## sauravshakya 3 years ago WHAT WILL THIS PERSON SEE FOR this extra one second. suppose there a man and an object seperated by some distance. we know that it will take some time for light to travel from that object to the man's eye. Now, when the object is at position (i) then it will take 5 seconds for the light to travel from the object to the man' eye. so the man will see the object after 5 seconds. Now, when the object is at position (F) then it will take 6 seconds. Now, at the third case the object is moving from the position (i) to position (F)... and the man is just looking at the object.

1. sauravshakya

Now since the object is moving away from the man it will take some more time for the light to travel and reach the man's eye. So will there be a case he will see the object for fraction of second and not see for another fraction of second. And this Fraction of second will add to 1. I hope i made my question clear.

2. sauravshakya

@experimentX PLZ HELP...

3. sauravshakya

add to 1 second since 6-5 is one second.

4. mathavraj

i dont understand it y will the man wont c the obj..the light is still shining on it rit..so im not sure i think when the obj has started moving from I he will see the obj at I for some time till the light reaches him...and so he sees everything delayed...similar to the stars we see the stars as they were before millions of yrs ago..im not sure abt my answer..so if im wrong pls dont scold me

5. ganeshie8

not an attempt to answer as i suck in these... just adding to the question : think how situation becomes if the object reflects/emits light ONLY when its moving. will it flicker... **** assume you have good high-speed camera that takes pics at Ghz... (discovery time-warp show)

6. sauravshakya

@mukushla CAN U HELP?

7. aish_premrenu

What is the question again? can you clarify?

8. Vaidehi09

i have to agree with @mathavraj. the observer will see the entire thing...just delayed by some time. first, he will see the object at I, 5 sec later. now the object takes 1 sec to travel from I to F. so in the 1 sec AFTER those 5 secs, he will see just that - the object moving from I to F. it won't be like he sees the object for the first half, and not for the second half. the object won't disappear from his sight at all times, in the given situation.

9. Vaidehi09

@ganeshie8 i'm not sure about this...so feel free to correct me: if the object reflects light ONLY when it is moving, then it will do so throughout its motion from I to F, right? so maybe, it'll be like this: we won't see the object when it is at rest at I. we'll see it when it moves from I to F. and then again, when it stops at F, it'll disappear from our sight. so something like, it appears out of thin air for some time and disappears again....that'll be a sight!

10. qpHalcy0n

This is starting to sound like length contraction... :]

11. Dennylama

You haven't specified how long the object takes to travel from I to F. If the travel time = 1 second, the object will be traveling at the speed of light, and according to the Lorentz-FitzGerald contraction formula the object would be invisible ( at velocity C of the object , length in the direction of travel = 0 to a stationary observer )

12. qpHalcy0n

^ This. Length contraction. The effect is virtually unnoticeable until the object's speed gets extremely close to the speed of light. Remember when you are "seeing" an object, you are seeing light. Not an object...

13. Dennylama

If the object travels from I to F in 1 second, it would have to be traveling at the speed of light! The contraction would render it invisible in that one second. .

14. ten

The light from the object would be red-shifted while it's moving away. Almost all galaxies are moving away from us, so the light constantly takes longer to get to us.The wavelengths stretch out a bit longer as the move away.

15. neharika

um..it sounds to me...as far as i can understand that there will only be a delay of 5 min in visualizing the object..after that..it will see the object moving from (i) to (ii)..provided that these points are both close enough to the observer and at enough distance from each other..

16. sauravshakya

@akhil009