A 25-cm-long rod lies along the optical axis of a converging lens...

Question

A 25-cm-long rod lies along the optical axis of a converging lens, perpendicular to the lens plane. The lens has a 30 cm focal length. The rod's real , along the optical axis on the other side of the lens, is also 25 cm long. What is the distance from the lens to the nearest end of the rod?

Accepted Answer

Hello, fellow physicists today, we want to solve the following practice problem together. So first off, let's read the problem and highlight all the key pieces of information that we need to use in order to solve this problem in a physics lab experiment on optics. A student uses a converging lens with a focal length of 15 centimeters to produce an image that is exactly matches the size of a thin stick. The stick is placed horizontally along the principal axis of the lens and has a size of 8. centimeters, determine the distance between the lens and the sticks farthest end. So that's our end goal is to determine the distance between the lens and the stick's farthest end. OK. So we're given some multiple choice answers here. They're all in the same units of centimeters. Let's read them off to see what our final answer might be. A is 12 B, is 26 C is 35 and D is 43. So first off, let us start by drawing a diagram to help us better visualize the conditions provided to us in this problem. So here we have our lens are converging lens and then we also have the matches or produces an image that I exactly matches the size of a thin stick. OK. And this is the lenses, principal axis or AC I should say. So let's call the S one and S two, the lens and S prime one and S prime two, the, the stick. So by using our diagram in using the information provided to us in the problem itself, we can write that that S one minus S two is equal to 8. centimeters. And let's call that equation one. And we can also say let's call this equation two, the S two prime minus S one prime is equal to 8.0 centimeters. So we can use this relationship between the object's distance and the image distance. And the focal length F or a lens can be written as and let's call it equation three. That one divided by S one plus one divided by S one prime is equal to one divided by F. OK. So we can go on to say, let's call it the equation for that one divided by S two plus one, divided by S two prime equals one divided by F which F is the focal length. So using the same logic from equation three, we can write equation four. So at this point, we have four different equations with four unknown variables. Our end goal is to determine the value for S one. So let us start by solving for S one by isolating and solving for S one prime and S two prime using the equations three and four. So one, S one prime is equal to one divided by F. So one divided by S one prime is equal to one divided by F minus one divided by S one which we can rearrange and isolate the sulfur S one prime which when we rearrange using a little algebra, we get that F multiplied by S one divided by S one minus F. So we could do the same method or S two prime. So S two prime equals one divided by. So one divided by S two prime is equal to one divided by F minus one divided by S two. And let's call it equation six. So doing the exact same thing isolating and solving for S two prime, we get that S two prime is equal to F multiplied by S two divided by S two minus F. So now we can use the equation one and plug it into equation six. So let's do that. So S two prime is equal to F multiplied by S one minus eight centimeters divided by S one minus eight centimeters minus F. And let's note, make a note here that S two is equal to S one minus eight centimeters. OK. So now we need to use equation two to help us manipulate the math, to get our equation to yield a simple quadratic equation. So let's do that. So S two prime and see is subtracted from. So S two prime minus S one prime is equal to eight centimeters. So we can write that F multiplied by S one minus eight centimeters or 8.0 centimeters. I should say divided by S one minus 8.0 centimeters minus F minus F multiplied by S one divided by S one minus F is equal to eight centimeters. OK. So now we can plug in our known variables and start to simplify down to a quadratic function. So let's do that. So the focal length is centimeters multiplied by S one minus eight centimeters divided by S one minus eight centimeters minus centimeters minus centimeters multiplied by S one divided by S one minus 15 centimeters is equal to eight centimeters. So at this stage, so that in order to save some time. And so this video isn't too long, we're gonna have to use a bunch of algebra to get the resulting quadratic function. So we're trying to manipulate this equation that we just wrote into a quadratic function. So when we do that, the simplified resulting quadratic function will be S one squared. So if you don't understand how to do the algebra, I would watch some algebra videos to practice. So you can practice manipulating the equation to get it down to a quadratic function. But in order to save time, I'm gonna skip over it. So S one squared minus three or S one squared minus 38 multiplied by S one plus equals zero. OK. So as we know, a quadratic function gives us two answers. So let's write them both down to see what our two answers will be. So S one equals mm equals negative multiplied by negative 38 minus. Let's start with minus or minus square root of negative 38 squared, which just in case is a quick little side note, remember that the quadratic function states that X equals opposite B plus or minus square root B squared minus four ac all over two A. OK, minus four multiplied by one multiplied by divided by two, multiplied by one and her second equation. Well, let's write down the answer. So when you plug in all that into a calculator, you'll get 3.47 centimeters. So now let's do s one for the plus condition. So we write everything down a second time except it's gonna be plus a set of minus. So minus or negative multiplied by negative 38 plus square root, negative 38 squared minus four, multiplied by one multiplied by 120 divided by two, multiplied by one. So when you plug that into a calculator, we should get 34. centimeters. But when we round up, we should get centimeters. OK? So looking at our 3.47 centimeter answer. This solution is impossible because the stick will pass through the lens and that isn't, that's invalid. We can't have that as an answer. So the distance between the lens and the stick at the farthest end is 35 centimeters is our final and correct answer and we did it. So let's look at our multiple choice answers to see what the correct answer is. The correct answer is the letter C 35 centimeters. Thank you so much for watching. Hopefully that helped and I can't wait to see you in the next video. Bye.

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A 25-cm-long rod lies along the optical axis of a converging lens, perpendicular to the lens plane. The lens has a 30 cm focal length. The rod's real , along the optical axis on the other side of the lens, is also 25 cm long. What is the distance from the lens to the nearest end of the rod?

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