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A photographer takes a photograph of a Boeing 747 airliner (length 70.7 $\mathrm{m}$ ) when it is flying directly overhead at an altitude of 9.50 $\mathrm{km}$ . The lens has a focal length of 5.00 $\mathrm{m} .$ How long is the image of the airliner on the film?

The image of the airliner on the film is $[0.0372 \mathrm{m}]$

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

McMaster University

Lectures

02:30

In optics, ray optics is a geometric optics method that uses ray tracing to model the propagation of light through an optical system. As in all geometric optics methods, the ray optics model assumes that light travels in straight lines and that the index of refraction of the optical material remains constant throughout the system.

10:00

In optics, reflection is the change in direction of a wavefront at an interface between two different media so that the wavefront returns into the medium from which it originated. Common examples include the reflection of light, sound and water waves. The law of reflection says that for specular reflection the angle at which the wave is incident on the surface equals the angle at which it is reflected. Reflection may also be referred to as "mirror image" or "specular reflection". Refraction is the change in direction of a wave due to a change in its speed. The refractive index of a material is a measure of its ability to change the direction of a wave. A material with a higher refractive index will change the direction of a wave to a greater degree than a material with a lower refractive index. When a wave crosses the boundary between two materials with different refractive indices, part of the wave is refracted; that is, it changes direction. The ratio of the speeds of propagation of the two waves determines the angle of refraction, which is the angle between the direction of the incident and the refractive rays.

05:38

A photographer takes a pho…

01:53

A movie camera with a (sin…

02:08

02:43

02:14

Suppose a 200 $\mathrm{mm}…

05:53

$\bullet$ It's all do…

01:30

PHOTOGRAPHY The angle of …

01:49

A camera with a $100 \math…

02:12

Suppose your 50.0 $\mathrm…

Okay, we have We have object distance of 9500 meters. Okay. And we have a focal length of five meters so immediately you can see that object. Distance ass is much, much greater than f Ah. So in this equation, one of her f equals one over s prime. Plus, remember ass the one of wrasse primes. This is approximately the one arrest prime goes to zero. Ah, because after summer, smaller than mess or one of rascals deserve. Excuse me because I felt I was song on us and therefore we can approximate its prime to the F eso image is formed at the focal length of five meters. Calm the lens therefore ah, magnification is just negative that over imaged of rhe object distance. So that's negative. Five over 9500 on DH. So that's negative. 5.26 times 10 to the negative four. Um um And so why prime over? Why are the absolute limit of the absolute value of y prime inside of our object height? It's just absolute value magnitude, something we know therefore y prime. The length of the image of the airplane is just 5.26 times 10 to the minus for um times the actual length of the of the aircraft. 70.7 meters. Therefore, the height of the images. Three point 772 times 10 of negative two meters, in other words, 3.72 centimeters and that's it.

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