hey i have another question if someone is willining to help

 

i can't work out how to use the c and v in this diagram ??

 

Thanks guys

 

Sarah

well actually i just get Sin(theta)/Beta

Draw the verticalline from the top end of the scope, and call this line LM (M on the ground). Call the vertex of angle \theta Q. You now 2 right angle triangles : LMP and LMQ, with base angles \theta' and \theta.

 

Note the following :

 

1. PQ = vt (where t is the time taken for the scope to move between the 2 positions)

2. LQ = ct

 

tan (\theta') = LM/PM = LM/(PQ+QM)

 

Plug and chug !

"PQ = vt"

"LQ = ct"

i don't see how these are the same times (t's)??

At time t=0

(i) the ray of light just enters the top of the scope, at L, and

(ii) the bottom of the scope is at P and moving to the right at speed v

 

At time t=T

(i) the ray of light reaches the bottom of the scope at Q, and hence

(ii) the bottom of the scope must be at Q

 

So, in the time it takes for the ray to travel down the scope, the scope bottom has moved from P to Q.

ok, but the distance from P to Q must be very very very small then?

t is going to be a couple of nanoseconds, depending on the size of the scope, and v is about 30 km/s

ok yep i got it now, thanks DQW