Question

When the light ray illustrated in the figure below passes through the glass block of index of refraction n = 1.50, it is shifted laterally by the distance d. (Let L = 2.20 cm and = 28.0°.)(a) Find the value of d. in cm(b) Find the time interval required for the light to pass through the glass block. in ps

Answer 1

Part (a)

According to Snell's law,

$n_a\sin \theta=n\sin r$

Substitute known values,

$\begin{gathered} (1)\sin 28.0^{\circ}=(1.50)\sin r \\ \sin r=\frac{0.469}{1.50} \\ r=\sin ^{-1}(0.313) \\ =18.2^{\circ} \end{gathered}$

The lateral shift can be given as,

$d=\frac{L}{\cos r}\sin (\theta-r)$

Substituting values,

$\begin{gathered} d=\frac{2.20\text{ cm}}{\cos18.2}\sin (28.0-18.2) \\ =\frac{2.20\text{ cm}}{0.95}(0.170) \\ =0.394\text{ cm} \end{gathered}$

Thus, the value of d is 0.394 cm.

Part (b)

The time taken to pass through the glass is,

$t=\frac{L}{c}$

Substituting values,

$\begin{gathered} t=\frac{(2.20\text{ cm)(}\frac{1\text{ m}}{100\text{ cm}})}{3\times10^8\text{ m/s}} \\ =(0.733\times10^{-10}s)(\frac{1\text{ ps}}{10^{-12}\text{ s}}) \\ =73.3\text{ ps} \end{gathered}$

Thus, the time taken by light to travel is 73.3 ps.