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when we switch on the electric light in the dark room, the light energy is radiated from source. Once we switch off the light, the radiated energy would be stopped from electric light, but what is the state of already radiated energy and why the room becomes dark again immediately.

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    $\begingroup$ Because the mean time between light being emitted and absorbed (rather than reflected) is tiny. $\endgroup$ – tfb Jan 17 '18 at 9:34
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The last photons from the lightbulb leave when the current stops.
(this is actually not completely true, you can still see the bulb fluorescing in the dark because the high temperature of the filament is still causing thermionc emissions).

To travel between the lightbuld and your eyes, these photons take $\frac{\Delta x}{c}$ seconds, where $c = 3 \cdot 10^8$ is the speed of light and $\Delta x$ is the distance between the bulb and your eyes - let's call it $100$ metres to be generous?
This gives $\sim 10^{-6}$ seconds, i.e. $ < 1 \mu s$.

The photons in transit (the radiated energy) is absorbed by the walls, or reflected for a couple of times before eventually being absorbed (and becoming heat).

The response of the human eye is around $1-10 \, \textrm{ms}$. If it were comparable to $\mu s$, you could probably see the parts of the room dark and parts bright, depending on where the light has already been absorbed.

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