Doubt for binary search

#1

why n/2^k =1 innthis binary search?

#2

Imagine the entire search space of n as consisting of n equal unit length segments. Now, how many iterations you need to make to reach a search space of 1 unit. That is equal to the value of entire search space divided by the branching factor = 1
Here branching factor is 2 , so hence the formula

#3

No, i m saying why is that =1. is that because here 1 refers to the end of iterations or something like that. i mean why=1 why not=2 or 3 or something?

#4

1 here refers to its worst case possibility. Remember that O(log n) is the worst case time complexity of binay search and is not the complexity for every case. There may exist cases in which we search for a number in just 1 iteration.