## 9.760 m

#### Topics

Electromagnetic Waves

Wave Optics

### Discussion

You must be signed in to discuss.
##### Christina K.

Rutgers, The State University of New Jersey

##### Aspen F.

University of Sheffield

### 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.

University of Wisconsin - Milwaukee

#### Topics

Electromagnetic Waves

Wave Optics

##### Christina K.

Rutgers, The State University of New Jersey

##### Aspen F.

University of Sheffield