Problem

A spy satellite circles Earth at an altitude of 2…

01:29

Problem 47 Medium Difficulty

Suppose a 5.00 -m-diameter telescope were constructed on the Moon, where the absence of atmospheric distortion would permit excellent viewing. If observations were made using 500 -nm light, what minimum separation between two objects could just be resolved on Mars at closest approach (when Mars is $8.0 \times 10^{7} \mathrm{km}$ from the Moon)?

Answer

9.760 m

Topics

Electromagnetic Waves

Wave Optics

Book Cover for College Physics 2017

College Physics 2017

Chapter 25

Optical Instruments

Discussion

You must be signed in to discuss.

Top Physics 103 Educators

Christina K.

Rutgers, The State University of New Jersey

Farnaz M.

Other Schools

Zachary M.

Hope College

Aspen F.

University of Sheffield

Watch More Solved Questions in Chapter 25

Problem 1

Problem 2

Problem 3

Problem 4

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

Problem 11

Problem 12

Problem 13

Problem 14

Problem 15

Problem 16

Problem 17

Problem 18

Problem 19

Problem 20

Problem 21

Problem 22

Problem 23

Problem 24

Problem 25

Problem 26

Problem 27

Problem 28

Problem 29

Problem 30

Problem 31

Problem 32

Problem 33

Problem 34

Problem 35

Problem 36

Problem 37

Problem 38

Problem 39

Problem 40

Problem 41

Problem 42

Problem 43

Problem 44

Problem 45

Problem 46

Problem 47

Problem 48

Problem 49

Problem 50

Problem 51

Problem 52

Problem 53

Problem 54

Problem 55

Problem 56

Problem 57

Problem 58

Problem 59

Problem 60

Problem 61

Problem 62

Problem 63

Problem 64

Problem 65

Problem 66

Problem 67

Video Transcript

in this problem. The diameter of the moon telescope is Deco's five meter. The wavelength of the light used to resolve to objects in Mars is 500 nanometer, which is 500 times 10 to the negative man meter. And the distance between the Mars and the moon is eight times to do 70 kilometer. So I'm converting this to eight times 10 to the 10 meters. Okay, so when the objects are just result than the angular separation of the object is, data is going to be that I mean, and by definition of this, it's going to be 1.22 Lambda over the Okay, So from here, we can actually get 1.22 times. The Web link is 500 times 10 to the negative, nine over five. And this will give me 1.22 times 10 to the negative seven. Grady in. All right now, we actually have to find the separation between the objects. So to do that, we're gonna use another one definition, which is for the Anglo resolution. If you remember, data is going to be called to separation between the objects and the distance between the observer to the object that I goes ass over are so from here you can actually write down as equals. Data are which is going to be 1.22 times 10 to the negative. Seven times eight times, 10 to the 10 and this is going to give me 97 60 meter, which is 9.76 kilometer.

Netra S.

University of Wisconsin - Milwaukee

Topics

Electromagnetic Waves

Wave Optics

Book Cover for College Physics 2017

College Physics 2017

Chapter 25

Optical Instruments

Top Physics 103 Educators

Christina K.

Rutgers, The State University of New Jersey

Farnaz M.

Other Schools

Zachary M.

Hope College

Aspen F.

University of Sheffield