The phrase “normal to the surface” frequently occurs in Optics. To understand this phase, we are going to observe a simpler case first, the plane mirror (or any plane surface). The normal is a line that is perpendicular to the part of the surface, where it crosses the surface. So, the angle between the surface and the normal is $90 text{textdegree}$. This is also applied to the curved surface. But, for curved surfaces, the normal lines are not parallel to each other, while for a plane surface the normal lines are parallel to each other. We can see both cases in the diagram below.

To find where is the image produced, we are going to use the mirror equation $frac{1}{d_{o}}+ frac{1}{d_{i}}= frac{1}{f}$ where $d_{o}$ is the position of the object, $d_{i}$ is the position of the image, and $f$ is the focal length. We know that the plane mirror is just a special case of a curved mirror. The plane mirror has radius of $infty$. We also know that te relation between the focal length and the radius is $2f=R$. So, we can write

$$
begin{align*}
frac{1}{f} &= frac{1}{d_{o}}+ frac{1}{d_{i}}\
frac{2}{R} &= frac{1}{d_{o}}+ frac{1}{d_{i}}\
frac{2}{ infty} &= frac{1}{d_{o}}+ frac{1}{d_{i}}\
0 &= frac{1}{d_{o}}+ frac{1}{d_{i}}\
frac{1}{d_{o}}&= – frac{1}{d_{i}}\
d_{i} &= – d_{o}
end{align*}
$$

The position of the object is always real (greater than zero), so the position of the image is lesser than zero. So, the image is always virtual (behind the mirror), and at the same distance from the mirror as the object. We can see this in the diagram below.

We know that for a concave mirror, the focal length is greater than zero, $f>0$. If the image is virtual that means the position of the image is lesser than zero, $d_{i}<0$. In this calculation, we are going to use the mirror equation $frac{1}{d_{o}}+ frac{1}{d_{i}}= frac{1}{f}$ Now, we can write

$$
begin{align*}
d_{i}&<0\
frac{1}{d_{i}}&<0 / + frac{1}{d_{o}}\
frac{1}{d_{o}}+ frac{1}{d_{i}} &< frac{1}{d_{o}}\
frac{1}{f}&< frac{1}{d_{o}} \
d_{o}&<f
end{align*}
$$

From the previous calculation, we can see that the image is virtual in the case of a concave mirror, only when the position of the object is lesser than the focal length.

$$
text{Mirror Formula : }dfrac{1}{f}=dfrac{1}{d_o}+dfrac{1}{d_i}
$$

Where $f$ is the focal length, it is negative for convex mirror and positive for concave mirror

$d_o$ is distance of object from mirror, It is always positive for real objects

$d_i$ is the distance of image from mirror, it is negative for virtual images and positive for real images

The length between the center of curvature and the mirror is called the radius, $r$. The focal point is always halfway between the center of curvature and the mirror. The length between the mirror and the focal point is called focal length, $f$. From this, we know that the radius is two times greater than the focal length $r=2f$

$$
r=2f
$$

$$
text{Magnification} = -dfrac{d_i}{d_o}
$$

If the magnification is negative, it means that the image is inverted

Recall that :
$$
dfrac{1}{d_o}+dfrac{1}{d_i}=dfrac{1}{f}
$$

Therefore
$$
dfrac{1}{d_i}=dfrac{1}{f}-dfrac{1}{d_o}
$$

We know that $d_o$ is always positive for real objects

We also know that $f$ is negative for convex mirrors

Therefore RHS is negative, which makes $d_i$ negative

Which means that the image is formed behind the mirror

Remember that: Only virtual images can be formed behind the mirror. Real images cannot be formed behind the mirror, It is because Light rays cannot meet behind the mirror

This proves that a convex mirror can only produce virtual images

Concave mirrors have a certain breaking point when it comes to magnification.
If the object is positioned on the center of curvature of the mirror, magnification will be $M=1$.
In order to achieve a smaller image, the object needs to be positioned **further** from the image than the **center of curvature**.

The angle between the incident ray and the normal is known as the angle of incidence $(theta_i)$

The angle between the reflected ray and the normal is known as the angle of reflection $(theta_r)$

$$
text{color{#c34632} By Law of reflection : $theta_i = theta_r$}
$$

Given that : $theta_i = 38text{textdegree}$

We are asked for $theta_r$

By Law of reflection, $theta_r = theta_i = 38text{textdegree}$

$$
text{color{#4257b2}$38text{textdegree}$}
$$

We know that the incident ray is at an angle of $53 textdegree$ with the normal. To solve this problem, we are going to use the Law of Reflection [ theta_{i}= theta_{r}]
$a)$ If the angle of incidence is $ theta_{i}=53 textdegree$ , then the reflected angle is
begin{align*}
theta_{i}&= theta_{r}\
theta_{r}&=53 textdegree
end{align*}
The reflected angle is [ framebox[1.1width]{$ therefore theta_{r}=53 textdegree $}]
\
$b)$ From the diagram below, we can see that the angle between the ray of incidence and the reflected ray is a sum of angles $ theta_{i}$ and $ theta_{r}$. So, we can write this as
begin{align*}
theta &= theta_{i}+ theta_{r}\
theta&= 53 textdegree +53 textdegree \
theta&= 106 textdegree
end{align*}
The angle between the incident and the reflected ray is [ framebox[1.1width]{$ therefore theta=106 textdegree $}]

$a)$ $theta_{r}= 53 text{textdegree}$

$b)$ $theta= 106 text{textdegree}$

Light ray enters the mirror at an angle of $alpha=36^o$ with the mirror. We need to determine the angle between an incident ray and reflected ray.

First we need to calculate the angle of incidence using simple geometry.
The angle of incidence is an angle between **normal** of the plane and the ray, which means that the angle of incidence $theta_i$ and the angle between the ray and **mirror** $alpha=36^o$ sum to an angle of $90^o$:
$$alpha + theta_i=90^o$$

From the previous equation we can get $theta_i$:
$$theta_i=90-alpha$$
$$theta_i=90-36$$
$$theta_i=54^o$$

The angle of incidence and reflection are equal:
$$theta_i=theta_r$$
Which means that:
$$theta_r=54^o$$

Finally, the angle between them is equal to their sum since they are both measured from the normal, but on opposite sides:
$$theta=theta_i+theta_r$$
$$theta=54+54$$
$$boxed{theta=108^o}$$

$$theta=108^o$$

Therefore the image is formed 1.2 m behind the mirror.

Distance between image and the camera is $1.2+1.2 = 2.4$ m

So the camera lens must be focused at 2.4 m

$a)$ In Figure 17-19, and also in the diagram below, we can see how light falls on the first mirror. To find the angle of reflection from the second mirror, we use the Law of Reflection and the geometric properties of the triangle. First, from the Law of Reflection we know [ theta_{i1}= theta_{r1}] where the angle of incidence is $ theta_{i1}=30 textdegree$, then the reflected angle from the first mirror is [ theta_{r1}=30 textdegree] To find the angle $ theta_{i2}$, first we need to find angles $ alpha $ and $ beta$. We can find from the property of a triangle that the sum of the angles in a triangle is $180 textdegree$. Also, the angle $ alpha$, we can calculate as
begin{align*}
theta_{r1}+ alpha &= 90 textdegree\
alpha&= 90 textdegree – theta_{r1}\
alpha&= 90 textdegree – 30 textdegree\
alpha&= 60 textdegree
end{align*}
Now, we can calculate $ beta$
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 the same way as we calculated the angle $ alpha$, we can calculate the angle of incidence on the second mirror
begin{align*}
theta_{i2}+ beta &= 90 textdegree\
theta_{i2} &= 90 textdegree -beta \
theta_{i2} &= 90 textdegree – 30 textdegree \
theta_{i2} &= 60 textdegree
end{align*}
Again, we use the Law of Reflection to find the angle of reflection [ theta_{i2}= theta_{r2}]
So, the angle of reflection from the second mirror is [ framebox[1.1width]{$ therefore theta_{r2}=60 textdegree $}]

$b)$ This is only a special case from part $a)$, when the angle $theta_{i1}$ is the same as $theta_{r2}$. In the same manner as before, we can find these angles. First, from the Law of Reflection we know

$$
begin{align*}
theta_{i1}&= theta_{r1}\
theta_{i2}&= theta_{r2}\
end{align*}
$$

So, we can say, they have the same value $theta_{i1}= theta_{r1}= theta_{i2}= theta_{r2} = theta$. Now, we can calculate the angles $alpha$ and $beta$

$$
begin{align*}
theta+ alpha &= 90 text{textdegree}\
alpha&= 90 text{textdegree} – theta\
end{align*}
$$

Also, we know that

$$
begin{align*}
alpha + beta + 90 text{textdegree} &= 180 text{textdegree} \
alpha + beta &=90 text{textdegree}\
beta &= 90 text{textdegree} – alpha \
beta &= 90 text{textdegree} – ( 90 text{textdegree} – theta)\
beta &= 180 text{textdegree} + theta\
end{align*}
$$

In the same way as we calculated the angle $alpha$, we can calculate the angle $beta$. And then we can find that the value of $theta$ is $45 text{textdegree}$. And we can draw this on the diagram.

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

$b)$ All of the angles are equal to $45 text{textdegree}$

From the figure you can see that :

$Delta$ ANC and $Delta$ AB$_1$A$_1$ are similar

By properties of similar triangles
$$
dfrac{A_1B_1}{MN} = dfrac{AA_1}{AC}
$$

$$
text{color{#4257b2}Note that : $A_1B_1 = AB$, Because Height of Image = Height of Object

$ $

And that : $AA_1 = 2AC$, Because the image and object are at the same distance from mirror }
$$

$$
dfrac{AB}{MN} = dfrac{2AC}{AC}
$$

$$
dfrac{AB}{MN} = 2
$$

$$
dfrac{AB}{2} = MN
$$

We conclude that the mirror size should be at least half of the man’s height.

We can see the situation from our problem in the diagram below. The mirrors form an angle of $45 text{textdegree}$, so from the property of a triangle, we know

$$
begin{align*}
alpha + beta + 45 text{textdegree} &= 180 text{textdegree} \
alpha + beta &=135 text{textdegree}
end{align*}
$$

Also, from the Law of Reflection, we know that the angle of incidence is the same as the angle of reflection. So, we can write

$$
begin{align*}
theta_{i1}&= theta_{r1}\
theta_{i2}&= theta_{r2}\
end{align*}
$$

We know that the angle of incidence on the first mirror is $theta_{i1}=30 text{textdegree}$, so the angle of reflection from the first mirror is $theta_{r1}=30 text{textdegree}$. Futhermore, the sum of the angles $alpha$ and $theta_{r1}$ is $90 text{textdegree}$, so we can write

$$
begin{align*}
theta_{r1}+ alpha &= 90 text{textdegree}\
alpha&= 90 text{textdegree} – theta_{r1}\
alpha&= 90 text{textdegree} – 30 text{textdegree}\
alpha&= 60 text{textdegree}
end{align*}
$$

Now, we can calculate $ beta$
begin{align*}
alpha + beta &=135 textdegree\
beta &= 135 textdegree – alpha \
beta&= 135 textdegree – 60 textdegree \
beta&= 75 textdegree
end{align*}
In the same manner as before, we can calculate the angles $ theta_{i2}$ and $ theta_{r2}$
begin{align*}
theta_{i2}+ beta &= 90 textdegree\
theta_{i2} &= 90 textdegree -beta \
theta_{i2} &= 90 textdegree -75 textdegree \
theta_{i2} &= 15 textdegree
end{align*}
So, the angle of reflection from the second mirror is [ framebox[1.1width]{$ therefore theta_{r2}=15 textdegree $}]

$$
theta_{r2}=15 text{textdegree}
$$

In the diagram below, we can see that the angle between the incident ray and the normal, $60 textdegree$. When the mirror is rotated for $18 textdegree$, it has a new normal line and the angle between the initial normal line, and the final normal line is $18 textdegree$ . Also, in the diagram, we can see that [ theta_{i1}= theta_{i2}+ 18 textdegree] From this, we can calculate the angle of incidence when the mirror is rotated $ theta_{i2}$
begin{align*}
theta_{i1}&= theta_{i2}+ 18 textdegree \
theta_{i2} &= theta_{i1} – 18 textdegree \
theta_{i2} &= 60 textdegree – 18 textdegree \
theta_{i2} &= 42 textdegree
end{align*}
From the Law of Reflection, we know that the angle of incidence is the same as the angle of reflection relative to the normal [ theta_{i}= theta_{r}= 42 textdegree ]
Now, we can calculate the angle $ theta$ between the reflected ray and the mirror as
begin{align*}
theta + theta_{r}&= 90 textdegree\
theta &= 90 textdegree – theta_{r} \
theta &= 90 textdegree – 42 textdegree \
theta &= 48 textdegree
end{align*}
So, the angle between the reflected ray and the mirror is [ framebox[1.1width]{$ therefore theta=48 textdegree $}]

$$
theta=48 text{textdegree}
$$

$$
text{color{#c34632}Recall that : $$text{Radius of Curvature} = 2times text{Focal Length}$$}
$$

Therefore

$$
text{Radius of Curvature} = 2times 10text{ cm} = 20text{ cm}
$$

$$
text{color{#4257b2}$$text{Radius of Curvature} = 20text{ cm}$$}
$$

$$
color{#4257b2}text{Recall that : Magnification} = -dfrac{d_i}{d_o}
$$

Substitute : $d_i = -9$ and $d_o = 18$

Notice the negative sign for $d_i$. It is Important to note that the image is virtual, which means that the image is formed behind the mirror

$$
text{ Magnification} = -dfrac{-9}{18} = 0.5
$$

Magnification of 0.5 means that image is upright and half the size of object

$$
text{color{#4257b2}Magnification = 0.5}
$$

In this problem we need to calculate the height of an object if the image appears to be $h_i=0.6,,rm{m}$ and the magnification of the mirror is $M=frac{1}{3}$.

Magnification of a mirror is defined as a ratio of dimensions of object’s image and its real dimensions:
$$M=frac{h_i}{h_o}$$

We need to calculate the $h_o$ from the previous equation:
$$h_o=frac{h_i}{M}$$
Inserting values we get:
$$h_o=frac{0.6}{frac{1}{3}}$$
Finally, the height of an object is:
$$boxed{h_o=1.8,,rm{m}}$$

$$h_o=1.8,,rm{m}$$

We can find the characteristic of the image with the diagram or with the mirror equation $frac{1}{d_{o}}+ frac{1}{d_{i}}= frac{1}{f}$
First, we are going to find the characteristic with equations. From Figure 17-21 we know

$$
begin{align*}
d_{o}>f \
frac{1}{d_{o}}< frac{1}{f}\
frac{1}{d_{o}} 0 \
d_{i} >0
end{align*}
$$

From this, we know that that the image is real. Now, we will observe relation between the radius $r=2f$ and the object position $d_{o}$

$$
begin{align*}
r>d_{o} \
2f>d_{o}\
frac{1}{d_{o}}> frac{1}{2f}\
frac{1}{f}- frac{1}{d_{i}} frac{1}{d_{i}} \
frac{1}{2f} > frac{1}{d_{i}}\
d_{i}> 2f\
d_{i}>r
end{align*}
$$

From this, we can see that the position of the image is greater than radius of mirror. And, to find if the image is inverted or upright, we are going to use the magnification equation

$$
begin{align*}
M&= -frac{d_{i}}{d_{o}} = frac{h_{i}}{h_{o}}\
h_{i}&= – h_{o} frac{d_{i}}{d_{o}}
end{align*}
$$

We can see that $d_{o}$, $d_{i}$, and $h_{o}$ are greater than zero, then the height of the image is lesser than zero, so the image is inverted.

Also, $d_{i}> d_{o}$, so the magnification is greater than $1$, and the image is larger than the object.

We need to determine the height and position of an image if an object is $h_o=3.8,,rm{cm}$ high and at a distance of $d_o=31,,rm{cm}$ away from a mirror that has a focus length of $f=16,,rm{cm}$

In order to calculate the distance of the image we can use the next equation:
$$frac{1}{d_o}+frac{1}{d_i}=frac{1}{f}$$

From the previous equation we can express $d_i$:
$$frac{1}{d_i}=frac{1}{f}-frac{1}{d_o}$$
$$frac{q}{d_i}=frac{d_o-f}{fd_o}$$
And finally:
$$d_i=frac{fd_o}{d_o-f}$$

Inserting values we get:
$$d_i=frac{16cdot 31}{31-16}$$
$$boxed{d_i=33.1,,rm{cm}}$$

Now we can calculate the height of an image using the equation for magnification of a mirror:
$$M=frac{h_i}{h_o}=-frac{d_i}{d_o}$$
$$h_i=h_ofrac{d_i}{d_o}$$

Inserting values we get:
$$h_i=-3.8cdot frac{33.1}{31}$$
$$boxed{h_o=-4.1,,rm{cm}}$$

$$d_i=33.1,,rm{cm}$$
$$h_o=-4.1,,rm{cm}$$

In this problem, we have a convex mirror, where we need to find the position of the image of the car. if we know the position of the car. To solve this, we are going to use the mirror equation $frac{1}{f} = frac{1}{d_{o}}+ frac{1}{d_{i}}$ where $f$ is the focal length of mirror, $d_{o}$ is the position of our object, and $d_{i}$ is the position of image.

From the problem, we know the focal length $f$, the position of the car $d_{o}$, and we need to find the position of the image of car $d_{i}$:

$$
begin{align*}
&f= – 6.00 hspace{0.5mm} mathrm{cm}\
&d_{o}= 10.00 hspace{0.5mm} mathrm{cm}
end{align*}
$$

To find the position of the image from the mirror equation, first we are going to subtract $frac{1}{d_{o}}$ on the both sides.

$$
begin{align*}
frac{1}{f} &= frac{1}{d_{o}}+ frac{1}{d_{i}} / – frac{1}{d_{o}} \
frac{1}{d_{i}}&= frac{1}{f} – frac{1}{d_{o}}
end{align*}
$$

In this step, we need to find the least common multipe for $frac{1}{f}$ and $frac{1}{d_{o}}$ hfill textcolor{white}{.}
begin{align*}
frac{1}{d_{i}}&= frac{1}{f} – frac{1}{d_{o}} \
frac{1}{d_{i}}&= frac{d_{o}- f}{f cdot d_{o}}
end{align*}

We have found the final expression for the image position, and now, we can substitute the values and calculate the position

$$
begin{align*}
d_{i}&= frac{f cdot d_{o}}{d_{o}- f}\
d_{i}&= frac{- 6.00 hspace{0.5mm} mathrm{cm} cdot 10.00 hspace{0.5mm} mathrm{cm}}{10.00 hspace{0.5mm} mathrm{cm}+ 6.00 hspace{0.5mm} mathrm{cm}}\
d_{i}&= frac{- 60.00 hspace{0.5mm} mathrm{cm^{2}}}{16.00 hspace{0.5mm} mathrm{cm}}\
d_{i}&=-3.75 hspace{0.5mm} mathrm{cm}
end{align*}
$$

Finally, we found the position of image hfill textcolor{white}{.}
[ framebox[1.1width]{$ therefore d_{i}=-3.75 hspace{0.5mm} mathrm{cm} $}]

$$
d_{i}=-3.75 hspace{0.5mm} mathrm{cm}
$$

In this problem, we have a concave mirror, where we need to find the position of the image, and it’s height. To solve this, we are going to use the mirror equation $frac{1}{f} = frac{1}{d_{o}}+ frac{1}{d_{i}}$ where $f$ is the focal length of mirror, $d_{o}$ is the position of our object, and $d_{i}$ is the position of image. Also, we are going to use the magnification equation to find the image height $M= -frac{d_{i}}{d_{o}}= frac{h_{i}}{h_{o}}$

From the problem, we know the focal length $f$, the position of the object $d_{o}$, and the object height $h_{o}$, and we need to find the position of the image $d_{i}$ and the image height $h_{i}$:

$$
begin{align*}
&f= 15.0 hspace{0.5mm} mathrm{cm}\
&d_{o}= 30.00 hspace{0.5mm} mathrm{cm}\
&h_{o}=1.8 hspace{0.5mm} mathrm{m}
end{align*}
$$

To find the position of the image from the mirror equation, first we are going to subtract $frac{1}{d_{o}}$ on the both sides.

$$
begin{align*}
frac{1}{f} &= frac{1}{d_{o}}+ frac{1}{d_{i}} / – frac{1}{d_{o}} \
frac{1}{d_{i}}&= frac{1}{f} – frac{1}{d_{o}}
end{align*}
$$

In this step, we need to find the least common multipe for $frac{1}{f}$ and $frac{1}{d_{o}}$ hfill textcolor{white}{.}
begin{align*}
frac{1}{d_{i}}&= frac{1}{f} – frac{1}{d_{o}} \
frac{1}{d_{i}}&= frac{d_{o}- f}{f cdot d_{o}}
end{align*}

We have found the final expression for the image position, and now, we can subsititue the values and calculate the position

$$
begin{align*}
d_{i}&= frac{f cdot d_{o}}{d_{o}- f}\
d_{i}&= frac{15.00 hspace{0.5mm} mathrm{cm} cdot 30.00 hspace{0.5mm} mathrm{cm}}{30.00 hspace{0.5mm} mathrm{cm}- 15.00 hspace{0.5mm} mathrm{cm}}\
d_{i}&= frac{450.00 hspace{0.5mm} mathrm{cm^{2}}}{15.00 hspace{0.5mm} mathrm{cm}}\
d_{i}&=30.0 hspace{0.5mm} mathrm{cm}
end{align*}
$$

Finally, we found the position of image hfill textcolor{white}{.}
[ framebox[1.1width]{$ therefore d_{i}=30.0 hspace{0.5mm} mathrm{cm} $}]

To find the image height, we need to multiply everything in the magnification equation with $h_{o}$ hfill textcolor{white}{.}
begin{align*}
– frac{d_{i}}{d_{o}} &= frac{h_{i}}{h_{o}} / cdot h_{o}\
h_{i}&= -h_{o}frac{d_{i}}{d_{o}}
end{align*}

So, the image height is hfill textcolor{white}{.}
[ framebox[1.1width]{$ therefore h_{i}=-1.8 hspace{0.5mm} mathrm{m} $}]

$d_{i}=30.0 hspace{0.5mm} mathrm{cm}$

$$
h_{i}=-1.8 hspace{0.5mm} mathrm{m}
$$

In this problem, we have a concave mirror, where we need to find the magnification of the image. To solve this, we are going to use the mirror equation $frac{1}{f} = frac{1}{d_{o}}+ frac{1}{d_{i}}$ where $f$ is the focal length of mirror, $d_{o}$ is the position of our object, and $d_{i}$ is the position of image. We are going to use the magnification equation to find the image height $M= -frac{d_{i}}{d_{o}}= frac{h_{i}}{h_{o}}$ Also, we need to find the focal length of the mirror, and for that we use the equation $f=frac{r}{2}$

From the problem, we know the radius $r$, and the position of the object $d_{o}$, and we need to find the position of the image $d_{i}$ and the magnification $M$:

$$
begin{align*}
&r= 40 hspace{0.5mm} mathrm{mm}\
&d_{o}= 16 hspace{0.5mm} mathrm{mm}
end{align*}
$$

First, we are going to find the focal length by using the equation $f=r/2$. Furthermore, we are going to substitute the values in the equation

$$
begin{align*}
f&= frac{r}{2}\
f&= frac{40 hspace{0.5mm} mathrm{mm}}{2}\
f&= 20 hspace{0.5mm} mathrm{mm}
end{align*}
$$

Now, when we have found the focal length, we are going to use it in the mirror equation.

To find the position of the image from the mirror equation, first we are going to subtract $frac{1}{d_{o}}$ on the both sides.

$$
begin{align*}
frac{1}{f} &= frac{1}{d_{o}}+ frac{1}{d_{i}} / – frac{1}{d_{o}} \
frac{1}{d_{i}}&= frac{1}{f} – frac{1}{d_{o}}
end{align*}
$$

In this step, we need to find the least common multiple for $frac{1}{f}$ and $frac{1}{d_{o}}$ hfill textcolor{white}{.}
begin{align*}
frac{1}{d_{i}}&= frac{1}{f} – frac{1}{d_{o}} \
frac{1}{d_{i}}&= frac{d_{o}- f}{f cdot d_{o}}
end{align*}

We have found the final expression for the image position, and now, we can substituue the values and calculate the position

$$
begin{align*}
d_{i}&= frac{f cdot d_{o}}{d_{o}- f}\
d_{i}&= frac{20 hspace{0.5mm} mathrm{mm} cdot 16 hspace{0.5mm} mathrm{mm}}{16 hspace{0.5mm} mathrm{mm}- 20 hspace{0.5mm} mathrm{mm}}\
d_{i}&= frac{320 hspace{0.5mm} mathrm{mm^{2}}}{-4 hspace{0.5mm} mathrm{mm}}\
d_{i}&=-80 hspace{0.5mm} mathrm{mm}
end{align*}
$$

Now, we use this information in the magnification equation.

To find the magnification of the image, we are going to use the magnification equation $M=- frac{d_{i}}{d_{o}}$. Also, in this step we are going to substitute the values, and calculate the magnification

$$
begin{align*}
M&=- frac{d_{i}}{d_{o}}\
M&=- frac{-80 hspace{0.5mm} mathrm{mm}}{16 hspace{0.5mm} mathrm{mm}}\
M&= 5
end{align*}
$$

Finally, the magnification of the image is hfill textcolor{white}{.}
[ framebox[1.1width]{$ therefore M=5 $}]

$$
M=5
$$

In this problem, we have a concave mirror, where we need to find the image position and height. To solve this, we are going to use the mirror equation $frac{1}{f} = frac{1}{d_{o}}+ frac{1}{d_{i}}$ where $f$ is the focal length of mirror, $d_{o}$ is the position of our object, and $d_{i}$ is the position of image. We are going to use the magnification equation to find the image height $M= -frac{d_{i}}{d_{o}}= frac{h_{i}}{h_{o}}$ Also, we need to find the focal length of the mirror, and for that we use the equation $f=frac{r}{2}$

From the problem, we know the radius $r$, the position of the object $d_{o}$, and the height of the object $h_{o}$, and we need to find the position of the image $d_{i}$ and the height of the image:

$$
begin{align*}
&r= 34.0 hspace{0.5mm} mathrm{cm}\
&d_{o}= 22.4 hspace{0.5mm} mathrm{cm}\
&h_{o}=3.0 hspace{0.5mm} mathrm{cm}
end{align*}
$$

First, we are going to find the focal length by using the equation $f=r/2$. Futhermore, we are goind to substitute the values in the equation

$$
begin{align*}
f&= frac{r}{2}\
f&= frac{34.0 hspace{0.5mm} mathrm{cm}}{2}\
f&= 17.0 hspace{0.5mm} mathrm{cm}
end{align*}
$$

Now, when we have found the focal length, we are going to use it in the mirror equation.

To find the position of the image from the mirror equation, first we are going to subtract $frac{1}{d_{o}}$ on the both sides.

$$
begin{align*}
frac{1}{f} &= frac{1}{d_{o}}+ frac{1}{d_{i}} / – frac{1}{d_{o}} \
frac{1}{d_{i}}&= frac{1}{f} – frac{1}{d_{o}}
end{align*}
$$

In this step, we need to find the least common multipe for $frac{1}{f}$ and $frac{1}{d_{o}}$ hfill textcolor{white}{.}
begin{align*}
frac{1}{d_{i}}&= frac{1}{f} – frac{1}{d_{o}} \
frac{1}{d_{i}}&= frac{d_{o}- f}{f cdot d_{o}}
end{align*}

We have found the final expression for the image position, and now, we can substitute the values and calculate the position

$$
begin{align*}
d_{i}&= frac{f cdot d_{o}}{d_{o}- f}\
d_{i}&= frac{17.0 hspace{0.5mm} mathrm{cm} cdot 22.4 hspace{0.5mm} mathrm{cm}}{22.4 hspace{0.5mm} mathrm{mm}- 17.0 hspace{0.5mm} mathrm{cm}}\
d_{i}&= frac{380.8 hspace{0.5mm} mathrm{cm^{2}}}{5.4 hspace{0.5mm} mathrm{cm}}\
d_{i}&= 70.5 hspace{0.5mm} mathrm{cm}
end{align*}
$$

Now, we use this information in the magnification equation.

Finally, we found the position of image hfill textcolor{white}{.}
[ framebox[1.1width]{$ therefore d_{i}=70.5 hspace{0.5mm} mathrm{cm} $}]

To fing the image height, we need to multiply everything in the magnification equation with $h_{o}$

$$
begin{align*}
– frac{d_{i}}{d_{o}} &= frac{h_{i}}{h_{o}} / cdot h_{o}\
h_{i}&= -h_{o}frac{d_{i}}{d_{o}}
end{align*}
$$

In this step, we are going to substitute the values, and calculate the image height

$$
begin{align*}
h_{i}&= -h_{o}frac{d_{i}}{d_{o}}\
h_{i}&= – 3.0 hspace{0.5mm} mathrm{cm} frac{70.5 hspace{0.5mm} mathrm{cm}}{22.4 hspace{0.5mm} mathrm{cm}}\
h_{i}&= – 3.0 hspace{0.5mm} mathrm{cm} cdot 3.15\
h_{i}&= – 9.4 hspace{0.5mm} mathrm{cm}\
end{align*}
$$

So, the image height is hfill textcolor{white}{.}
[ framebox[1.1width]{$ therefore h_{i}=-9.4 hspace{0.5mm} mathrm{cm} $}]

$d_{i}=70.5 hspace{0.5mm} mathrm{cm}$

$$
h_{i}=-9.4 hspace{0.5mm} mathrm{cm}
$$

In this problem, we have a concave mirror, where we need to find the image position (part $a)$) and the diameter of the image (part $b)$). To solve this, we are going to use the mirror equation $frac{1}{f} = frac{1}{d_{o}}+ frac{1}{d_{i}}$ where $f$ is the focal length of mirror, $d_{o}$ is the position of our object, and $d_{i}$ is the position of image. Also,e are going to use the magnification equation to find the image height $M= -frac{d_{i}}{d_{o}}= frac{h_{i}}{h_{o}}$

From the problem, we know the focal length $f$, the position of the object $d_{o}$, and the diameter of the object $D_{o}$, and we need to find the position of the image $d_{i}$ and the diameter of the image $D_{i}$:

$$
begin{align*}
&f= 12.0 hspace{0.5mm} mathrm{cm}\
&d_{o}= 8.0 hspace{0.5mm} mathrm{cm}\
&D_{o}=3.0 hspace{0.5mm} mathrm{cm}
end{align*}
$$

$a)$ To find the position of the image from the mirror equation, first we are going to subtract $frac{1}{d_{o}}$ on the both sides.

$$
begin{align*}
frac{1}{f} &= frac{1}{d_{o}}+ frac{1}{d_{i}} / – frac{1}{d_{o}} \
frac{1}{d_{i}}&= frac{1}{f} – frac{1}{d_{o}}
end{align*}
$$

In this step, we need to find the least common multipe for $frac{1}{f}$ and $frac{1}{d_{o}}$ hfill textcolor{white}{.}
begin{align*}
frac{1}{d_{i}}&= frac{1}{f} – frac{1}{d_{o}} \
frac{1}{d_{i}}&= frac{d_{o}- f}{f cdot d_{o}}
end{align*}

We have found the final expression for the image position, and now, we can substitute the values and calculate the position

$$
begin{align*}
d_{i}&= frac{f cdot d_{o}}{d_{o}- f}\
d_{i}&= frac{12.0 hspace{0.5mm} mathrm{cm} cdot 8.0 hspace{0.5mm} mathrm{cm}}{8.0 hspace{0.5mm} mathrm{mm}- 12.0 hspace{0.5mm} mathrm{cm}}\
d_{i}&= frac{96.0 hspace{0.5mm} mathrm{cm^2}}{-4.0 hspace{0.5mm} mathrm{cm}}\
d_{i}&= -24.0 hspace{0.5mm} mathrm{cm}
end{align*}
$$

Finally, we found the position of image hfill textcolor{white}{.}
[ framebox[1.1width]{$ therefore d_{i}=-24.0 hspace{0.5mm} mathrm{cm} $}]

$b)$ To find the image diameter, we need to multiply everything in the magnification equation with $D_{o}$ hfill textcolor{white}{.}
begin{align*}
– frac{d_{i}}{d_{o}} &= frac{D_{i}}{D_{o}} / cdot D_{o}\
D_{i}&= -D_{o}frac{d_{i}}{d_{o}}
end{align*}

So, the image diameter is hfill textcolor{white}{.}
[ framebox[1.1width]{$ therefore D_{i}=9.0 hspace{0.5mm} mathrm{cm} $}]

$a)$ $d_{i}=-24.0 hspace{0.5mm} mathrm{cm}$

$b)$ $D_{i}= 9.0 hspace{0.5mm} mathrm{cm}$

This problem has two parts. The first one, where we have the sunlight, and where we need to determine the focal length of the concave mirror. Another part is where we need to calculate the position of the image and the height of the image. We are going to use two equations, the mirror equation $frac{1}{f} = frac{1}{d_{o}}+ frac{1}{d_{i}}$ where $f$ is the focal length of mirror, $d_{o}$ is the position of our object, and $d_{i}$ is the position of image. Also, we are going to use the magnification equation to find the image height $M= -frac{d_{i}}{d_{o}}= frac{h_{i}}{h_{o}}$

We can assume that the sunlight comes from $infty$, so the light falls on the mirror parallel to the optical axis. All these rays intersect in the focal point, and at that point, they form the image. So, we can conclude that the focal length is $f=3.0 hspace{0.5mm} mathrm{cm}$

Now, we know the focal length $f$, the position of the object $d_{o}$, and the height of the object $h_{o}$, and we need to find the position of the image $d_{i}$ and the height of the image $h_{i}$:

$$
begin{align*}
&f= 3.0 hspace{0.5mm} mathrm{cm}\
&d_{o}= 12.0 hspace{0.5mm} mathrm{cm}\
&h_{o}=24 hspace{0.5mm} mathrm{mm}
end{align*}
$$

$a)$ In this part, we are going to draw the diagram to find where is the position of the image, and the height of the image. Also, we are going to use information above.

In this step, we need to find the least common multipe for $frac{1}{f}$ and $frac{1}{d_{o}}$ hfill textcolor{white}{.}
begin{align*}
frac{1}{d_{i}}&= frac{1}{f} – frac{1}{d_{o}} \
frac{1}{d_{i}}&= frac{d_{o}- f}{f cdot d_{o}}
end{align*}

We have found the final expression for the image position, and now, we can substitute the values and calculate the position

$$
begin{align*}
d_{i}&= frac{f cdot d_{o}}{d_{o}- f}\
d_{i}&= frac{3.0 hspace{0.5mm} mathrm{cm} cdot 12.0 hspace{0.5mm} mathrm{cm}}{12.0 hspace{0.5mm} mathrm{mm}- 3.0 hspace{0.5mm} mathrm{cm}}\
d_{i}&= frac{36.0 hspace{0.5mm} mathrm{cm^{2}}}{9.0 hspace{0.5mm} mathrm{cm}}\
d_{i}&= 4.0 hspace{0.5mm} mathrm{cm}
end{align*}
$$

Finally, we found the position of image hfill textcolor{white}{.}
[ framebox[1.1width]{$ therefore d_{i}=4.0 hspace{0.5mm} mathrm{cm} $}]

$c)$ In this part, we are going to find the image height. To find the image height, we need to multiply everything in the magnification equation with $h_{o}$

$$
begin{align*}
– frac{d_{i}}{d_{o}} &= frac{h_{i}}{h_{o}} / cdot h_{o}\
h_{i}&= -h_{o}frac{d_{i}}{d_{o}}
end{align*}
$$

So, the image height is hfill textcolor{white}{.}
[ framebox[1.1width]{$ therefore h_{i}=-8 hspace{0.5mm} mathrm{mm} $}]

$b)$ $d_{i}= 4.0 hspace{0.5mm} mathrm{cm}$

$c)$ $h_{i}= – 8 hspace{0.5mm} mathrm{mm}$

In this problem, we have a convex mirror, where we need to find the image position and the height of the image. To solve this, we are going to use the mirror equation $frac{1}{f} = frac{1}{d_{o}}+ frac{1}{d_{i}}$ where $f$ is the focal length of mirror, $d_{o}$ is the position of our object, and $d_{i}$ is the position of image. Also, we are going to use the magnification equation to find the image height $M= -frac{d_{i}}{d_{o}}= frac{h_{i}}{h_{o}}$ Also, we need to find the focal length of the mirror, and for that we use the equations $f=frac{r}{2}$ $r=frac{d}{2}$

Now, we know the diameter $D$, the position of the object $d_{o}$, and the height of the object $h_{o}$, and we need to find the position of the image $d_{i}$ and the height of the image $h_{i}$:

$$
begin{align*}
&D= 40.0 hspace{0.5mm} mathrm{cm}\
&d_{o}= 1.5 hspace{0.5mm} mathrm{m}\
&h_{o}=12.0 hspace{0.5mm} mathrm{cm}
end{align*}
$$

First, we are going to find the focal length by using the equation $f=r/2$ and $r=d/2$. Furthermore, we are going to substitute the values in the equation

$$
begin{align*}
f&= frac{d}{4}\
f&= frac{40.0 hspace{0.5mm} mathrm{cm}}{4}\
f&= 10.0 hspace{0.5mm} mathrm{cm}
end{align*}
$$

But, the mirror is convex, so the focal length is lesser than zero $f=-10.0 hspace{0.5mm} mathrm{cm}$ Now, when we have found the focal length, we are going to use it in the mirror equation.

To find the position of the image from the mirror equation, first we are going to subtract $frac{1}{d_{o}}$ on the both sides.

$$
begin{align*}
frac{1}{f} &= frac{1}{d_{o}}+ frac{1}{d_{i}} / – frac{1}{d_{o}} \
frac{1}{d_{i}}&= frac{1}{f} – frac{1}{d_{o}}
end{align*}
$$

In this step, we need to find the least common multipe for $frac{1}{f}$ and $frac{1}{d_{o}}$ hfill textcolor{white}{.}
begin{align*}
frac{1}{d_{i}}&= frac{1}{f} – frac{1}{d_{o}} \
frac{1}{d_{i}}&= frac{d_{o}- f}{f cdot d_{o}}
end{align*}

We have found the final expression for the image position, and now, we can substitute the values and calculate the position

$$
begin{align*}
d_{i}&= frac{f cdot d_{o}}{d_{o}- f}\
d_{i}&= frac{-0.1 hspace{0.5mm} mathrm{m} cdot 1.5 hspace{0.5mm} mathrm{m}}{1.5 hspace{0.5mm} mathrm{m}+ 0.1 hspace{0.5mm} mathrm{m}}\
d_{i}&= frac{-0.15 hspace{0.5mm} mathrm{m^{2}}}{1.6 hspace{0.5mm} mathrm{m}}\
d_{i}&= -9.4 hspace{0.5mm} mathrm{cm}
end{align*}
$$

To find the image height, we need to multiply everything in the magnification equation with $h_{o}$ hfill textcolor{white}{.}
begin{align*}
– frac{d_{i}}{d_{o}} &= frac{h_{i}}{h_{o}} / cdot h_{o}\
h_{i}&= -h_{o}frac{d_{i}}{d_{o}}
end{align*}

So, the height of the robin is hfill textcolor{white}{.}
[ framebox[1.1width]{$ therefore h_{i}=0.75 hspace{0.5mm} mathrm{cm} $}]

$d_{i}=-9.4 hspace{0.5mm} mathrm{cm}$

$$
h_{i}=0.75 hspace{0.5mm} mathrm{cm}
$$

In this problem, we have a plane mirror, and we need to find the angle of reflection. To solve this problem, we are going to use the law of reflection $theta_{i}= theta_{r}$ where $theta_{i}$ is the angle of incidence, and $theta_{r}$ is the angle of reflection. Also, the light source is moving, so we have to find the angle of incidence for that case.

We know the initial angle of incidence $theta_{i1}$, and how much the angle of incidence is increased $Delta theta_{i}$, and we need to find the final angle of incidence $theta_{i2}$, and the angle of reflection $theta_{r}$:

$$
begin{align*}
& theta_{i1}= 28 text{textdegree}\
& Delta theta_{i}=34 text{textdegree}
end{align*}
$$

First, we need to find the final angle of incidence. To do that, we can find the sum of the initial angle of incidence and the $Delta theta_{i}$

$$
begin{align*}
theta_{i2}&= theta_{i1} + Delta theta_{i}\
theta_{i2}&= 28 text{textdegree} + 34 text{textdegree}\
theta_{i2}&= 62 text{textdegree}
end{align*}
$$

Now we know the final angle of incidence, and we can find the angle of reflection with the law of reflection

$$
begin{align*}
theta_{i2}&= theta_{r}\
theta_{r}&= 62 text{textdegree}
end{align*}
$$

Finally, the angle of reflection is hfill textcolor{white}{.}
[ framebox[1.1width]{$ therefore theta_{r}= 62 textdegree $}]

$$
theta_{r}= 62 text{textdegree}
$$

In this problem, we have a convex mirror, and we need to find the height and the position of the image from the diagram. We are going to use two rays from one point (top of the arrow), to find the top of the image. Also, we are going to use squares of $1 hspace{0.5mm} mathrm{cm}$ sides. Our diagram is in the scale $1:2$

From the diagram, we can see that the position of the image is $frac{2}{3}f$. The focal length is $f=-4 hspace{0.5mm} mathrm{cm}$, so the position of the image is $d_{i}=-2.67 hspace{0.5mm} mathrm{cm}$.

Also, we can see that the height of the image is one third of the height of the object, $frac{1}{3}h_{o}$. The height of the object is $h_{o}=3 hspace{0.5mm} mathrm{cm}$, so the height of the image is $h_{i}=1 hspace{0.5mm} mathrm{cm}$.

$d_{i}=-2.67 hspace{0.5mm} mathrm{cm}$

$$
h_{i}=1 hspace{0.5mm} mathrm{cm}
$$

In this problem, we have a concave mirror, and we need to find the position of the image. We are going to use two equations, the mirror equation $frac{1}{f} = frac{1}{d_{o}}+ frac{1}{d_{i}}$ where $f$ is the focal length of mirror, $d_{o}$ is the position of our object, and $d_{i}$ is the position of image. Also, we need to find the focal length of the mirror, and for that we use the equation $f=frac{r}{2}$

In this problem, we know the radius of the mirror $r$, and the position of the object $d_{o}$, and we need to find the position of the image $d_{i}$:

$$
begin{align*}
&r=24.0 hspace{0.5mm} mathrm{cm}\
&d_{o}=4.4 hspace{0.5mm} mathrm{cm}\
end{align*}
$$

First, we are going to find the focal length by using the equation $f=r/2$. Furthermore, we are going to substitute the values in the equation

$$
begin{align*}
f&= frac{r}{2}\
f&= frac{24.0 hspace{0.5mm} mathrm{cm}}{2}\
f&= 12.0 hspace{0.5mm} mathrm{cm}
end{align*}
$$

Now, when we have found the focal length, we are going to use it in the mirror equation.

To find the position of the image from the mirror equation, first we are going to subtract $frac{1}{d_{o}}$ on the both sides.

$$
begin{align*}
frac{1}{f} &= frac{1}{d_{o}}+ frac{1}{d_{i}} / – frac{1}{d_{o}} \
frac{1}{d_{i}}&= frac{1}{f} – frac{1}{d_{o}}
end{align*}
$$

In this step, we need to find the least common multipe for $frac{1}{f}$ and $frac{1}{d_{o}}$ hfill textcolor{white}{.}
begin{align*}
frac{1}{d_{i}}&= frac{1}{f} – frac{1}{d_{o}} \
frac{1}{d_{i}}&= frac{d_{o}- f}{f cdot d_{o}}
end{align*}

We have found the final expression for the image position, and now, we can substitue the values and calculate the position

$$
begin{align*}
d_{i}&= frac{f cdot d_{o}}{d_{o}- f}\
d_{i}&= frac{12.0 hspace{0.5mm} mathrm{cm} cdot 4.4 hspace{0.5mm} mathrm{cm}}{4.4 hspace{0.5mm} mathrm{cm}- 12.0 hspace{0.5mm} mathrm{cm}}\
d_{i}&= frac{52.8 hspace{0.5mm} mathrm{cm^{2}}}{-7.6 hspace{0.5mm} mathrm{m}}\
d_{i}&= -6.95 hspace{0.5mm} mathrm{cm}
end{align*}
$$

Finally, we found the position of image hfill textcolor{white}{.}
[ framebox[1.1width]{$ therefore d_{i}=-6.95 hspace{0.5mm} mathrm{cm} $}]

$$
d_{i}=-6.95 hspace{0.5mm} mathrm{cm}
$$

In this problem, we have a concave mirror, and we need to find the position of the image (part $a$), and the height of the image (part $b$). We are going to use two equations, the mirror equation $frac{1}{f} = frac{1}{d_{o}}+ frac{1}{d_{i}}$ where $f$ is the focal length of mirror, $d_{o}$ is the position of our object, and $d_{i}$ is the position of image. Also, we need to find the focal length of the mirror, and for that we use the equation $f=frac{r}{2}$ Furthermore, we are going to use the magnification equation to find the image height $M= -frac{d_{i}}{d_{o}}= frac{h_{i}}{h_{o}}$

In this problem, we know the radius of the mirror $r$, the position of the object $d_{o}$, and the height of the object $h_{o}$, and we need to find the position of the image $d_{i}$, and the height of the image $h_{i}$:

$$
begin{align*}
&r=26.0 hspace{0.5mm} mathrm{cm}\
&d_{o}=30.0 hspace{0.5mm} mathrm{cm}\
&h_{o}=2.4 hspace{0.5mm} mathrm{cm}
end{align*}
$$

First, we are going to find the focal length by using the equation $f=r/2$. Furthermore, we are going to substitute the values in the equation

$$
begin{align*}
f&= frac{r}{2}\
f&= frac{26.0 hspace{0.5mm} mathrm{cm}}{2}\
f&= 13.0 hspace{0.5mm} mathrm{cm}
end{align*}
$$

Now, when we have found the focal length, we are going to use it in the mirror equation.

$$
end{document}
$$

$a)$ In this part, we are going to find the position of the image. To find the position of the image from the mirror equation, first we are going to subtract $frac{1}{d_{o}}$ on the both sides.

$$
begin{align*}
frac{1}{f} &= frac{1}{d_{o}}+ frac{1}{d_{i}} / – frac{1}{d_{o}} \
frac{1}{d_{i}}&= frac{1}{f} – frac{1}{d_{o}}
end{align*}
$$

In this step, we need to find the least common multipe for $frac{1}{f}$ and $frac{1}{d_{o}}$ hfill textcolor{white}{.}
begin{align*}
frac{1}{d_{i}}&= frac{1}{f} – frac{1}{d_{o}} \
frac{1}{d_{i}}&= frac{d_{o}- f}{f cdot d_{o}}
end{align*}

We have found the final expression for the image position, and now, we can substitue the values and calculate the position

$$
begin{align*}
d_{i}&= frac{f cdot d_{o}}{d_{o}- f}\
d_{i}&= frac{13.0 hspace{0.5mm} mathrm{cm} cdot 30.0 hspace{0.5mm} mathrm{cm}}{30.0 hspace{0.5mm} mathrm{cm}- 13.0 hspace{0.5mm} mathrm{cm}}\
d_{i}&= frac{390.0 hspace{0.5mm} mathrm{cm^{2}}}{17.0 hspace{0.5mm} mathrm{m}}\
d_{i}&= 22.94 hspace{0.5mm} mathrm{m}
end{align*}
$$

So, we found the position of image hfill textcolor{white}{.}
[ framebox[1.1width]{$ therefore d_{i}=22.94 hspace{0.5mm} mathrm{cm} $}]

$b)$ In this part, we are going to find the height of the image. To find the image height, we need to multiply everything in the magnification equation with $h_{o}$

$$
begin{align*}
– frac{d_{i}}{d_{o}} &= frac{h_{i}}{h_{o}} / cdot h_{o}\
h_{i}&= -h_{o}frac{d_{i}}{d_{o}}
end{align*}
$$

In this step, we are going to substitute the values, and calculate the image height

$$
begin{align*}
h_{i}&= -h_{o}frac{d_{i}}{d_{o}}\
h_{i}&= – 2.4 hspace{0.5mm} mathrm{cm} frac{22.94 hspace{0.5mm} mathrm{cm}}{30 hspace{0.5mm} mathrm{cm}}\
h_{i}&= -2.4 hspace{0.5mm} mathrm{cm} cdot 0.765\
h_{i}&= -1.84 hspace{0.5mm} mathrm{cm}\
end{align*}
$$

So, the height of the image is hfill textcolor{white}{.}
[ framebox[1.1width]{$ therefore h_{i}=-1.84 hspace{0.5mm} mathrm{cm} $}]

$a)$ $d_{i}=22.94 hspace{0.5mm} mathrm{cm}$

$b)$ $h_{i}=-1.84 hspace{0.5mm} mathrm{cm}$

In this problem, we have a concave mirror, and we need to find the radius of the mirror. We are going to use three equations, the mirror equation $frac{1}{f} = frac{1}{d_{o}}+ frac{1}{d_{i}}$ where $f$ is the focal length of mirror, $d_{o}$ is the position of our object, and $d_{i}$ is the position of image. Also, we are going to use the magnification equation to find the image height $M= -frac{d_{i}}{d_{o}}= frac{h_{i}}{h_{o}}$ Furthermore, we are going to find the radius from the equation $r=2f$

In this problem, we know the position of the object $d_{o}$, and the magnification $M$, and we need to find the radius of the mirror:

$$
begin{align*}
&d_{o}=20 hspace{0.5mm} mathrm{cm}\
&M=+3.2
end{align*}
$$

First, we are going to find the position of the image. We use the magnification equation $M= – frac{d_{i}}{d_{o}}$. We are going to multiply both sides with $-d_{o}$

$$
begin{align*}
M&=- frac{d_{i}}{d_{o}} / cdot (-d_{o})\
d_{i}&=-M d_{o}\
end{align*}
$$

Now, we are going to substitute the values in previous equation, and calculate the position of the image

$$
begin{align*}
d_{i}&=-M d_{o}\
d_{i}&=-(+3.2) cdot 20hspace{0.5mm} mathrm{cm}\
d_{i}&=-64hspace{0.5mm} mathrm{cm}
end{align*}
$$

To find the radius of the mirror from the mirror equation, first we are going to find the least common multipe for $frac{1}{d_{i}}$ and $frac{1}{d_{o}}$ hfill textcolor{white}{.}
begin{align*}
frac{1}{f} &= frac{1}{d_{o}}+ frac{1}{d_{i}} \
frac{1}{f} &= frac{d_{i}+d_{o}}{d_{o}d_{i}}
end{align*}

We have found the final expression for the focal length, now, we can substitute the values and calculate the focal length

$$
begin{align*}
f&= frac{d_{o}d_{i}}{d_{i}+d_{o}}\
f&=frac{-64 hspace{0.5mm} mathrm{cm} cdot 20 hspace{0.5mm} mathrm{cm}}{-64 hspace{0.5mm} mathrm{cm}+20 hspace{0.5mm} mathrm{cm}}\
f&=frac{-1280 hspace{0.5mm} mathrm{cm^{2}}}{-44 hspace{0.5mm} mathrm{cm}}\
f&=29.1 hspace{0.5mm} mathrm{cm}
end{align*}
$$

Now, we can find the radius of the mirror by using the equation $r=2f$ hfill textcolor{white}{.}
begin{align*}
r&=2f\
r&=2 cdot 29.1 hspace{0.5mm} mathrm{cm}\
r&=58.2 hspace{0.5mm} mathrm{cm}
end{align*}

Finally, the radius is hfill textcolor{white}{.}
[ framebox[1.1width]{$ therefore r=58.2 hspace{0.5mm} mathrm{cm} $}]

$$
58.2 hspace{0.5mm} mathrm{cm}
$$

In this problem, we have a convex mirror, where we need to find the focal length of the mirror. To solve this, we are going to use the mirror equation $frac{1}{f} = frac{1}{d_{o}}+ frac{1}{d_{i}}$ where $f$ is the focal length of mirror, $d_{o}$ is the position of our object, and $d_{i}$ is the position of image. Also, we are going to use the magnification equation to find the image height $M= -frac{d_{i}}{d_{o}}= frac{h_{i}}{h_{o}}$

In this problem, we know the position of the image $d_{i}$, and the magnification $M$, and we need to find the focal length:

$$
begin{align*}
&d_{o}=-36 hspace{0.5mm} mathrm{cm}\
&M=+0.5
end{align*}
$$

First, we are going to find the position of the object. We use the magnification equation $M= – frac{d_{i}}{d_{o}}$. We are going to multiply both sides with $-frac{d_{o}}{M}$

$$
begin{align*}
M&=- frac{d_{i}}{d_{o}} / cdot (-frac{d_{o}}{M})\
d_{o}&=-frac{d_{i}}{M}\
end{align*}
$$

Now, we are going to substitute the values in previous equation, and calculate the position of the image

$$
begin{align*}
d_{o}&=-frac{d_{i}}{M}\
d_{o}&=-frac{-36 hspace{0.5mm} mathrm{cm}}{0.5}\
d_{o}&=72 hspace{0.5mm} mathrm{cm}
end{align*}
$$

To find the focal length, we are going to find the least common multipe for $frac{1}{d_{i}}$ and $frac{1}{d_{o}}$ hfill textcolor{white}{.}
begin{align*}
frac{1}{f} &= frac{1}{d_{o}}+ frac{1}{d_{i}} \
frac{1}{f}&= frac{d_{i}+d_{o}}{d_{o}d_{i}}
end{align*}

We have found the final expression for the focal length, and now, we can substitute the values and calculate it

$$
begin{align*}
f&= frac{d_{i} d_{o}}{d_{i}+d_{o}}\
f&= frac{(-36hspace{0.5mm} mathrm{cm}) cdot 72 hspace{0.5mm} mathrm{cm}}{- 36hspace{0.5mm} mathrm{cm}+72 hspace{0.5mm} mathrm{cm}}\
f&=frac{(- 36hspace{0.5mm} mathrm{cm}) cdot 72 hspace{0.5mm} mathrm{cm}}{36 hspace{0.5mm} mathrm{cm}}\
f&=- 72hspace{0.5mm} mathrm{cm}
end{align*}
$$

So, we found the focal length hfill textcolor{white}{.}
[ framebox[1.1width]{$ therefore f=-72 hspace{0.5mm} mathrm{cm} $}]

$$
f=-72 hspace{0.5mm} mathrm{cm}
$$

The surveillance mirror is a convex mirror because it helps to eliminate blind spots. So, in this problem, we have a convex mirror, where we need to find the position of the image (part $a$), and the height of the image (part $b$). To solve this, we are going to use the mirror equation $frac{1}{f} = frac{1}{d_{o}}+ frac{1}{d_{i}}$ where $f$ is the focal length of mirror, $d_{o}$ is the position of our object, and $d_{i}$ is the position of image. Also, we are going to use the magnification equation to find the image height $M= -frac{d_{i}}{d_{o}}= frac{h_{i}}{h_{o}}$ Furthermore, we need to find the focal length of the mirror, and for that we use the equation $f=frac{r}{2}$

In this problem, we know the radius $r$, the position of the object $d_{o}$, and the object height $h_{o}$, and we need to find the focal length:

$$
begin{align*}
&r=3.8 hspace{0.5mm} mathrm{m}\
&d_{o}=6.5 hspace{0.5mm} mathrm{m}\
&h_{o}=1.7 hspace{0.5mm} mathrm{m}
end{align*}
$$

Now, we can find the focal length by using the equation $f=r/2$. Furthermore, we are going to substitute the values in the equation

$$
begin{align*}
f&= frac{r}{2}\
f&= frac{3.8 hspace{0.5mm} mathrm{m}}{2}\
f&= 1.9 hspace{0.5mm} mathrm{m}
end{align*}
$$

But the mirror is convex, so the focal length is negative $f=-1.9 hspace{0.5mm} mathrm{m}$

$a)$ In this part, we need to find the position of the image. To find the position of the image from the mirror equation, first we are going to subtract $frac{1}{d_{o}}$ on the both sides.

$$
begin{align*}
frac{1}{f} &= frac{1}{d_{o}}+ frac{1}{d_{i}} / – frac{1}{d_{o}} \
frac{1}{d_{i}}&= frac{1}{f} – frac{1}{d_{o}}
end{align*}
$$

In this step, we need to find the least common multipe for $frac{1}{f}$ and $frac{1}{d_{o}}$ hfill textcolor{white}{.}
begin{align*}
frac{1}{d_{i}}&= frac{1}{f} – frac{1}{d_{o}} \
frac{1}{d_{i}}&= frac{d_{o}- f}{f cdot d_{o}}
end{align*}

We have found the final expression for the image position, and now, we can substitue the values and calculate the position

$$
begin{align*}
d_{i}&= frac{f cdot d_{o}}{d_{o}- f}\
d_{i}&= frac{-1.9 hspace{0.5mm} mathrm{m} cdot 6.5 hspace{0.5mm} mathrm{m}}{6.5 hspace{0.5mm} mathrm{m}+ 1.9 hspace{0.5mm} mathrm{m}}\
d_{i}&= frac{-12.35 hspace{0.5mm} mathrm{m^{2}}}{8.4 hspace{0.5mm} mathrm{m}}\
d_{i}&= -1.47 hspace{0.5mm} mathrm{m}
end{align*}
$$

So, the position of the image is hfill textcolor{white}{.}
[ framebox[1.1width]{$ therefore d_{i}=-1.47 hspace{0.5mm} mathrm{m} $}]

$b)$ In this part, we need to find the height of the image. To find the image height, we need to multiply everything in the magnification equation with $h_{o}$

$$
begin{align*}
– frac{d_{i}}{d_{o}} &= frac{h_{i}}{h_{o}} / cdot h_{o}\
h_{i}&= -h_{o}frac{d_{i}}{d_{o}}
end{align*}
$$

In this step, we are going to substitute the values, and calculate the image height

$$
begin{align*}
h_{i}&= -h_{o}frac{d_{i}}{d_{o}}\
h_{i}&= – 1.7 hspace{0.5mm} mathrm{m} frac{-1.47hspace{0.5mm} mathrm{m}}{6.5 hspace{0.5mm} mathrm{m}}\
h_{i}&= -1.7 hspace{0.5mm} mathrm{m} cdot (-0.23)\
h_{i}&= 0.38 hspace{0.5mm} mathrm{m}\
end{align*}
$$

So, the height of the image is hfill textcolor{white}{.}
[ framebox[1.1width]{$ therefore h_{i}=0.38 hspace{0.5mm} mathrm{m} $}]

$a)$ $d_{i}=-1.47 hspace{0.5mm} mathrm{m}$

$b)$ $h_{i}=0.38 hspace{0.5mm} mathrm{m}$

In this problem, we need to find the type of mirror (concave $f>0$ or convex $f0$). Also, we know the magnification and the position of the object. To solve the problem, we are going to use several equations, the mirror equation $frac{1}{f}= frac{1}{d_{o}}+ frac{1}{d_{i}}$, the magnification equation $M= -frac{d_{i}}{d_{o}}= frac{h_{i}}{h_{o}}$ and the relation between the focal length and the radius $r=2f$

In this problem, we know the magnification $M$, the position of the object $d_{o}$, and that the image if upright $h_{i}>0$, and we need to find the type of mirror, and its radius:

$$
begin{align*}
&M=7.5\
&d_{o}=14.0 hspace{0.5mm} mathrm{mm}\
&h_{i}>0
end{align*}
$$

$a)$ In this part, we are going to determine the type of mirror, and for that, we are going to use the mirror equation, and the magnification equation. First, we know that $d_{o}>0$, and $M>0$, and we are going to multiply both sides with $-d_{o}$

$$
begin{align*}
M&= – frac{d_{i}}{d_{o}} / cdot (-d_{o})\
d_{i}&=-M d_{o}
end{align*}
$$

Now, we know that the product of $M d_{o}$ is greater than zero, so the left side is lesser than zero hfill textcolor{white}{..}
begin{align*}
d_{i}&=-M d_{o}\
M d_{o}&>0\
d_{i}&<0
end{align*}
So, now we know that the image is virtual.

Also, from magnification equation, we can determine the relation of the absolute value of $d_{i}$ and $d_{o}$. We know that the magnification is greater than one, $M>1$ hfill textcolor{white}{..}
begin{align*}
frac{|d_{i}|}{|d_{o}|}&=M\
frac{|d_{i}|}{|d_{o}|}&>1\
|d_{i}|&>|d_{o}|
end{align*}

Now, we use the mirror equation, to determine is the focal length greater or less than zero. we need to find the least common multipe for $frac{1}{d_{i}}$ and $frac{1}{d_{o}}$ hfill textcolor{white}{..}
begin{align*}
frac{1}{f}&= frac{1}{d_{i}}+frac{1}{d_{o}}\
frac{1}{f}&= frac{d_{o}+d_{i}}{d_{i} d_{o}}
end{align*}

We know that $|d_{i}|>|d_{o}|$ and $d_{i}<0$, so their sum is less than zero, and their product is less than zero. So, the focal length is greater than zero. If the focal length is greater than zero, the mirror is concave.

$b)$ To find the radius, we need to find the position of the image, and the focal length. We are going to use equations from previous part, $d_{i}=-M d_{o}$ $f= frac{d_{i} d_{o}}{d_{o}+ d_{i}}$

Finally, we can find the radius hfill textcolor{white}{.}
begin{align*}
r&=2f\
r&= 2 cdot 16.2 hspace{0.5mm} mathrm{mm}\
r&= 32.4 hspace{0.5mm} mathrm{mm}
end{align*}
So, the radius of the curvature is [ framebox[1.1width]{$ therefore r=32.4 hspace{0.5mm} mathrm{mm} $}]

$a)$ Concave mirror

$b)$ $r=32.4 hspace{0.5mm} mathrm{mm}$

In this problem, we have a concave mirror. Also, we know the focal length and the position of the object in points 1 and 2. We need to observe images from the arrows in points 1 and 2. To solve this problem, we are going to draw rays from the tops of the arrows. Also, the diagram below is on a scale of $1:20$.

First, we observe the image of the arrow in position 1. The arrow is positioned at $2.5 hspace{0.5mm} mathrm{m}$ from the mirror. Also, it is behind the center of the mirror. We can see that the image is smaller than the object, inverted, real, and it is between the center and the focal point.

Now, we observe the image of the arrow in position 2. The arrow is positioned at $1.5 hspace{0.5mm} mathrm{m}$ from the mirror. In the diagram, we see that the arrow is halfway between the center and the focal point. For this case, the image is also real and inverter, but it is bigger than the object, and it is behind the center.

In this problem, we need to find the type of mirror (concave or convex). We can do that by observing the focal length of the mirror. We know where the ball is positioned, and the mirror forms the virtual image, $d_{i}<0$. Also, we know that if we replace the spherical mirror with a plane the image is closer for $12 hspace{0.5mm} mathrm{cm}$.

To solve this problem, we are going to use the mirror equation $frac{1}{f}= frac{1}{d_{o}}+ frac{1}{d_{i}}$ where $f$ is the focal length of mirror, $d_{o}$ is the position of our object, and $d_{i}$ is the position of image. Also, we are going to use the information that the image in the plane mirror is $12 hspace{0.5mm} mathrm{cm}$ closer than the image in the spherical mirror, and the position of the object and the image for a plane mirror is the same $d_{o_{p}}=d_{i_{p}}$

First, we are going to calculate the position of the image for a spherical mirror. In order to calculate that, we have to know the position of the image for a plane mirror. We know the position of the object, $d_{o}$, so we know the position of the image in the plane mirror

$$
begin{align*}
d_{o}&=22 hspace{0.5mm} mathrm{cm}\
d_{o_{p}}&=d_{o}\
d_{i_{p}}&=d_{o_{p}}\
d_{i_{p}}&=22 hspace{0.5mm} mathrm{cm}
end{align*}
$$

In this step, we can find the position of the image for a spherical mirror. We know that the image in a plane mirror is for $12 hspace{0.5mm} mathrm{cm}$ closer than the image in a spherical mirror, so we can write

$$
begin{align*}
d_{i}&=d_{i_{p}}+12 hspace{0.5mm} mathrm{cm}\
d_{i}&=22 hspace{0.5mm} mathrm{cm}+12 hspace{0.5mm} mathrm{cm}\
d_{i}&=34 hspace{0.5mm} mathrm{cm}
end{align*}
$$

In this step, we need to find the least common multipe for $frac{1}{d_{i}}$ and $frac{1}{d_{o}}$ hfill textcolor{white}{.}
begin{align*}
frac{1}{f}&= frac{1}{d_{o}} + frac{1}{d_{i}} \
frac{1}{f}&= frac{d_{i} + d_{o}}{d_{o} cdot d_{i}}
end{align*}

We have found the final expression for the focal length, and now, we can substitute the values and calculate it

$$
begin{align*}
f&= frac{d_{o} cdot d_{i}}{d_{i}+ d_{o}}\
f&= frac{22 hspace{0.5mm} mathrm{cm} cdot (-34 hspace{0.5mm} mathrm{cm})}{-34 hspace{0.5mm} mathrm{cm}+ 22 hspace{0.5mm} mathrm{cm}}\
f&=frac{-748 hspace{0.5mm} mathrm{cm^2}}{-12 hspace{0.5mm} mathrm{cm}}\
f&=62.3 hspace{0.5mm} mathrm{cm}
end{align*}
$$

In this problem, we have a convex mirror, where we need to find the focal length of the mirror. To solve this, we are going to use the mirror equation $frac{1}{f} = frac{1}{d_{o}}+ frac{1}{d_{i}}$ where $f$ is the focal length of mirror, $d_{o}$ is the position of our object, and $d_{i}$ is the position of image. Also, we are going to use the magnification equation to find the image height $M= -frac{d_{i}}{d_{o}}= frac{h_{i}}{h_{o}}$

In this problem, we know the height of the object $h_{o}$, the position of the object $d_{o}$, and the height of the image $h_{i}$, and we need to find the focal length:

$$
begin{align*}
&h_{o}=1.6 hspace{0.5mm} mathrm{m}\
&d_{o}=3.2 hspace{0.5mm} mathrm{m}\
&h_{i}=0.28 hspace{0.5mm} mathrm{m}
end{align*}
$$

First, we are going to find the position of the image. We use the magnification equation $M= – frac{d_{i}}{d_{o}}= frac{h_{i}}{h_{o}}$. We are going to multiply both sides with $-d_{o}$

$$
begin{align*}
– frac{d_{i}}{d_{o}}&= frac{h_{i}}{h_{o}} / cdot (-d_{o})\
d_{i}&=-d_{o} frac{h_{i}}{h_{o}}\
end{align*}
$$

Now, we are going to substitute the values in previous equation, and calculate the position of the image

$$
begin{align*}
d_{i}&=-d_{o} frac{h_{i}}{h_{o}}\
d_{i}&=-3.2 hspace{0.5mm} mathrm{m} frac{0.28 hspace{0.5mm} mathrm{m}}{1.6 hspace{0.5mm} mathrm{m}}\
d_{i}&=-3.2 hspace{0.5mm} mathrm{m} cdot 0.175\
d_{i}&=-0.56 hspace{0.5mm} mathrm{m}
end{align*}
$$

In this step, we need to find the least common multipe for $frac{1}{d_{i}}$ and $frac{1}{d_{o}}$ hfill textcolor{white}{.}
begin{align*}
frac{1}{f}&= frac{1}{d_{i}} + frac{1}{d_{o}} \
frac{1}{f}&= frac{d_{o}+d_{i}}{d_{i} d_{o}}
end{align*}

We have found the final expression for the focal length, and now, we can substitue the values and calculate the focal length

$$
begin{align*}
f&=frac{d_{i} d_{o}}{d_{o}+d_{i}}\
f&= frac{-0.56 hspace{0.5mm} mathrm{m} cdot 3.2 hspace{0.5mm} mathrm{m}}{3.2 hspace{0.5mm} mathrm{m}+(-0.56 hspace{0.5mm} mathrm{m})}\
f&=frac{-1.792 hspace{0.5mm} mathrm{m^2}}{2.64 hspace{0.5mm} mathrm{m}}\
f&=-0.68 hspace{0.5mm} mathrm{m}
end{align*}
$$
.

So, we found the focal length hfill textcolor{white}{.}
[ framebox[1.1width]{$ therefore f=-0.68 hspace{0.5mm} mathrm{m} $}]

$$
f=-0.68 hspace{0.5mm} mathrm{m}
$$

In this problem, we have a concave mirror, and we need to find the position of the image and the image height. We are going to use two rays from one point (top of the arrow), to find the top of the image. Also, we are going to use squares of $1 hspace{0.5mm} mathrm{cm}$ sides. The diagram is on a scale of $1:300$.

In this step, we are going to find the position of the image. From the diagram, we can see that the image is $4.7 hspace{0.5mm} mathrm{cm}$ from the mirror. Or in real size, the image is $14.1 hspace{0.5mm} mathrm{m}$.

Furthermore, the height of the image is less than one centimeter. It is about $0.8 hspace{0.5mm} mathrm{cm}$, or in real size $2.4 hspace{0.5mm} mathrm{m}$. The image is inverted, so the height is $-2.4 hspace{0.5mm} mathrm{m}$.

In this problem, we have a convex mirror, where we need to find the focal length of the mirror. First, we are going to draw a diagram. After that, we are going to use the mirror equation $frac{1}{f} = frac{1}{d_{o}}+ frac{1}{d_{i}}$ where $f$ is the focal length of mirror, $d_{o}$ is the position of our object, and $d_{i}$ is the position of image. Also, we are going to use the magnification equation to find the image height $M= -frac{d_{i}}{d_{o}}= frac{h_{i}}{h_{o}}$

In this step, we are going to draw a diagram. We know the position of the object $d_{o}$, and the image $d_{i}$. Also, we know the height of the object $h_{o}$, and the image $h_{i}$:

$$
begin{align*}
&d_{o}=12.0 hspace{0.5mm} mathrm{cm}\
&d_{i}=-6.0 hspace{0.5mm} mathrm{cm}\
&h_{o}=4.0 hspace{0.5mm} mathrm{cm}\
&h_{i}=2.0 hspace{0.5mm} mathrm{cm}
end{align*}
$$

To find the focal length, we draw two rays, one to the center of the mirror, and the another one to the center of the curvature. Also, we are going to use squares of $1 hspace{0.5mm} mathrm{cm}$ sides.

We know that $f=r/2$. Also, we can see from the diagram that the radius of the curvature is $24.0 hspace{0.5mm} mathrm{cm}$. The mirror is convex, and the focal length is less than zero. So, the focal length is $f=-12.0 hspace{0.5mm} mathrm{cm}$

Now, we are going to verify our result. First, we are going to use the magnification equation, to verify positions and heights $M= -frac{d_{i}}{d_{o}}= frac{h_{i}}{h_{o}}$

The magnification, when we use $d_{o}$ and $d_{i}$, is hfilltextcolor{white}{.}
begin{align*}
M&= – frac{d_{i}}{d_{o}}\
M&= -frac{-6.0 hspace{0.5mm} mathrm{cm}}{12.0 hspace{0.5mm} mathrm{cm}}\
M&=0.5
end{align*}

The magnification, when we use $h_{o}$ and $h_{i}$, is hfilltextcolor{white}{.}
begin{align*}
M&= frac{h_{i}}{h_{o}}\
M&= frac{2.0 hspace{0.5mm} mathrm{cm}}{4.0 hspace{0.5mm} mathrm{cm}}\
M&=0.5
end{align*}

In this step, we need to find the least common multipe for $frac{1}{d_{i}}$ and $frac{1}{d_{o}}$ hfill textcolor{white}{.}
begin{align*}
frac{1}{f}&= frac{1}{d_{i}} + frac{1}{d_{o}} \
frac{1}{f}&= frac{d_{o}+d_{i}}{d_{i} cdot d_{o}}
end{align*}

We have found the final expression for the focal length, and now, we can substitute the values and calculate the position

$$
begin{align*}
f&=frac{d_{i} d_{o}}{d_{o}+d_{i}}\
f&= frac{-6.0 hspace{0.5mm} mathrm{cm} cdot 12.0 hspace{0.5mm} mathrm{cm}}{12.0 hspace{0.5mm} mathrm{cm}- 6.0 hspace{0.5mm} mathrm{cm}}\
f&= frac{-72.0 hspace{0.5mm} mathrm{cm^{2}}}{6.0 hspace{0.5mm} mathrm{cm}}\
f&= -12 hspace{0.5mm} mathrm{cm}
end{align*}
$$

So, we found the focal length hfill textcolor{white}{.}
[ framebox[1.1width]{$ therefore f=-12.0 hspace{0.5mm} mathrm{cm} $}]

$$
f=-12.0 hspace{0.5mm} mathrm{cm}
$$

In this problem, we have a concave mirror and the object that is approaching the mirror. To describe how the size of the image is changing as it rolls along, we are going to use the mirror equation $frac{1}{f}= frac{1}{d_{i}}+ frac{1}{d_{o}}$ where $f$ is the focal length of mirror, $d_{o}$ is the position of our object, and $d_{i}$ is the position of the image, and the magnification equation $M= – frac{d_{i}}{d_{o}}$ Also, we are going to observe some cases (when the ball is behind the center, when the ball is between the center and the focal point, when the ball is between the mirror and the focal point, when the ball is in the center, and when the ball is in the focal point). Furthermore, we need to find the focal length of the mirror, and for that we use the equation $f=frac{r}{2}$

First, we are going to find the expression for the position of the image from the mirror equation. To find the position of the image from the mirror equation, first we are going to subtract $frac{1}{d_{o}}$ on the both sides.

$$
begin{align*}
frac{1}{f} &= frac{1}{d_{o}}+ frac{1}{d_{i}} / – frac{1}{d_{o}} \
frac{1}{d_{i}}&= frac{1}{f} – frac{1}{d_{o}}
end{align*}
$$

In this step, we need to find the least common multipe for $frac{1}{f}$ and $frac{1}{d_{o}}$ hfill textcolor{white}{.}
begin{align*}
frac{1}{d_{i}}&= frac{1}{f} – frac{1}{d_{o}} \
frac{1}{d_{i}}&= frac{d_{o}- f}{f cdot d_{o}}
end{align*}

The first case is when the ball is behind the center, $d_{o}>r$, or $d_{o}>2f$, so we can write this as $d_{o}=C cdot 2f$, where $C$ is the constant that is greater than one. Now, we can write the position of the image

$$
begin{align*}
d_{i}&= frac{f cdot d_{o}}{d_{o}- f}\
d_{i}&= frac{f cdot 2Cf}{2Cf- f}\
d_{i}&= frac{ 2Cf^{2}}{(2C-1) f}\
end{align*}
$$

In this step, we can divide everything with $f$, and replace $2Cf$ with $d_{o}$ hfill textcolor{white}{.}
begin{align*}
d_{i}&= frac{ 2Cf^{2}}{(2C-1) f}\
d_{i}&= frac{ 2Cf}{2C-1}\
d_{i}&= frac{ d_{o}}{2C-1}\
end{align*}

The part $2C-1$ is greater than $1$, so we can write $d_{i}<d_{o}$, but also, we can conclude that the image is real. When we use this in the magnification equation, we get that the magnification is less than $1$. So, we can say that the image is smaller. Also, we have a minus sign in the magnification equation, so the image is inverted.

Now, the case when the ball is between the center and the focal point hfill textcolor{white}{.} [2f>d_{o}>f] We can solve this as before, we write $d_{o}$ as [d_{o}=Cf]

In this step, we replace $Cf$ with $d_{o}$ hfill textcolor{white}{.}
begin{align*}
d_{i}&= frac{ Cf^{2}}{(C-1) f}\
d_{i}&= frac{ Cf}{C-1}\
d_{i}&= frac{ d_{o}}{C-1}\
end{align*}

The part $C-1$ is greater than $0$, but less than $1$, so we can write $d_{i}>d_{o}$, but also, we can conclude that the image is real. When we use this in the magnification equation, we get that the magnification is greater than $1$. So, we can say that the image is bigger. Also, we have a minus sign in the magnification equation, so the image is inverted.

And finally, the case when the ball is between the focal point and the mirror hfill textcolor{white}{.} [f>d_{o}] We can solve this like in previous cases, to write $d_{o}$ as [d_{o}= frac{f}{C}] where $C$ is a constant that is greater than $1$.

In this step, we replace $f/C$ with $d_{o}$ hfill textcolor{white}{.}
begin{align*}
d_{i}&= frac{ frac{f^{2}}{C}}{(frac{1}{C}-1) f}\
d_{i}&= frac{ frac{f}{C}}{frac{1}{C}-1}\
d_{i}&= frac{ d_{o}}{frac{1}{C}-1}\
end{align*}

The part $frac{1}{C}-1$ is less than $0$, so the image is virtual. Also, we can see $|d_{i}|>|d_{o}|$. When we use this in the magnification equation, we get that the magnification is greater than $1$. So, we can say that the image is bigger. Also, we have a minus sign in the magnification equation, but we have a minus sign from the image position, so the image is upright.

When the ball is in the center $d_{o}=2f$, we can write the image position as hfill textcolor{white}{.}
begin{align*}
d_{i}&= frac{f cdot d_{o}}{d_{o}- f}\
d_{i}&= frac{f cdot 2f}{2f- f}\
d_{i}&= frac{ 2f^{2}}{f}\
d_{i}&=2f\
d_{i}&=d_{o}
end{align*}

We can see the position of the image is the same as the position of the object. Also, the magnification is $1$, so the image has the same height like the object, but there is a minus sign, so the image is inverted.

When the ball is in the focal point $d_{o}=f$, we can write the image position as hfill textcolor{white}{.}
begin{align*}
d_{i}&= frac{f cdot d_{o}}{d_{o}- f}\
d_{i}&= frac{f cdot f}{f- f}\
d_{i}&= frac{ f^{2}}{ 0}
end{align*}

For this case, there is no image, because the position of the image is in $infty$.

In this problem, we have a concave mirror, and we need to find the focal length. We are going to use the mirror equation $frac{1}{f} = frac{1}{d_{o}}+ frac{1}{d_{i}}$ where $f$ is the focal length of mirror, $d_{o}$ is the position of our object, and $d_{i}$ is the position of image.

In this step, we need to find the least common multipe for $frac{1}{d_{i}}$ and $frac{1}{d_{o}}$ hfill textcolor{white}{.}
begin{align*}
frac{1}{f}&= frac{1}{d_{i}} + frac{1}{d_{o}} \
frac{1}{f}&= frac{d_{o}+d_{i}}{d_{i} cdot d_{o}}
end{align*}

We have found the final expression for the focal length, and now, we can substitute the values and calculate the position

$$
begin{align*}
f&= frac{d_{i} cdot d_{o}}{d_{o}+d_{i}}\
f&= frac{22 hspace{0.5mm} mathrm{cm} cdot 22 hspace{0.5mm} mathrm{cm}}{22 hspace{0.5mm} mathrm{cm}+22 hspace{0.5mm} mathrm{cm}}\
f&= frac{22^{2} hspace{0.5mm} mathrm{cm^{2}}}{2 cdot 22 hspace{0.5mm} mathrm{cm}}\
f&= 11 hspace{0.5mm} mathrm{m}
end{align*}
$$

So, we found the focal length hfill textcolor{white}{.}
[ framebox[1.1width]{$ therefore f=11 hspace{0.5mm} mathrm{cm} $}]

$$
f=11 hspace{0.5mm} mathrm{cm}
$$

In this problem, we have a concave mirror. To solve this problem, we are going to use two equations, the mirror equation $frac{1}{f} = frac{1}{d_{o}}+ frac{1}{d_{i}}$ where $f$ is the focal length of mirror, $d_{o}$ is the position of our object, and $d_{i}$ is the position of the image. Also, our final expression should be the relationship between the object position and the image position for a plane mirror $d_{o}=-d_{i}$

The radius of curvature increases to $infty$, so the focal length increases to $infty$, because hfill textcolor{white}{.}
begin{align*}
f&=frac{r}{2}\
f&=frac{infty}{2}\
f&=infty
end{align*}

In this step, we are going to subtract $ frac{1}{d_{o}}$ on the both sides hfill textcolor{white}{.}
begin{align*}
0 &= frac{1}{d_{o}}+ frac{1}{d_{i}} / – frac{1}{d_{o}} \
frac{1}{d_{i}}&= – frac{1}{d_{o}}
end{align*}

Finally, we find the reciprocal value of the previous expression, and we get the relationship between the object position and the image position for a plane mirror $d_{o}=-d_{i}$

Plane mirror that is $d_o=6,,rm{cm}$ away from the object is substituted with a concave mirror. This causes the image to be $delta_i=8,,rm{cm}$ farther behind the mirror. We need to determine the focal length of the mirror if an object is between mirror and the focal point.

When talking about plane mirrors, the object and image are at the equal distance from the mirror. This means that new distance of image is:
$$d_i=d_o+delta_i$$
$$d_i=-6-8$$
$$d_i=-14,,rm{cm}$$
Where we took negative values because the distance is behind mirror.

The next equation that we need is:
$$frac{1}{d_o}+frac{1}{d_i}=frac{1}{f}$$

From the previous equation we can express $f$:
$$frac{1}{f}=frac{d_i+d_o}{d_od_i}$$
And $f$ is:
$$f=frac{d_od_i}{d_o+d_i}$$

Inserting values we get:
$$f=frac{6cdot (-14)}{6-14}$$
$$boxed{f=10.5,,rm{cm}}$$

$$f=10.5,,rm{cm}$$

First, we are going to look at the characteristics of the first image. That image is inverted, $h_{i1}<0$. The image of the first mirror is the object of the second mirror, so the height of the object for the second mirror is less than zero, $h_{o2}0$, then the relation between $h_{o2}$ and $h_{i2}$ is less than zero $frac{h_{i2}}{h_{o2}}<0$

If we compare previous expression with the magnification equation, we get that the position of the object and the position of the image should be greater than zero $d_{o2}>0 hspace{2.5mm} d_{i2}>0$

If we use this in the mirror equation $f_{2}=frac{d_{o2}d_{i2}}{d_{o2}+d_{i2}}$ we get that the focal length is greater than zero, $f_{2}>0$. So the mirror has to be concave.

In this problem, we have a system of mirrors, one concave and one convex mirror. In the first part (part $a$), we want to find the explanation why the convex mirror is used, and in the second part (part $b$), we want to find is the final image inverted or upright. To solve this problem, we are going to use the mirror equation $frac{1}{f}= frac{1}{d_{o}}+ frac{1}{d_{i}}$ where $f$ is the focal length of mirror, $d_{o}$ is the position of our object, and $d_{i}$ is the position of the image. Also, we are going to use the magnification equation to find the image height $M= -frac{d_{i}}{d_{o}}= frac{h_{i}}{h_{o}}$

$a)$ The image from the first mirror is virtual $d_{i1}<0$. That image is the object for the second mirror, so $d_{o2}0$

To find the relation between the position of the image and the position of the object, we use the magnification equation and the information that the image is smaller than the object $M<1$. So, we can write

$$
begin{align*}
M&<1\
-frac{d_{i2}}{d_{o2}}&-d_{o2}
end{align*}
$$

We have already seen that $|d_{i2}|>|d_{o2}|$, so the focal length is less than zero, $f_{2}<0$. So, we saw that the mirror should be convex, if the final image is real.

$b)$ It was shown that the magnification is greater than zero, $M>0$. We are going to use that in this part $M=frac{h_{i2}}{h_{o2}}$ The image from the first mirror is inverted, so the object of the second mirror is inverted, $h_{o2}<0$.

If the image is inverted $h_{o2}0$, then the image from the second mirror is also inverted, $h_{i2}<0$.

$a)$ There are several types of coated mirrors, some of them are metal coated mirrors and dielectric coated mirrors. Dielectric mirrors consist of many layers of dielectrics, and each of them has a different index of diffraction. The most commonly used mirrors are metal coated mirrors. These are mirrors that have a layer of metal on them. We will now look at the pros and cons of these mirrors. The advantages of dielectric coated mirrors are that they have high reflectivity, while the disadvantage of metal coated mirrors is that they have lower reflectivity than dielectric ones. Also, the advantage of dielectric mirrors is the high damage threshold and high scratch resistance. Metal coated mirrors have little resistance to scratches, but they can also have a protective layer, so they are more resistant. The advantage of metal coated mirrors is that they are cheaper than dielectric ones and that the reflection does not depend much on the angle.

$b)$ Aluminum mirrors are very practical and accurate mirrors. They are accurate up to one-quarter of the wavelength of light. With conventional methods, it is possible to make an aluminum mirror. Usually, $99.5 %$ of pure aluminum is taken, because it gives the best results. When making mirrors, optical flat aluminum is taken and chemically polished to obtain an aluminum mirror. It is necessary to put on a protective layer because aluminum reacts with the atmosphere. Usually, they put the thin film of nickel, because it is good for protection. The aluminum mirrors are much lighter than glass mirrors, so they can be used in various areas, such as aerospace, defense, and medical optical systems. Also, it is easier to make numerous shapes from aluminum mirrors. Aluminum mirrors are made to have high reflectivity at any incident angle.

In this problem, we have a child that decelerates from the initial speed to zero on the path of $6.2 hspace{0.5mm} mathrm{m}$. We need to find the coefficient of friction of the waxed floor.

To solve this problem, we are going to use equations hfill textcolor{white}{.} [a=frac{v_{f}^{2}-v^{2}_{i}}{2s}] and [m cdot a =- F_{friction}] where $F_{friction}$ is the force of friction $F_{friction}= mu N$, $mu$ is the coefficient of friction, and $N$ is the normal force.

We are going to divide previous expression with the mass $m$ hfill textcolor{white}{.}
begin{align*}
m cdot a& =- mu m g/: m\
a&=- mu g
end{align*}

We are going to divide everything with the $-g$, to get the expression for the coefficient of friction hfill textcolor{white}{.}
begin{align*}
a&=- mu g / : (-g)\
mu&=- frac{a}{g}
end{align*}

In this step, we are going to calculate the acceleration. In the equation $a=frac{v_{f}^{2}-v^{2}_{i}}{2s}$, we are going to substitute the value, so we write

$$
begin{align*}
a&=frac{v_{f}^{2}-v^{2}_{i}}{2s}\
a&=frac{(0 hspace{0.5mm} mathrm{m/s})^{2}-(4.7 hspace{0.5mm} mathrm{m/s})^{2}}{2 cdot 6.2 hspace{0.5mm} mathrm{m}}\
a&=-1.8 hspace{0.5mm} mathrm{m/s^2}
end{align*}
$$

Finally, we can calculate the coefficient of friction. We are going to substitute values for $a$ and $g$ hfill textcolor{white}{.}
begin{align*}
mu&=- frac{-1.8 hspace{0.5mm} mathrm{m/s^2}}{9.8 hspace{0.5mm} mathrm{m/s^2}}\
mu&=0.2
end{align*}

So, we found the coefficient of friction hfill textcolor{white}{.}
[ framebox[1.1width]{$ therefore mu=0.2 $}]

$$
mu = 0.2
$$

In this problem, we have a piece of cooper of $1.0 hspace{0.5mm} mathrm{g}$, that is falling from the height. While it was falling, it lost half of its energy. We need to calculate how much did it heat during the fall.

To solve this problem, we are going to use the law of energy conservation $E_{i}=E_{f}$ where $E_{i}$ is an initial energy, in our case the potential energy $mgh$, and $E_{f}$ si a final energy, in our case the sum of kinetic energy and thermal loss $frac{1}{2}mv^{2}+E_{loss}$.

First, we are going to calculate a potential energy $E_{o}$ by substituting the values hfill textcolor{white}{.}
begin{align*}
E_{p}&=mgh\
E_{p}&=1.0 cdot 10^{-3} hspace{0.5mm} mathrm{kg} cdot 9.81cdot 10^{4} hspace{0.5mm} mathrm{m/s^{2}} cdot 1.0 cdot 10^{4} hspace{0.5mm} mathrm{m}\
E_{p}&=98.1 hspace{0.5mm} mathrm{J}
end{align*}

In this step, we are going to calculate a kinetic energy $E_{k}$ by substituting the values hfill textcolor{white}{.}
begin{align*}
E_{k}&=frac{1}{2} mv^{2}\
E_{k}&=frac{1}{2} cdot 1.0 cdot 10^{-3} hspace{0.5mm} mathrm{kg} cdot (70.0 hspace{0.5mm} mathrm{m/s})^{2}\
E_{k}&=2.4 hspace{0.5mm} mathrm{J}
end{align*}

Now, we can write the law of energy conservation $E_{i}=E_{f}$ as hfill textcolor{white}{.}
begin{align*}
E_{i}&=E_{f}\
E_{p}&=E_{k}+E_{loss}
end{align*}

To find the expression for the energy loss, we need to subtract $E_{k}$ on both sides hfill textcolor{white}{.}
begin{align*}
E_{p}&=E_{k}+E_{loss} / – E_{k} \
E_{loss}&=E_{p}-E_{k}
end{align*}

Finally, we can calculate how much the piece heat during the fall $Q$ hfill textcolor{white}{.}
begin{align*}
Q&= frac{E_{loss}}{2}\
Q&=frac{95.7hspace{0.5mm} mathrm{J}}{2}\
Q&=47.8 hspace{0.5mm} mathrm{J}
end{align*}

So, we found the heat energy hfill textcolor{white}{.}
[ framebox[1.1width]{$ therefore Q=47.8 hspace{0.5mm} mathrm{J} $}]

$$
Q=47.8 hspace{0.5mm} mathrm{J}
$$

In this problem, we have the body that exerts pressure, with the force of $600 hspace{0.5mm} mathrm{N}$, on the cross-sectional area of $0.25 hspace{0.5mm} mathrm{m^{2}}$. We need to find the pressure that we need to blow into the straw to lift the body.

To solve this problem, we are going to use the equation hfill textcolor{white}{.} [P= frac{F}{A}] where $P$ is the exerted pressure, $F$ is the force, and $A$ is an area on which the force acts. We are going to compare the pressure from the body, and the pressure that we need to blow into the straw to lift the body. The minimal pressure that we need to blow into the straw to lift the body is the same as the pressure from the body, so we can write [P_{1}=P_{2}]

How the pressure that we need to blow into the straw to lift the body is equal to the pressure from the body, then our pressure is [ framebox[1.1width]{$ therefore P_{2}= 2400 hspace{0.5mm} mathrm{Pa} $}]

$$
P_{2}= 2400 hspace{0.5mm} mathrm{Pa}
$$

To solve this problem, we are going to use the equation hfill textcolor{white}{.} [T=2 pi sqrt{frac{l}{g}}] where $T$ is the period, $g$ is the gravitational acceleration, and $l$ is a length of a pendulum. To find the period on the Moon, we need to find gravitational acceleration on the Moon. We can do this if we compare the equations [F_{g}= G frac{mM_{Moon}}{R_{Moon}^{2}}] and [F_{g}=mg_{Moon}]

First, we are going to calculate a gravitational acceleration on the Moon. We have two equations, $F_{g}= G frac{mM_{Moon}}{R_{Moon}^{2}}$ and $F_{g}=mg_{Moon}$, that are equal to each other

$$
begin{align*}
mg_{Moon}&=G frac{mM_{Moon}}{R_{Moon}^{2}}
end{align*}
$$

Now, we are going to divide both sides in previous equation with a mass $m$ hfill textcolor{white}{.}
begin{align*}
mg_{Moon}&=G frac{mM_{Moon}}{R_{Moon}^{2}} /: m\
g_{Moon}&=G frac{M_{Moon}}{R_{Moon}^{2}}
end{align*}

Now, we can find the period of pendulum on the Moon by substituting the values in the equation $T=2 pi sqrt{frac{l}{g}}$ hfill textcolor{white}{.}
begin{align*}
T_{Moon}&=2 pi sqrt{frac{l}{g_{Moon}}}\
T_{Moon}&=2 pi sqrt{frac{2.0 hspace{0.5mm} mathrm{m}}{1.62 hspace{0.5mm} mathrm{m/s^{2}}}}\
T_{Moon}&=6.98hspace{0.5mm} mathrm{s}
end{align*}

The period of pendulum on the Moon is hfill textcolor{white}{.} [ framebox[1.1width]{$ therefore T_{Moon}=6.98 hspace{0.5mm} mathrm{s} $}]

The period of pendulum on Earth is hfill textcolor{white}{.} [ framebox[1.1width]{$ therefore T_{Earth}=2.84 hspace{0.5mm} mathrm{s} $}]

$T_{Moon}=6.98hspace{0.5mm} mathrm{s}$

$$
T_{Earth}=2.84 hspace{0.5mm} mathrm{s}
$$

$a)$ In Figures 15-12 and 15-13, we can see the behavior of the frequencies relative if the pipe is open or closed. To find if the pipe should be open or closed, we can observe the resonant frequency in both cases. To find the shorter pipe, we can compare the same frequency from the different pipes.

For closed pipe, the frequency is $f_{1}=frac{v}{4L_{closed}}$, and for open pipe the frequency is $f_{1}=frac{v}{2L_{open}}$.

How the frequencies are the same, we can write them. Also, we can divide both sides with $v/2$

$$
begin{align*}
frac{v}{4L_{closed}}&=frac{v}{2L_{open}}/ : v/2\
L_{open}&=2L_{closed}
end{align*}
$$

$b)$ The sound will not be the same. Even if the fundamental frequency $f_{1}$ is the same as the frequencies, higher frequencies are different. Higher frequencies for closed pipe are odd-number, like $3f_{1}, 5f_{1},…$. And, the higher frequencies for open pipe are $2f_{1}, 3f_{1}…$.

$A_{G}$ = $Asin(omega_{G}t)$;
$A_{R}$ = $Asin(omega_{GR}t)$;

Let the two two light rays consist of sinusoidal waves with $A_{G}$ and $A_{R}$ be the amplitudes of green and red light waves respectively at a given point. Both the waves are considered to have the same phase and same maximum amplitude for simplicity.

$A_{O}$ = $A_{G}$ + $A_{R}$

$A_{O}$ = $2Asin((omega_{G}+omega_{R})/2)cos((omega_{G}-omega_{R})/2)$

Wave frequency =$( omega_{G}+omega_{R})/2$;
Beat ferquency = $(omega_{G}-omega_{R})/2$

$(omega_{G}+omega_{R})/2$ = $2pi(560+455)/2$ = 3188.72 rad/s

Resultant frequency is (3188.72)/2$pi$ = 507.50 Hz

$f=omega/2pi$;
This resultant is the frequency of yellow light