Home Page Textbooks Physics: Principles and Problems Page 460: Practice Problems

Physics: Principles and Problems

9th Edition

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

ISBN: 9780078458132

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Page 460: Practice Problems

Exercise 1

Solution 1

Solution 2

Step 1

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Diffuse reflection is the reflection of light from a surface such that an incident ray is reflected at many angles rather than at just one angle as in the case of specular reflection.

Step 2

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Rough surfaces create diffuse reflection, and smooth surfaces create specular reflection.

When water is spilled on ground glass, it fills the rough areas and makes the surface smooth, which can create specular reflection

Result

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

Step 1

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**Types of reflection**

We can define two types of reflection of light when it hits the surface:

– Specular reflection: Reflection that obeys the law of reflection, where the angle of incidence is the same as the angle of the reflected ray.
– Diffuse reflection: Reflection that does not obey the law of reflection, meaning that the incident light wave scatters along many directions after it hits the surface.

Step 2

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Diffuse reflection is the consequence of the reflection on a rough (non-smooth) surface.

The surface may appear to be smooth when we look at it, but the roughness can be seen microscopically. This is the case with the ground glass. So, the light hitting a ground glass will have a diffuse reflection.

Step 3

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**Why does the reflection turn to specular?**

After the water is spilled on the ground glass, the microscopic rough surface is leveled by the water filling in the imperfect shapes. Thus, we have a real smooth surface, which will by definition have a specular reflection!

Exercise 2

Step 1

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When the light falls on a mirror, highly reflective surface, it reflects by the Law of Reflection. The Law of Reflection 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. In our case, the angle of incidence is $ 42 textdegree$
$a)$ By Law of Reflection, we can calculate the angle of reflection
begin{align*}
theta_{i}&= theta_{r} \
theta_{r}&= 42 textdegree
end{align*}
So, the angle of reflection is [ framebox[1.1width]{$ therefore theta_{r}= 42 textdegree $}]
$b)$ On the diagram below, we can see how to calculate the angle between the incident ray and the mirror. Also, we know that the angle between the mirror and the normal is $90 textdegree$, so we can write
begin{align*}
theta + theta_{i} &= 90 textdegree \
theta &= 90 textdegree -theta_{i} \
theta &= 90 textdegree – 42 textdegree \
theta &= 48 textdegree
end{align*}
The angle between the mirror and the incidence ray is [ framebox[1.1width]{$ therefore theta= 48 textdegree $}]
$c)$ Also, on the diagram below, we can see how to calculate the angle between the incident ray and the reflected ray, so we write
begin{align*}
theta_{r} + theta_{i} &= theta_{tot} \
theta_{tot} &=2 theta_{i} \
theta_{tot} &=2 cdot 42 textdegree\
theta_{tot} &=84 textdegree
end{align*}
The angle between the reflected ray and the incidence ray is [ framebox[1.1width]{$ therefore theta_{tot}= 84 textdegree $}]

Step 2

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Result

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$a)$ $theta_{r}= 42 text{textdegree}$

$b)$ $theta= 48 text{textdegree}$

$c)$ $theta_{tot}= 84 text{textdegree}$

Exercise 3

Step 1

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By Law of Reflection

The angle that a reflected ray makes as measured from the normal to a reflective surface equals the angle that the incident ray makes as measured from the same normal.

$$
fbox$color{#4257b2}theta_i= theta_r$ = 35text{textdegree}
$$

Result

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$$
text{color{#4257b2}$ 35text{textdegree}$}
$$

Exercise 4

Step 1

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If the laser is moved so that the angle of incidence increases by $13text{textdegree}$, Then
$$
theta_{itext{, New}} = theta_{itext{, Initial}} +13text{textdegree} = 38text{textdegree}+13text{textdegree} = 51text{textdegree}
$$

So the new angle of Incidence is $51text{textdegree}$

Step 2

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By Law of Reflection

The angle that a reflected ray makes as measured from the normal to a reflective surface equals the angle that the incident ray makes as measured from the same normal.

$$
fbox$color{#4257b2}theta_i= theta_r$
$$

Step 3

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So the new angle of reflection is $51text{textdegree}$

Result

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$$
text{color{#4257b2}New angle of reflection is $51text{textdegree}$}
$$

Exercise 5

Step 1

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On the diagram delow, we can see a situation from our problem. The angles of incidence and of the reflection on the first mirror are [ theta_{i1}= theta_{r1}= 30 textdegree] From diagram, we can see that the angle between the first reflected ray and the first mirror $ alpha $ can be calculated as
begin{align*}
alpha + theta_{r1} &= 90 textdegree \
alpha &= 90 textdegree -theta_{r1} \
alpha &= 90 textdegree – 30 textdegree \
alpha &= 60 textdegree
end{align*}
Also, we can see that angles $ alpha$ and $ beta$ are angles of right angle triangle. We can calculate the angle between the second mirror and the second incidence ray, $beta$, as
begin{align*}
alpha + beta + 90 textdegree &= 180 textdegree \
alpha + beta &= 90 textdegree \
beta &= 90 textdegree – alpha \
beta &= 90 textdegree – 60 textdegree \
beta &= 30 textdegree
end{align*}
In similar way as for the angle $ alpha$, we can calculate the angle of second incidence ray $ theta_{i2}$
begin{align*}
beta + theta_{i2} &= 90 textdegree \
theta_{i2} &= 90 textdegree -beta \
theta_{i2} &= 90 textdegree – 30 textdegree \
theta_{i2} &= 60 textdegree
end{align*}
By the Law of Reflection, for the second mirror we have
[ theta_{i2}= theta_{r2}= 60 textdegree]
So, the angle of reflection on the second mirror is [ framebox[1.1width]{$ therefore theta_{r2}= 60 textdegree $}]

Step 2

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Result

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$$
theta_{r2}= 60 text{textdegree}
$$

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