Question

More than 200 years later, Albert A. Michelson sent a beam of light from a revolving mirror to a stationary mirror 15 km away. Show that the time interval between light leaving and returning to the revolving mirror was 0.0001 s.

Answer 1

Explanation:

distance in meter= 15000 m

velocity of light v=3×10^8 m/s

time interval t= d/v

=$\frac{15000}{ 3\times10^8}$

= 0.00005 sec

This is the time for the path in one direction.  If you are looking for the time from the revolving mirror to the stationary mirror back to the revolving mirror, then it is twice the above value, which would be

2× 0.00005 =  0.0001 s

Answer 2

Answer

given,

speed of light = 3 x 10⁸ m/s

distance of the mirror = 15 km

                                    = 15000 m

time interval between light leaving and returning = ?

we know,

distance = speed  x time

distance traveled by the light = 2 x 15000 = 30000 m

$t = \dfrac{distance}{speed}$

$t = \dfrac{30000}{3 \times 10^8}$

      t = 0.0001 s

hence, it is shown that the time interval is t = 0.0001 s