Chapter 22, Problem 18P

A ray of light strikes a flat, 2.00-cm-thick block of glass (n = 1.50) at ail angle of 30.0° with respect to the normal (Fig. P22.18). (a) Find the angle of refraction at the lop surface. (b) Find the angle of incidence at the bottom surface and the refracted angle. (c) Find the lateral distance d by which the light beam is shifted. (d) Calculate the speed of light in the glass and (e) the time required for the light to pass through the glass block. (f) Is the travel time through the block affected by the angle of incidence? Explain.

To determine

The angle of refraction at the top surface.

Answer to Problem 18P

The angle of refraction at the top surface is $19.5°$.

Explanation of Solution

From Snell’s law, at the first surface,

${\theta }_2={\sin }^{-1}\left(\frac{n_{air}\sin {\theta }_1}{n_{glass}}\right)$

Here,

$n_{air}$ is the refractive index of air

$n_{glass}$ is the refractive index of glass

${\theta }_1$ is the angle of incidence

${\theta }_2$ is the angle of refraction

Substitute $1.00$ for $n_{air}$, $1.50$ for $n_{glass}$ and $30.0°$ for ${\theta }_1$.

$\begin{array}{c}{\theta }_2={\sin }^{-1}\left(\frac{\left(1.00\right)\sin 30.0°}{1.50}\right)\\ =19.5°\end{array}$

Conclusion:

Thus, the angle of refraction at the top surface is $19.5°$.

To determine

The angle of incidence at the bottom surface and the refracted angle.

Answer to Problem 18P

The angle of incidence at the bottom surface is $19.5°$ and the refracted angle is $30.0°$.

Explanation of Solution

The upper surface and the lower surface are parallel, the angle of incidence at the lower surface will be

${\theta }_2=19.5°$

The angle of refraction is,

${\theta }_3={\sin }^{-1}\left(\frac{n_{glass}\sin {\theta }_{glass}}{n_{air}}\right)$

Substitute $1.00$ for $n_{air}$, $1.50$ for $n_{glass}$ and $19.5°$ for ${\theta }_{glass}$.

$\begin{array}{c}{\theta }_3={\sin }^{-1}\left(\frac{\left(1.50\right)\sin 19.5°}{1.00}\right)\\ =30.0°\end{array}$

Conclusion:

Thus, the angle of incidence at the bottom surface is $19.5°$ and the refracted angle is $30.0°$.

To determine

The lateral distance by which the light beam is shifted.

Answer to Problem 18P

The lateral distance by which the light beam is shifted is $0.386\text{cm}$.

Explanation of Solution

The following diagram shows the sketch of the path of the ray.

From the figure,

$\begin{array}{c}h=\frac{2.00\text{cm}}{\cos {\theta }_2}\\ =\frac{2.00\text{cm}}{\cos 19.5°}\\ =2.12\text{cm}\end{array}$

Also,

$\begin{array}{c}\alpha ={\theta }_1-{\theta }_2\\ =30.0°-19.5°\\ =10.5°\end{array}$

Thus, the lateral distance by which the light beam is shifted is,

$\begin{array}{c}d=h\sin \alpha \\ =\left(2.12\text{cm}\right)\sin 10.5°\\ =0.386\text{cm}\end{array}$

Conclusion:

Thus, the lateral distance by which the light beam is shifted is $0.386\text{cm}$.

To determine

The speed of light in the glass.

Answer to Problem 18P

The speed of light in the glass is $2.00\times {10}^8\text{m/s}$.

Explanation of Solution

The equation for the speed of light in glass is,

$v=\frac{c}{n_{glass}}$

Substitute $3.00\times {10}^8\text{m/s}$ for $c$ and $1.50$ for $n_{glass}$.

$\begin{array}{c}v=\frac{3.00\times {10}^8\text{m/s}}{1.50}\\ =2.00\times {10}^8\text{m/s}\end{array}$

Conclusion:

Thus, the speed of light in the glass is $2.00\times {10}^8\text{m/s}$.

To determine

The time required for the light to travel through the glass.

Answer to Problem 18P

The time required for the light to travel through the glass is $1.06\times {10}^{-10}\text{s}$.

Explanation of Solution

The equation for time required for the light to travel through the glass is,

$t=\frac{d}{v}$

Substitute $2.12\text{cm}$ for $d$ and $2.00\times {10}^8\text{m/s}$ for $v$.

$\begin{array}{c}t=\frac{2.12\text{cm}}{2.00\times {10}^8\text{m/s}}\left(\frac{1\text{m}}{{10}^2\text{cm}}\right)\\ =1.06\times {10}^{-10}\text{s}\end{array}$

Conclusion:

Thus, the time required for the light to travel through the glass is $1.06\times {10}^{-10}\text{s}$.

To determine

Is the travel time through the block affected by the angle of incidence.

Answer to Problem 18P

Yes, the travel time through the block is affected by the angle of incidence.

Explanation of Solution

A change in the angle of incidence will cause a change in the angle of refraction. Thus, the distance the distance travelled by the light also changes. So, the travel time will also change.

Conclusion:

Yes, the travel time through the block is affected by the angle of incidence.