Chapter 25, Problem 25P

A ray of light strikes the midpoint of one face of an equiangular (60°−60°−60°) glass prism (n = 1.5) at an angle of incidence of 30°. (a) Trace the path of the light ray through the glass and find the angles of incidence and refraction at each surface. (b) If a small fraction of light is also reflected at each surface, what are the angles of reflection at the surfaces?

To determine

The trace of path of the light ray through the glass and the angles of incidence and refraction at each surface.

Answer to Problem 25P

The angles of incidence and refraction at surface $1$ are $30°$ and $19.47°$ respectively and the angles of incidence and refraction at surface $2$ are $40.53°$ and $77.1°$ respectively and the traced path of the light ray through the glass is shown below.

Explanation of Solution

Introduction:

When a ray incidents on the glass then the rays get refracted through angle of refraction $r_1$ at surface $1$ and then refracted through angle of refraction $r_2$ at surface $2$.

Given info: The angle of incidence at surface $1$ is $30°$ and the refractive index of the glass is $1.5$.

The path of the light ray through the glass is shown below,

Figure (1)

Write the Snell’s law, for surface $1$,

$n_1\sin i_1=n_2\sin r_1$

Here,

$n_1$ is the refractive index of air.

$i_1$ is the incidence angle of surface $1$.

$n_2$ is the refractive index of glass.

$r_1$ is the refracted angle of surface $1$

Rearrange the above expression for $r_1$.

$r_1={\sin }^{-1}\left[\frac{n_1}{n_2}\sin i_1\right]$

Substitute $1$ for $n_1$, $1.5$ for $n_2$ and $30°$ for $i_1$ in the above equation to find the value of $r_1$.

$\begin{array}{c}{r}_1={\sin }^{-1}\left[\frac{1}{1.5}\sin 30°\right]\\ =19.47°\end{array}$

Thus the incidence angle at surface $1$ is $30°$ and the refracted angle at surface $1$ is $19.47°$.

From the figure, the corresponding angles are equal.

$\angle QRC=\angle ABC$

Substitute $60°$ for $\angle ABC$ in the above equation.

$\angle QRC=60°$

From the figure, the alternate angles are equal.

$\angle PRC=r_1$

Substitute $19.47°$ for $r_1$ in the above equation.

$\angle PRC=19.47°$

From the figure

$i_2=\angle QRC-\angle PRC$

Substitute $60°$ for $\angle QRC$ and $19.47°$ for $\angle PRC$ in the above equation.

$\begin{array}{c}{i}_2=60°-19.47°\\ =40.53°\end{array}$

Thus the angle of incidence at surface $2$ is $40.53°$.

Write the Snell’s law, for surface $2$,

$n_2\sin i_2=n_1\sin r_2$

Here,

$n_1$ is the refractive index of air.

$i_2$ is the incidence angle of surface $2$.

$n_2$ is the refractive index of glass.

$r_2$ is the refracted angle of surface $2$

Rearrange the above expression for $r_2$.

$r_2={\sin }^{-1}\left[\frac{n_2}{n_1}\sin i_2\right]$

Substitute $1$ for $n_1$, $1.5$ for $n_2$ and $40.53°$ for $i_2$ in the above equation to find the value of $r_2$.

$\begin{array}{c}{r}_2={\sin }^{-1}\left[\frac{1.5}{1}\sin 40.53°\right]\\ =77.1°\end{array}$

Thus the angle of refraction at surface $2$ is $77.1°$.

Conclusion:

Therefore, the angles of incidence and refraction at surface $1$ are $30°$ and $19.47°$ respectively and the angles of incidence and refraction at surface $2$ are $40.53°$ and $77.1°$ respectively.

To determine

The angles of reflection at each surface.

Answer to Problem 25P

The angle of reflection at surface $1$ is $30°$ and the angle of reflection at surface $2$ is $40.53°$.

Explanation of Solution

Given info: The angle of incidence at surface $1$ is $30°$ and the refractive index of the glass is $1.5$.

From the angle of reflection at surface $1$ is,

$i_1={r^{\prime }}_1$

Here,

$i_1$ is the angle of incidence at surface $1$.

${r^{\prime }}_1$ is the angle of reflection at surface $1$.

Rearrange the above expression for ${r^{\prime }}_1$.

${r^{\prime }}_1=i_1$

Substitute $30°$ for $i_1$ in the above equation to find the value of ${r^{\prime }}_1$.

${r^{\prime }}_1=30°$

Thus the angle of reflection at surface $1$ is $30°$.

From the angle of reflection at surface $2$ is,

$i_2={r^{\prime }}_2$

Here,

$i_2$ is the angle of incidence at surface $2$.

${r^{\prime }}_2$ is the angle of reflection at surface $2$.

Rearrange the above expression for ${r^{\prime }}_2$.

${r^{\prime }}_2=i_2$

Substitute $40.53°$ for $i_2$ in the above equation to find the value of ${r^{\prime }}_2$.

${r^{\prime }}_2=40.53°$

Thus the angle of reflection at surface $2$ is $40.53°$.

Conclusion:

Therefore, the angle of reflection at surface $1$ is $30°$ and the angle of reflection at surface $2$ is $40.53°$.