Answer & explanation
Correct answer: option 2
$ I = I_0 (1 - e^{\frac{-t}{\tau}}) = \frac{3}{4} I_0$
$\Rightarrow e^{\frac{-t}{\tau}} = \frac{1}{4}$
$\Rightarrow \tau = \frac{t}{2ln2} = \frac{4}{2ln2} = \frac{2}{ln2} $
Correct answer: option 2
$ I = I_0 (1 - e^{\frac{-t}{\tau}}) = \frac{3}{4} I_0$
$\Rightarrow e^{\frac{-t}{\tau}} = \frac{1}{4}$
$\Rightarrow \tau = \frac{t}{2ln2} = \frac{4}{2ln2} = \frac{2}{ln2} $