A 1.0-cm-tall object is 10 cm in front of a converging lens th...

Question

A 1.0-cm-tall object is 10 cm in front of a converging lens that has a 30 cm focal length. Calculate the image position and height. Compare with your ray-tracing answers in part a.

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 lepidoptera uses a magnifying glass, a type of converging lens that has a focal length of 10.0 centimeters to inspect a moth that is 2.0 centimeters long. The moth is placed 9.0 centimeters away from the lens, calculate the numerical values for I, the position S prime and I I the height H prime of the image. So that's our end goal is we're trying to find two separate answers. We're trying to find an answer for part I and I, I, for I, we're trying to find the position S prime and for I, I, we're trying to find the height H prime of the image. Awesome. We're also given some multiple choice answers for parts I and I I. So for both parts for both S prime and H prime, they're all in units of centimeters. So let's read them off to see what our final answer pair might be. So for A S prime is negative 90 H prime is 20. For BS prime is negative 90 H prime is 10. For CS prime is 90 H prime is 20. And finally, for DS prime is 90 and H prime is 10. OK. So first off, let us write down all of our known and unknown variables. So H is equal to 2.0 centimeters. And H represents the objects height S is equal to 9.0 centimeters. And this represents the object distance. F our focal length is given to us as 10 centimeters, 10.0 centimeters. We do not know what S prime is. We do not know what the image distance is and we do not know what the image height is, which is H prime. So H prime and S prime, we do not know what those two values are. So to solve for part I let us start to solve this problem by finding the image distance S prime. So to solve for this value, we need to recall and use the thin lens equation since the relationship between the object distance and the image distance and the focal length is given by the following one, divided by F is equal to one divided by S plus one divided by S prime. So now we could substitute in all of our known variables and solve for S prime by rearranging the equation. So S prime is equal to. And after we rearrange and simplify, we'll get that and plug in our known variables. So once we simplify and plug in our known variables, we'll find out that S prime is equal to one divided by one, divided by 10.0 centimeters minus one, divided by 9.0 centimeters. So when we plug that into a calculator, we will get that S prime is equal to 90 centimeters, negative 90 centimeters, which we found our first answer hooray. So for part I it's equal to negative 90 centimeters. So S prime is equal to negative 90 centimeters. And let's note that the negative sign indicates that the image is virtual. Now we could start solving for part I I. So now we need to focus on the images height. So to find this value, we must recall and use the magnification equation. Note that the magnification of a lens is given by the ratio of the image height to the object's height and is also equal to the ratio of the images distance to the images sorry from the images distance to the object's distance written as. So we can write these ratios as M is equal to negative S prime divided by S is equal to H prime divided by H. So now we need to rearrange this equation to isolate and solve for H prime. So when we do that, when we rearrange and plug in all of our known variables, we will get that H prime is equal to negative negative 90.0 centimeters, which we could take our simple N negative 90 centimeters and just add a 0.0 on there to keep it consistent decimals when we're solving. And we need to multiply our negative 90.0 centimeters by 2.0 centimeters. And we need to divide everything by 9.0 centimeters. And when we do that, when we plug it into a calculator, we will get an even 20 centimeters. So that means that R H prime, which is our height of the image is equal to 20 centimeters. And this is also our answer for part I I, so therefore, the image will be 10 times larger than the object. So looking at our multiple choice answers, the correct answer has to be the letter A I is S prime is equal to negative 90 centimeters and IIH prime is equal to 20 centimeters. Thank you so much for watching. Hopefully, that helped and I can't wait to see you in the next video. Bye.

A 1.0-cm-tall object is 10 cm in front of a converging lens that has a 30 cm focal length. Calculate the image position and height. Compare with your ray-tracing answers in part a.

Key Concepts

Lens Formula

Magnification

Ray Tracing

Watch next