So how much has the exposure to light darkened the photographic plate? As much as expected for 301 ns or 33 ns ?
For 33 ns. That is the time that the clock inside the ship-muon indicates. We measure 301 ns of exposure, but a much shorter time was experienced by the moving photographic plate. This question is tricky also because the light will be red-shifted by <PERSON> effect, but lets forget about this detail.
These questions about "what the other observer would observe that I observe" are complicated. I suggest to follow the approach of the invariants: there are quantities that, measured by any observer, give the same result. One of them is the proper-distance between events. An event is any time and position (in this example, height) coordinate. There are only two important events in this problem: the muon enters the atmosphere at 15 km above the ground (lets called it $E_1$), and the muon reaching the ground ($E_2$). The specific coordinates given to the events depend on the observer. For example, for the muon, assuming that it sets its clock to zero when it enters the atmosphere:
$E_1: (t,z)_M=(0 s,z_0)$
Is $z_0=0$ for the muon? well, it could be $z_0=15$ km, or $z_0=100$ km, it really does not matter. What it matters is that for the event 2, for the muon:
$E_2: (t,z)_M=(t_M,z_0)$,
that is, both events, for the muon, happen at its same spatial position. What about us from the ground? well, lets assume we set our clock to zero when the muon reach the top of the atmosphere (but it does not matter really):
$E_1: (\tau,\zeta){\rm Earth}=(0,15~{\rm km}),\quad E_2: (\tau,\zeta){\rm Earth}=(\tau_M,0~{\rm km})$.
I am using Greek letters for my coordinates. 272
Now, special relativity tells us that the proper distance between these events is the same as measured by any observer. The proper distance is the squared difference of the time (times $c^2$) minus the spatial distance squared. That is: $$c^2(t_M-0)^2-(z_0-z_0)^2=c^2(0-\tau_M)^2-(15~{\rm km}-0)^2\tag{1}$$ And that's it. Forget about that one will be contracted measured by the other, and one time is dilated, etc...Lets divide Eq. (1) by $c^2\tau_M^2$: $$ \left(\frac{t_M}{\tau_M}\right)^2=1-\left(\frac{15~{\rm km}}{c\tau_M}\right)^2$$ but we know that $(15~{\rm km}/\tau_M)=0.994 c$, and therefore you have it, the time dilation. Since $\tau'_M=301~{\rm ns}$, therefore $t'_M=301~{\rm ns}\,\sqrt{1-0.994^2}=33~{\rm ns}$. I use a $'$ to denote the fact that we are not talking of the first event at 15 km from the ground, but from a much closer distance. The time dilation factor is the same.
Ok, lets solve the original problem that has three events:
1. Muon enters the atmosphere at 15 km of height.
2. The ray of light that the muon emitted toward the ground reaches the ground.
3. The muon reaches the ground.
Again, latin letters for the muon coordinates and greek for the ground observer. For the first event: \begin{align} E_1: &(0,0) \quad\text{ for the muon.}\ &(0,15~{\rm km}) \quad\text{ for the ground observer.} \end{align}
For the second event: \begin{align} E_1: &(t_2,z_2) \quad\text{ for the muon.}\ &(15{\rm km}/c,0) \quad\text{ for the ground observer. 272