Chapter 17.1, Problem 3PP

a.

To determine

Angle of reflection.

Answer to Problem 3PP

  ${\theta }_r=42°$

Explanation of Solution

Given:

Angle of incidence, ${\theta }_i=42°$

Formula used:

From the law of reflection,

  ${\theta }_i={\theta }_r$

Where, ${\theta }_i$ is angle of incidence and ${\theta }_r$ is angle of reflection.

Calculation:

As,

  ${\theta }_i={\theta }_r$

So,

  ${\theta }_r={\theta }_i=42°$

Conclusion:

Therefore, the angle of incidence is ${\theta }_r=42°$ .

b.

To determine

Angle of incidence with mirror.

Answer to Problem 3PP

  ${\theta }_{i\text{mirror}}=48°$

Explanation of Solution

Given:

Angle of incidence, ${\theta }_i=42°$

Formula used:

From the law of reflection,

  ${\theta }_{i\text{mirror}}=90°-{\theta }_i$

Where, ${\theta }_i$ is angle of incidence with normal and ${\theta }_{i\text{mirror}}$ is angle of incidence with mirror.

Calculation:

On putting the value,

  $\begin{array}{c}{\theta }_{i\text{mirror}}=90°-42°\\ =48°\end{array}$

Conclusion:

Therefore, the angle of incidence with mirror is ${\theta }_{i\text{mirror}}=48°$ .

c.

To determine

Angle between incidence ray and reflected ray.

Answer to Problem 3PP

  ${\theta }_i+{\theta }_r=84°$

Explanation of Solution

Given:

Angle of incidence, ${\theta }_i=42°$

Formula used:

From the law of reflection,

  ${\theta }_i={\theta }_r$

Where, ${\theta }_i$ is angle of incidence and ${\theta }_r$ is angle of reflection.

Calculation:

The angle between incident and reflected ray is ${\theta }_i+{\theta }_r$

As,

  ${\theta }_i={\theta }_r$

So,

  $\begin{array}{c}{\theta }_i+{\theta }_r={\theta }_i+{\theta }_i\\ =2{\theta }_i\\ =2\times 42°\\ =84°\end{array}$

Conclusion:

Therefore, angle between incidence ray and reflected ray is ${\theta }_i+{\theta }_r=84°$ .