Question

Two light pulses are emitted simultaneously from a source. The pulses take parallel paths to a detector 7.65 m away, but one moves through air and the other through a block of ice. Determine the difference in the pulses times of arrival at the detector. (Assume the index of refraction of ice is 1.309.) Answer in units of ns.

Answer 1

Answer:

The difference in the pulses times of arrival at the detector is  $\Delta t = 0.79*10^{-8} \ s$

Explanation:

From the question we are told that

   The distance of the detector from the source is $d = 7.65 \ m$

    The index of refraction of ice is  $n_i = 1.309$

   

Generally the speed of light is  a constant with a value  $c = 3. *10^{8} \ m/ s$

So the time taken for the first light source through air is  

      $t_a = \frac{d}{c}$

substituting value

       $t_a = \frac{7.65}{3.0 *10^{8}}$

       $t_a = 2.55 *10^{8} \ s$

The time taken to travel through ice is

      $t_i = \frac{d}{\frac{c}{n_i} }$

substituting values

      $t_i = \frac{7.65}{\frac{3.0*10^{8}}{1.309} }$

       $t_i = 3.34 *10^{-8}$

The in pulses time arrival is mathematically evaluated as

      $\Delta t = t_2 - t_1$

 substituting values

     $\Delta t = (3.34 - 2.55)*10^{-8}$

     $\Delta t = 0.79*10^{-8} \ s$