 m. So now we can plug in all of our known variables and solve for D prime prime. So let's do that. So D prime prime is equal to negative 1.52 divided by 1.00 multiplied by 0.66 m plus 3.0 multiplied by 10 to the power of negative 3 m which is equal to negative 1.0076 which is approximately equal to 1.0 m. Hooray, we did it. So the observer outside of the greenhouse will see the plant at an apparent distance of 1.0 m. In this value. We write it as positive because we can't have a negative distance, the negative sign in this case, just indicates the orientation of the image that's produced, which we're not concerned with this, with that particular piece of information for this problem. So we can just ignore it. So our final answer is 1.0 m and this corresponds with the multiple choice answer letter B and that's it. Thank you so much for watching. Hopefully, that helped and I can't wait to see you in the next video. Bye.

Key Concepts

Refraction

Index of Refraction

Optical Path Length

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