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.

Respuesta :

Answer:

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

Explanation:

From the question we are told that

   The distance of the detector from the source is [tex]d = 7.65 \ m[/tex]

    The index of refraction of ice is  [tex]n_i = 1.309[/tex]

   

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

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

      [tex]t_a = \frac{d}{c}[/tex]

substituting value

       [tex]t_a = \frac{7.65}{3.0 *10^{8}}[/tex]

       [tex]t_a = 2.55 *10^{8} \ s[/tex]

The time taken to travel through ice is

      [tex]t_i = \frac{d}{\frac{c}{n_i} }[/tex]

substituting values

      [tex]t_i = \frac{7.65}{\frac{3.0*10^{8}}{1.309} }[/tex]

       [tex]t_i = 3.34 *10^{-8}[/tex]

The in pulses time arrival is mathematically evaluated as

      [tex]\Delta t = t_2 - t_1[/tex]

 substituting values

     [tex]\Delta t = (3.34 - 2.55)*10^{-8}[/tex]

     [tex]\Delta t = 0.79*10^{-8} \ s[/tex]