Answer
(a) $506ns$
(b) $510ns$
Work Step by Step
(a) We know that
$t=\frac{n_1d_1+n_2d_2}{c}$
We plug in the known values to obtain:
$t=\frac{(1.00)\sqrt{(50.0m)^2+(50.0m)^2})+1.40(\frac{50.0m}{cos30.3^{\circ}})}{3.00\times 10^8m/s}$
$t=506\times 10^{-9}s=506ns$
(b) The time taken for the straight line path is given as
$t^{\prime}=\frac{n_1d_1+n_2d_2}{c}$
$t^{\prime}=\frac{(n_1+n_2)d}{c}$
We plug in the known values to obtain:
$t^{\prime}=\frac{(1.00+1.40)\sqrt{(100.0m)^2+(50.0m+(50.0m)tan30.3)^2}}{3.00\times 10^8m/s}$
$t^{\prime}=510\times 10^{-9}s=510ns$
