A 4 cm tall object is placed 15 cm in front of a concave mirror with a focal length of 5 cm. Where is the image produced? Is this image real or virtual? Is it upright or inverted? What is the height of the image?

s_{i} = 7.5cm; Real; Inverted, 2cm

A 4 cm tall object is placed 15 cm in front of a concave mirror with a focal length of 5 cm. Where is the image produced? Is this image real or virtual? Is it upright or inverted? What is the height of the image?

Hey guys. I hope you're able to solve this problem on your own. If not, here's a little bit of help. A 4 centimeter tall object is placed 15 centimeters in front of a concave mirror with a focal length of 5 centimeters. Where is the image produced? Is the image real or virtual? Is it upright or inverted? And what is the height of the image? Okay. So pretty much the same round of questions that you're always gonna get asked about the mirror equation. And the first three are all gonna be answered by the same bit of information, the image distance. The image distance is obviously gonna tell us where the image is located. It's It's also gonna tell us whether the image is real or inverted based on the sign of the image distance, and it's gonna tell us whether it's upright or inverted based on whether the image is real or virtual. Okay? So those first three questions all answered by the same piece of information. Now we wanna use the mirror equation, 1 over the image object distance plus 1 over the image distance equals one over the focal length, we wanna use this to find the image distance. So I'm gonna isolate one over si. It's 1 over f minus 1 over so. Now the question is what's the focal length? We're told that it's 5 centimeters but is it positive or negative? This is a concave mirror. A concave mirror converges light so it has a positive focal length. So this is going to be 1 over 5 minus the object is placed 15 centimeters in front of the mirror. So minus 1 over 15, that's the object distance. Let's simplify this. This is going to be 3 over 15 minuteus 1 over 15, which is 2 over 15. I always like to do this because it's quick and it makes the calculation easier, and you can see right away that the image distance is 15 over 2 centimeters, which is 7 and a half centimeters. So once again guys, do not forget this is not your answer. This is the reciprocal of your answer because this is one over s I. This is one of the most common mistakes I see students make is that they forget to reciprocate that answer. Okay. So take the reciprocal, 15 over 2 is 7 and a half centimeters. This is a positive image distance. So is it a real image or a virtual image? With a positive distance it is real. And since it's real, is it upright or inverted? Real images are always inverted. Okay? Those 3 always go together. Positive image distance, real image, inverted. Okay? Now what's the height of the image? For that, we need to use the magnification equation. Once again, I'm not gonna bother with the sign. We already know that this image is inverted, so the magnification, if it's negative, is just gonna tell us that. I'm just gonna plug in 7.5 as the image distance divided by 15 as the object distance, which is half. Right? So that means that the image height is half the object height. The object is 4 centimeters tall, so the image is 2 centimeters tall. Okay? That's everything we know about this image. It's located 7 and a half centimeters in front of the mirror. It's a real image. It's inverted and it has a height of 2 centimeters. Alright guys that wraps up this problem. Thanks for watching.

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A 4 cm tall object is placed 15 cm in front of a concave mirror with a focal length of 5 cm. Where is the image produced? Is this image real or virtual? Is it upright or inverted? What is the height of the image?

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