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$\bullet$ It's all done with mirrors. A photographer standing 0.750 $\mathrm{m}$ in front of a plane mirror is taking a photograph of her image in the mirror, using a digital camera having a lens with afocal length of 19.5 $\mathrm{mm}$ (a) How far is the lens from the light sensors of the camera? (b) If the camera is 8.0 $\mathrm{cm}$ high, how high is its image on the sensors?

a. 19.8 $\mathrm{mm}$b. 1.06 $\mathrm{mm}$

Physics 102 Electricity and Magnetism

Physics 103

Chapter 25

Optical Instruments

Electromagnetic Waves

Reflection and Refraction of Light

Cornell University

Rutgers, The State University of New Jersey

University of Michigan - Ann Arbor

University of Sheffield

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All right. So for the first question, he was asking us all parcelled lands for a life sensors off camera. Okay, so is asking us with this the image distance to people find out average distance. Okay, so we know that while having people along with you plus one were you fogle, ants, bees, that image distance used the object business. So if we do, some arrangement will get while over me is equal to one over after minus one over you. Okay. And this can be equal to, um you see, uh, u minus f over you are. Okay, so we can be inquisitive. U f over u minus at. Okay. So v zero to you wear over you mine's huh? And the object is since in his case is two times 0.75 meters, which is one point by leader. Okay. Why? I said, Well, it's because the distance between the mirror and the persons he still points on my meter. And since you look in a mirror so the image of the person is for at a distance off 0.7 0.7 find me to be young here because on in a mirror The image of you and yourself is symmetric. So the actual options this is, is two times holding 75 year, which is 1.5 meters. Okay. And the focal lens is giving us 9 19.5 millimeter, which is 19.5 times tens of pollen. 93 Peter. So we'll get we is equal to 1.5 meter times, 19.5 times potential power. Thank three meter over one point by meter, minus 19.5 has tens of power factory meter and this will give us the image of distance. It's a bow. Um e it was about 19 point a millimeter. Okay, so for question be it was asking us how high is the image on the sensors? So you want us to find out the hydro in beach on window the formula off a magnification which is m equal to minus V over u. He's also legal age crying over h. He's the image. Distant you is the object Sisters. Ace prime is the actual heights. Um, sorry finds the hydro image. H is the actual heights. And the minus sign here means image. Missing bird. Okay, so since we were looking for to find out the height of image so we can design a random in here to put HPAI on the left side, which is a strike him in equal to minus three over you. Times age. Okay? And we know leave on the previous question, which is minus 19 going a times sent to a power 93 meter over you with just one point by meaner test H. It's just given. You were saying that a camera was a senior high. It was just 0.0 a meter. Okay. And this will give us H crimes. He goto negative zero point 00 106 meter. Okay. Just like what I said. That gang sign here, it doesn't mean, uh, naked wearing it just means that image is inverted. Okay, So if we are looking for how high is it? Actually, so we don't need a value off the price, which is absolutely, absolutely here, too. Reduce the confusion here. Okay. We'll get absolute, uh, negative point. Oh, while the six meter okay. And he's will give us point. Oh, while six meter, which is 1.6 now. Meter. Okay, So the image on the sensor, it will be 1.6 millimeter height. Okay, A long point. 06 millimeter tall. So these are the solutions for the passion. Thank you.

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