# Shining light on a retreating object

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1. Feb 1, 2017

### MackBlanch

1. The problem statement, all variables and given/known data

I came up with this thought experiment last night, but I'm not confident in my solution--mostly because I forgo the time values I thought would be necessary to complete it.

• A reflective object moves past a lamp at time T0 with constant (non-relativistic) velocity V.

• Some time later, at time time T1 the lamp turns on.

• The lamp turns off at time T2 when struck by the light reflected back by the object.

• The lamp stops being struck by reflected light at time T3.

Are the following durations of equal length? If not, then order them by length of duration:

1) The amount of time the lamp is 'on' (i.e. How long the lamp emits light): tLampEmits
2) The amount of time the object is reflecting (i.e. How long the object 'sees' the lamp): tObjectReflects
3) The amount of time the lamp gets struck by reflected light. (i.e. How long the lamp 'sees' the object): tReflectionsStrike

2. Relevant equations

'c' is defined as the speed of light.

Light will travel a distance D in time D/c.

3. The attempt at a solution

The first light ray emitted by the lamp will strike the object and be reflected when the object is some distance, D0, away. This first ray will then travel a total distance to and from the object of:

2 * D0

The final ray emitted by the lamp will travel some distance, D1, to and from the object for a total distance of:

2 * D1

Because the object is moving away from the lamp,

D0 < D1.​

The lamp is on for the amount of time it takes light to go to and from the object,

tLampEmits = 2 * D0 / c​

The object reflects from the time light first strikes it at distance D0 until light stops striking it at distance D1. During this time, light will travel a distance, D0, from the object to the lamp plus a distance, D1, from the lamp to the object (with no overlap as a condition in the question). That is,

tObjectReflects = (D0 + D1) / c​

The lamp will be struck by light for as long as it takes its final emission to travel to the object at D1, and back.

tReflectionsStrike = 2 * D1 / c​

So, the durations will be different, and since D1 > D0, ordered as follows:

{ tReflectionsStrike, tObjectReflects, tLampEmits }​

In prose:

The lamp will see the object for longer than it reflects, while the object will reflect for longer than the lamp is 'on'.

It seems reasonable then, that for a stationary object these durations are equal, while for an approaching object, the order is reversed.​

2. Feb 1, 2017

### Staff: Mentor

Correct - at least as long as we can neglect relativistic effects for the object.