Home Page Textbooks Physics: Principles and Problems Page 463: Section Review

Physics: Principles and Problems

9th Edition

Elliott, Haase, Harper, Herzog, Margaret Zorn, Nelson, Schuler, Zitzewitz

ISBN: 9780078458132

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Page 463: Section Review

Exercise 6

Solution 1

Solution 2

Step 1

1 of 5

We need to determine the angle between the reflected ray and the mirror if the light ray strikes a flat, smooth surface at an angle of $80^o$ to the normal.

Step 2

2 of 5

This problem cann be solved using The Law of Reflection:
$$theta_i=theta_r$$
Where $theta_i$ is an angle of incidence and $theta_r$ is angle of reflection, both relative to the **normal of the surface**.
This means that the angle of the reflected ray is $80^o$ relative to the normal of the surface.

Step 3

3 of 5

Since the normal of the surface and surface are at an angle of $90^o$, we can write down equation, we know that the sum of angle between surface and the reflected ray, $alpha$ and the angle between the reflected ray and normal $theta_r$ needs to be equal to $90^o$. Therefore:
$$alpha+theta_r=90^o$$

Step 4

4 of 5

Finally we can calculate $alpha$:
$$alpha=90^o-theta_r$$
$$alpha=90^o-80^o$$
$$boxed{alpha=10^o}$$

Result

5 of 5

$$alpha=10^o$$

Step 1

1 of 3

In our problem, the angle between the incidence ray and the normal of the mirror is $80 textdegree$, which can be seen in the diagram below. To solve this problem, we are going to use The Law of Reflection, which says [ theta_{i}= theta_{r}] where $ theta_{i}$ is an angle of incidence ray relative to the normal of the surface, and $ theta_{r}$ is an angle of reflected ray relative to the normal of the surface. So, our angle of reflected ray is $80 textdegree$. From diagram, we can see that the angle between the reflected ray and the mirror $ alpha $ can be calculated as
begin{align*}
alpha + theta_{r} &= 90 textdegree \
alpha &= 90 textdegree -theta_{r} \
alpha &= 90 textdegree – 80 textdegree \
alpha &= 10 textdegree
end{align*}
The angle betweem the reflected ray and the mirror is [ framebox[1.1width]{$ therefore alpha= 10 textdegree $}]

Step 2

2 of 3

Result

3 of 3

$$
alpha= 10 text{textdegree}
$$

Exercise 7

Solution 1

Solution 2

Step 1

1 of 5

A lightbeam consits of a number of individual light rays that travel parallel to each other and hit the surface at the same angle.

Step 2

2 of 5

If the light beam is incident to a smooth surface it is hitting, its light rays will reflect and remain parallel to each other.

Step 3

3 of 5

If the surface is not smooth, but has **microscopic roughness** the light rays of the light beam will disperse in different directions.

Step 4

4 of 5

This means that the roughness of the surface acts for each beam as a smooth surface but at **many different angles**

Step 5

5 of 5

This finally means that light rays of a light beam are incident before hitting the rough surface and diffused upon reflection due to the different agles of the the smaller part of the surface.

Step 1

1 of 3

A light beam can be thought of as a bundle of individual light rays which are traveling parallel to each other.

Each individual light ray of the bundle follows the law of reflection.

If the bundle of light rays is incident upon a smooth surface, then the light rays reflect and remain concentrated in a bundle upon leaving the surface.

On the other hand, if the surface is microscopically rough, the light rays will reflect and diffuse in many different directions.

Step 2

2 of 3

The roughness of the material means that each individual ray meets a surface which has a different orientation.

The normal line at the point of incidence is different for different rays. Subsequently, when the individual rays reflect off the rough surface according to the law of reflection, they scatter in different directions.

The result is that the rays of light are incident upon the surface in a concentrated bundle and are diffused upon reflection.

Result

3 of 3

$bf{}$$text{color{#4257b2}SEE EXPLANATION}$

Exercise 8

Step 1

1 of 3

$$
text{color{#c34632}Perfectly smooth surfaces create specular reflection.

$ $

Rough surfaces create diffuse reflection.}
$$

Step 2

2 of 3

Window glass, smooth water and polished metal will create specular reflection.

Note that the objects which create specular reflection can act like mirrors

The following metals will create diffuse reflection :

paper, rough metal, ground glass, plastic milk jug

Result

3 of 3

$bf{}$$text{color{#4257b2}SEE ANSWER}$

Exercise 9

Solution 1

Solution 2

Step 1

1 of 2

$$
text{color{#c34632}For a real object, the virtual mirror will form a virtual image behind the mirror.

$ $

The size of the image will be same as that of the object

$ $

Distance of image from mirror is same as the object’s distance from mirror}
$$

Step 2

2 of 2

So the Image is Virtual

It is formed 3 metres behind the mirror

Height of the Image is 50 cm

Step 1

1 of 3

To see what is the image of a dog on the plane mirror, we have to choose two points on the dog and reflect them. The intersection of the optical axis and the reflected ray is the point of the image. We can see this in the diagram below. Also, we see that for the plane mirror, the image has the same height as the original, and the same distance from the mirror. But, there is also a difference between them. The original, or in our case the dog, is real, and the image in the mirror is virtual. So, our image is $50 hspace{0.5mm} mathrm{cm}$ high, at the distance of $3 hspace{0.5mm} mathrm{m}$ from the mirror, and it is virtual.

Step 2

2 of 3

Result

3 of 3

The image of the dog is $50 hspace{0.5mm} mathrm{cm}$ high, at the distance of $3 hspace{0.5mm} mathrm{m}$ from the mirror, and it is virtual.

Exercise 10

Step 1

1 of 3

We will mark the first car with the letter A, the second car with the letter B, and the Sun with the letter S. We know that the angle $alpha$ is $45 text{textdegree}$. From the diagram, we can see that the angles $alpha$ and $beta$ are the corresponding angles, which means that $beta= 45 text{textdegree}$ Also from the diagram, we can calculate the reflected angle

$$
begin{align*}
beta + theta_{r}&=90 text{textdegree}\
theta_{r}&=90 text{textdegree} – beta\
theta_{r}&=90 text{textdegree} – 45 text{textdegree}\
theta_{r}&=45 text{textdegree}\
end{align*}
$$

From The Law of Reflection can calculate the incident angle $theta_{i}= theta_{r} Rightarrow theta_{i}= 45 text{textdegree}$ So, we can calculate, but also we can see from the diagram, that the Sun is right above the first car. To find the position of the Sun relative to the first car, we can observe the angle between the incident angle and horizontal axis $theta_{i}+ theta_{r} = 90 text{textdegree}$

Step 2

2 of 3

Result

3 of 3

The Sun is above the first car. The angle of incidence of the Sun and the horizontal axis is $90 text{textdegree}$

Exercise 11

Step 1

1 of 2

You can view an object only if the light reflected from that object enters your eyes.

In case of diffuse reflection, the incident light is reflected in all directions.

This enables you to view the object from any angle.

Result

2 of 2

$bf{}$$text{color{#4257b2}SEE ANSWER}$

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