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. Zoom lens, I. A zoom lens is a lens that varies in focal length. The zoom lens on a certain digital camera varies in focal length from 6.50 $\mathrm{mm}$ to 19.5 $\mathrm{mm}$ . This camera is focused on an object 2.00 $\mathrm{m}$ tall that is 1.50 $\mathrm{m}$ from the camera. Find the distance between the lens and the photo sensors and the height of the image (a) when the zoom is set to 6.50 $\mathrm{mm}$ focal length and (b) when it is at 19.5 $\mathrm{mm}$ . (c) Which is the telephoto focal length, 6.50 $\mathrm{mm}$ or 19.5 $\mathrm{mm} ?$

a) 0.871 $\mathrm{cm}$b) 2.64 $\mathrm{cm}$c) 19.5 $\mathrm{mm}$

Physics 102 Electricity and Magnetism

Physics 103

Chapter 25

Optical Instruments

Electromagnetic Waves

Reflection and Refraction of Light

Cornell University

Hope College

University of Sheffield

Lectures

02:30

In optics, ray optics is a…

10:00

In optics, reflection is t…

06:07

A camera lens has a fixed …

08:18

Zoom Lens. Consider the si…

03:02

Photography. A 35 -mm came…

07:59

A zoom lens system is a co…

01:51

A lens for a digital camer…

08:54

An object is located at $x…

01:14

A camera lens used for tak…

08:07

CamerasYou would like …

05:19

A simple camera telephoto …

03:04

(II) A 105 -mm-focal-lengt…

(II) A 105-mm-focal-length…

02:35

Focusing Distance For a ca…

02:22

Camera lenses are describe…

06:25

A telephoto lens system ob…

02:12

Suppose your 50.0 $\mathrm…

04:53

$\cdot$ A certain digital …

05:48

Choosing a Camera Lens. Th…

02:58

A thin lens with a focal l…

05:25

(II) An 80 -mm-focal-lengt…

00:44

$\cdot$ A thin lens with a…

Okay, so you have two parts and part of it. The focal length is 0.65 centimetres are 6.5 millimeters. Ah, and you want to find ah, image distance as prime. So one of grass prime, remember, is just one of her after minus one over object distance. So this is one of her 10.65 centimetres minus object whatever object systems, which is 1 50 centimetres. So you work that out and you get us prime of six point 53 millimetres. In other words, 0.653 centimeters. And so you confined magnification here from negative image distance over object distance. So that's negative 0.653 over 1 50 about centimeters. And so that's negative. 4.35 times 10 to the late of three. By the way, I'm using centimeters has my units of link here because it is a happy medium between the millimeter and the meter that we have. Ah, in the question. Ah, and so the image height, which we just want the magnitude. And so that's equal to the magnitude of the magnification times object hype. And so that's 4.35 times 10 to the negative three times object had us two meters, so that's 200 centimetres. So image hi ISS less than a centimetre, it's actually 0.87 centimeters or a 0.7 millimeters. In Part b, you have basically the same thing, but here one of OK and one of her f is is what's different? Ah f is come f is 19.5 millimeters sea of one of references, one point 95 centimeters minus again 1 50 centimetres for one of 1 50 centimeters for the image height or the image of object distance excusing. So as prime works out to be 1.98 centimeters or 19.8 millimetres and in the same way we find magnification. So this is 1.98 over 1 50 both centimetres, so those units cancel and that's negative 1.32 times total. Negative, too, is a magnification, and again we find the magnitude of the ah, an image height and so that's one point through to tear a negative two times 200 giving us ah 2.64 centimeters or to 60 26.4 millimeters and so and part C. We want the We want the telephoto locally, the telephone of focal length is which at wherever I am is larger. So so the answer is 19.5 millimeters. It's telephoto focal length, and this is because magnification is larger. And as you can see, the magnitude of magnification is close to one order of magnitude larger in the 19.5 millimeters case, and that's it.

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