## 2.53 \times 10^{5} m

#### Topics

Electromagnetic Waves

Wave Optics

### Discussion

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##### Christina K.

Rutgers, The State University of New Jersey

LB
##### Aspen F.

University of Sheffield

### Video Transcript

in this problem. The image of an object off size H O in the moon formed by the objective of the telescope, is shown in the figure. The focal length of the objective off the telescope is 1500 centimeters. The distance between the Earth and moon, which is also the object distance, is 3.8 times to do the eight meter. The height of the image formed by the objective H is one centimeter Now to find the H O, which is the height of the the size of the object in the moon. We used the fact that this that since the object is distant object, the image formed by the objective would be at the focal length of the objective. So what that means is actually d I is going to be half of the objective is going to be cool, too. 15 meter. Okay, so now we can look into figure again and meth medically, you can see that the triangle that performed on the object side and the emails side are actually the similar triangle. So mathematically we can also see that h I over Asia. Oh, because d I over d'oh, which is the angle. Tater. What? So from here, h o is going to be go to H I times d o over d I. And this is going to be equal to zero point 01 times 3.8 times 10 to the eight. 3.8 times 10 to the eight over 15. And this is going to be colder 2.53 times 10 to the five meter now one mile equals 1609 meters. So in terms of mile, this is going to be 2.53 times 10 to the five over 161609 mile. What? She's actually he called to 1 57 Let me write down on the top here. So this actually is going to be you called Thio 1 57 0.44 miles.

University of Wisconsin - Milwaukee

#### Topics

Electromagnetic Waves

Wave Optics

##### Christina K.

Rutgers, The State University of New Jersey

LB
##### Aspen F.

University of Sheffield