Resolving Power and minimum seperation

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The discussion revolves around calculating the minimum separation of two stars observed through a telescope with a 2.42 m diameter aperture and a light wavelength of 547 nm. The formula used is theta = 1.22 * lambda / D, leading to a proposed minimum separation of 2.818 x 10^-4 m. Participants clarify that a distance of 1022 m for the stars is unrealistic, suggesting it should be 10^22 m instead. This adjustment significantly changes the minimum separation calculation to 2.758 x 10^15 m. The conversation emphasizes the importance of accurate distance measurements in astronomical observations.
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(a) Two stars are photographed utilizing a telescope with a circular aperture of diameter of 2.42 m and light with a wavelength of 547 nm. If both stars are 1022 m from us, what is their minimum separation so that we can recognize them as two stars (instead of just one)?

theta=1.22* lambda/D

(b) A car passes you on the highway and you notice the taillights of the car are 1.27 m apart. Assume that the pupils of your eyes have a diameter of 6.7 mm and index of refraction of 1.36. Given that the car is 14.2 km away when the taillights appear to merge into a single spot of light because of the effects of diffraction, what wavelength of light does the car emit from its taillights (what would the wavelength be in vacuum)?
 
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What do you think?
 
Just to check my own understanding. Is the answer to part a) 2.818*10^-4m? Come to think off it, can the star be only 1022m away from us?

Delzac
 
Delzac said:
Just to check my own understanding. Is the answer to part a) 2.818*10^-4m? Come to think off it, can the star be only 1022m away from us?
No, if you were only around 1 km from the center of a star, you would be inside the star's core. Your body would be crushed to a size of a small stone, under such extreme temperature and pressure that the atoms in your body would lose their electrons and the nuclei would fuse to form new elements. You would become an extraordinarily dense, hot mishmash that is barely even comprehensible by most humans.

Perhaps you mean the distance of the stars is 1 x 1022 m away from us?
 
If the distance is 10^22m, then will be answer be 2.758*10^15m?

delzac
 
ok:) thank you:)
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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