## 1.05 \mathrm{m} \text { to } 6.30 \mathrm{m}

#### Topics

Electromagnetic Waves

Wave Optics

### Discussion

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##### Top Physics 103 Educators ##### Andy C.

University of Michigan - Ann Arbor ##### Marshall S.

University of Washington ##### Aspen F.

University of Sheffield ### Video Transcript

So in this problem we have a camera with the lens of focal length 175 millimeter in which the female's distance can be actually adjusted from 180 millimeter to 210 millimeter. So let's say Q one equals 180 millimeter and Que Tu goes 210 millimeter. So based on that, we can easily find the object Distance ridge. All right, so all we have to do is just use the lens equation. So from less equation, Lindsay equation is won over F. It costs one overpay plus one of our cue. So from here, we can actually write down P. It calls cute times half over Q minus F. So if we wanna find out two different cases. So let's say we want to find p one p one is going to be equal to Q one times, half over Q one minus s. So we have the value of Q one and F. You just plug it in here, which is going to give you 6300 millimeter or 6.3 meter, and P two is going to be Q two times F over cute, too, minus F all right, so just plug in Q two and f here, and you're going to get 1050 millimeter, which is 1.5 meter. So the object distance actually ranges from 1.5 millimeters. 1.5 meter, too. 6.3 meter. University of Wisconsin - Milwaukee

#### Topics

Electromagnetic Waves

Wave Optics

##### Top Physics 103 Educators ##### Andy C.

University of Michigan - Ann Arbor ##### Marshall S.

University of Washington ##### Aspen F.

University of Sheffield 