Chapter 37, Problem 15PQ

Light rays strike a plane mirror at an angle of 45.0° as shown in Figure P37.15. At what angle should a second mirror be placed so that the reflected rays are parallel to the first mirror?

To determine

The angle at which the second mirror be placed so that the reflected rays are parallel to the first mirror.

Answer to Problem 15PQ

The angle at which the second mirror be placed so that the reflected rays are parallel to the first mirror is $67.50°$.

Explanation of Solution

The following figure shows the reflection diagram.

Figure-(1)

Here, ${\theta }_{\text{i}}$ is the incident angle, ${\theta }_{\text{r}}$ is the refracted angle, $\theta$ is the angle of the second mirror with the first mirror, ${\varphi }_{\text{i}}$ is the angle of reflected light ray from the first mirror with the second mirror, and ${\varphi }_{\text{r}}$ is the angle of the reflected light ray from the second mirror with the second mirror.

Calculate the angle from the above diagram.

    $\begin{array}{c}{\theta }_{\text{i}}+45.0°=90.0°\\ {\theta }_{\text{i}}=90.0°-45.0°\\ =45.0°\end{array}$

According to law of reflection ${\theta }_{\text{i}}={\theta }_{\text{r}}$. Hence,

    ${\theta }_r=45.0°$

Calculate the other angles from the above diagram.

    ${\theta }_{\text{r}}+{\gamma }_{\text{r}}=90.0°$                                                                                      (I)

From the above figure, the total angle in the triangle $ABC$ is equal to $180°$.

    ${\theta }_{\text{r}}+2{\theta }_{\text{i}}+90.0°=180.0°$                                                                          (II)

For the second mirror, the total angle in a straight line is equal to $180°$.

    $2{\theta }_{\text{i}}+{\varphi }_{\text{i}}+{\varphi }_{\text{r}}=180.0°$                                                                           (III)

From the above figure, the total angle in the triangle $ACD$ is equal to $180°$.

    $\theta +{\varphi }_{\text{i}}+{\gamma }_{\text{r}}=180.0°$                                                                               (IV)

Conclusion:

Substitute $45.0°$ for ${\theta }_{\text{r}}$ in equation (I) to calculate ${\gamma }_{\text{r}}$.

    $\begin{array}{c}45.0°+{\gamma }_{\text{r}}=90.0°\\ {\gamma }_{\text{r}}=90.0°-45.0°\\ {\gamma }_{\text{r}}=45.0°\end{array}$

Substitute $45.0°$ for ${\theta }_{\text{r}}$ in equation (II) and solve for $2{\theta }_{\text{i}}$.

    $\begin{array}{c}45.0°+2{\theta }_{\text{i}}+90.0°=180.0°\\ 2{\theta }_{\text{i}}=180.0°-90.0°-45.0°\\ 2{\theta }_{\text{i}}=45.0°\end{array}$

Substitute $45.0°$ for $2{\theta }_{\text{i}}$ and ${\varphi }_{\text{i}}$ for ${\varphi }_{\text{r}}$ in equation (III) to calculate ${\varphi }_{\text{i}}$.

    $\begin{array}{c}45.0°+{\varphi }_{\text{i}}+{\varphi }_{\text{i}}=180.0°\\ 2{\varphi }_{\text{i}}=180.0°-45.0°\\ {\varphi }_{\text{i}}=\frac{180.0°-45.0°}{2}\\ {\varphi }_{\text{i}}=67.50°\end{array}$

Substitute $45.0°$ for ${\gamma }_{\text{r}}$, and $67.50°$ for ${\varphi }_{\text{i}}$ in equation (IV) to calculate $\theta$.

    $\begin{array}{c}45.0°+\theta +67.50°=180.0°\\ \theta =180.0°-67.50°-45.0°\\ \theta =67.50°\end{array}$

Therefore, the angle at which the second mirror be placed so that the reflected rays are parallel to the first mirror is $67.50°$.