Physics: Principles and Problems
9th Edition
ISBN: 9780078458132
Textbook solutions
All Solutions
Page 469: Practice Problems
Exercise 12
Step 1
1 of 3
In the diagram below, we can see the picture from the Example 2, but in the scale $1:4$, for example, the height of the original is $2.0 hspace{0.5mm} mathrm{cm}$, and in the diagram is $h_{0}= 0.5 hspace{0.5mm} mathrm{cm}$. To find the image height and the image position, we need two rays from one point. The intersection of these rays is the image of that point. We need two points, one is the top of the arrow, and the other one is the bottom of the arrow. We can compare values in the real scenario and the diagram ($1:4$)
begin{center}
begin{tabular}{| c | c | c | }
hline
& Real case & The diagram\
hline
The original height & $2.0 hspace{0.5mm} mathrm{cm}$ & $0.5 hspace{0.5mm} mathrm{cm}$ \
The image height & $-1.0 hspace{0.5mm} mathrm{cm}$ & $-0.25 hspace{0.5mm} mathrm{cm}$ \
The original distance & $30.0 hspace{0.5mm} mathrm{cm}$ & $7.5 hspace{0.5mm} mathrm{cm}$ \
The image distance & $15.0 hspace{0.5mm} mathrm{cm}$ & $3.75 hspace{0.5mm} mathrm{cm}$ \
The radius & $20.0 hspace{0.5mm} mathrm{cm}$ & $5.0 hspace{0.5mm} mathrm{cm}$ \
The focal length & $10.0 hspace{0.5mm} mathrm{cm}$ & $2.5 hspace{0.5mm} mathrm{cm}$ \
hline
end{tabular}
end{center}
Also, in the diagram below, one square has length side of $1.0 hspace{0.5mm} mathrm{cm}$.
begin{center}
begin{tabular}{| c | c | c | }
hline
& Real case & The diagram\
hline
The original height & $2.0 hspace{0.5mm} mathrm{cm}$ & $0.5 hspace{0.5mm} mathrm{cm}$ \
The image height & $-1.0 hspace{0.5mm} mathrm{cm}$ & $-0.25 hspace{0.5mm} mathrm{cm}$ \
The original distance & $30.0 hspace{0.5mm} mathrm{cm}$ & $7.5 hspace{0.5mm} mathrm{cm}$ \
The image distance & $15.0 hspace{0.5mm} mathrm{cm}$ & $3.75 hspace{0.5mm} mathrm{cm}$ \
The radius & $20.0 hspace{0.5mm} mathrm{cm}$ & $5.0 hspace{0.5mm} mathrm{cm}$ \
The focal length & $10.0 hspace{0.5mm} mathrm{cm}$ & $2.5 hspace{0.5mm} mathrm{cm}$ \
hline
end{tabular}
end{center}
Also, in the diagram below, one square has length side of $1.0 hspace{0.5mm} mathrm{cm}$.
Step 2
2 of 3
Result
3 of 3
We draw a diagram with the scale of $1:4$ and got the values like in Example 2.
Exercise 13
Solution 1
Solution 2
Step 1
1 of 5
We need to determine the position of the image if an object is $36,,rm{cm}$ in front of a concave mirror that hase a focal length of $16,,rm{cm}$
Step 2
2 of 5
In order to solve this problem, we need to use an equation for concave mirrors:
$$frac{1}{d_o}+frac{1}{d_i}=frac{1}{f}$$
Where $d_o$ is distance of an object, $d_i$ distance of the image and $f$ a focal length of the mirror.
$$frac{1}{d_o}+frac{1}{d_i}=frac{1}{f}$$
Where $d_o$ is distance of an object, $d_i$ distance of the image and $f$ a focal length of the mirror.
Step 3
3 of 5
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:
$$d_i=frac{fd_o}{d_o-f}$$
$$frac{1}{d_i}=frac{1}{f}-frac{1}{d_o}$$
$$frac{q}{d_i}=frac{d_o-f}{fd_o}$$
And finally:
$$d_i=frac{fd_o}{d_o-f}$$
Step 4
4 of 5
Now we can input values to get:
$$d_i=frac{0.16cdot 0.36}{0.36-0.16}$$
Finally:
$$boxed{d_i=28.8,,rm{cm}}$$
$$d_i=frac{0.16cdot 0.36}{0.36-0.16}$$
Finally:
$$boxed{d_i=28.8,,rm{cm}}$$
Result
5 of 5
$$d_i=28.8,,rm{cm}$$
Step 1
1 of 3
Our object is at the distance of $d_{o}=36.0 hspace{0.5mm} mathrm{cm}$ from the center of a concave mirror. From diagram below, we can see what are the properties of the image in the concave mirror, and the image is real, smaller than original and reverse. To calculate the position of the image, we use the equation [ frac{1}{d_{o}}+ frac{1}{d_{i}}= frac{1}{f}] where $d_{i}$ is the position of the image, and $f=16.0 hspace{0.5mm} mathrm{cm}$ is the focal length of the mirror. Now, we can write
begin{align*}
frac{1}{d_{o}}+ frac{1}{d_{i}}&= frac{1}{f}\
frac{1}{d_{i}}&= frac{1}{f} – frac{1}{d_{o}}\
frac{1}{d_{i}}&= frac{d_{o} – f}{f hspace{0.5mm} d_{o}} \
d_{i}&= frac{f hspace{0.5mm} d_{o}}{d_{o} – f}\
d_{i}&= frac{16.0 hspace{0.5mm} mathrm{cm} cdot 36.0 hspace{0.5mm} mathrm{cm}}{36.0 hspace{0.5mm} mathrm{cm} – 16.0 hspace{0.5mm} mathrm{cm}}\
d_{i}&= 28.8 hspace{0.5mm} mathrm{cm}
end{align*}
The image position is [ framebox[1.1width]{$ therefore d_{i}= 28.8 hspace{0.5mm} mathrm{cm} $}]
begin{align*}
frac{1}{d_{o}}+ frac{1}{d_{i}}&= frac{1}{f}\
frac{1}{d_{i}}&= frac{1}{f} – frac{1}{d_{o}}\
frac{1}{d_{i}}&= frac{d_{o} – f}{f hspace{0.5mm} d_{o}} \
d_{i}&= frac{f hspace{0.5mm} d_{o}}{d_{o} – f}\
d_{i}&= frac{16.0 hspace{0.5mm} mathrm{cm} cdot 36.0 hspace{0.5mm} mathrm{cm}}{36.0 hspace{0.5mm} mathrm{cm} – 16.0 hspace{0.5mm} mathrm{cm}}\
d_{i}&= 28.8 hspace{0.5mm} mathrm{cm}
end{align*}
The image position is [ framebox[1.1width]{$ therefore d_{i}= 28.8 hspace{0.5mm} mathrm{cm} $}]
Step 2
2 of 3
Result
3 of 3
$$
d_{i}= 28.8 hspace{0.5mm} mathrm{cm}
$$
d_{i}= 28.8 hspace{0.5mm} mathrm{cm}
$$
Exercise 14
Solution 1
Solution 2
Step 1
1 of 8
We need to determine the position and height of the image if a $3,,rm{cm}$ tall object is standing $20,,rm{cm}$ away from the concave mirror with a radius of $16,,rm{cm}$
Step 2
2 of 8
In order to solve this problem, we need to know a couple of properties of the image in the concave mirror:
– the image is real
– the image is smaller than original
– the image is inverted
– the image is real
– the image is smaller than original
– the image is inverted
Step 3
3 of 8
Next, we need to calculate the focal length of the mirror. This can be done by using a simple equation that connects radius of the mirror and its focal length:
$$f=frac{r}{2}$$
Inserting values we get:
$$f=frac{0.16}{2}$$
$$f=8,,rm{cm}$$
$$f=frac{r}{2}$$
Inserting values we get:
$$f=frac{0.16}{2}$$
$$f=8,,rm{cm}$$
Step 4
4 of 8
Now we can calculate the position of the image:
$$frac{1}{d_o}+frac{1}{d_i}=frac{1}{f}$$
From this equation we can express $d_i$:
$$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}$$
$$frac{1}{d_o}+frac{1}{d_i}=frac{1}{f}$$
From this equation we can express $d_i$:
$$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}$$
Step 5
5 of 8
Now we can insert values:
$$d_i=frac{0.08cdot 0.2}{0.2-0.08}$$
$$boxed{d_i=13.3,,rm{cm}}$$
$$d_i=frac{0.08cdot 0.2}{0.2-0.08}$$
$$boxed{d_i=13.3,,rm{cm}}$$
Step 6
6 of 8
With known distance, we can calculate the height of the image:
$$m=-frac{d_i}{d_o}=frac{h_i}{h_o}$$
Where $m$ is magnification and $h_o$ is height of an object.
From the previous equation we can get:
$$h_i=-h_ofrac{d_i}{d_o}$$
$$m=-frac{d_i}{d_o}=frac{h_i}{h_o}$$
Where $m$ is magnification and $h_o$ is height of an object.
From the previous equation we can get:
$$h_i=-h_ofrac{d_i}{d_o}$$
Step 7
7 of 8
Inserting values we get:
$$h_i=-0.03cdot frac{0.133}{0.2}$$
$$boxed{h_i=-2,,rm{cm}}$$
$$h_i=-0.03cdot frac{0.133}{0.2}$$
$$boxed{h_i=-2,,rm{cm}}$$
Result
8 of 8
$$d_i=13.3,,rm{cm}$$
$$h_i=-2,,rm{cm}$$
$$h_i=-2,,rm{cm}$$
Step 1
1 of 4
Our object is at the distance of $d_{o}=20.0 hspace{0.5mm} mathrm{cm}$ from the center of a concave mirror. From diagram below, we can see what are the properties of the image in the concave mirror, and the image is real, smaller than original and inverted. We know the the radius of a concave mirror is $r=16.0 hspace{0.5mm} mathrm{cm}$, and from that we know the focal length of a mirror
$$
begin{align*}
r&= 2f\
f&= frac{r}{2}\
f&= frac{16.0 hspace{0.5mm} mathrm{cm}}{2}\
f&= 8.0 hspace{0.5mm} mathrm{cm}
end{align*}
$$
Now, we can find the position of the image
$$
begin{align*}
frac{1}{d_{o}}+ frac{1}{d_{i}}&= frac{1}{f}\
frac{1}{d_{i}}&= frac{1}{f} – frac{1}{d_{o}}\
frac{1}{d_{i}}&= frac{d_{o} – f}{f hspace{0.5mm} d_{o}} \
d_{i}&= frac{f hspace{0.5mm} d_{o}}{d_{o} – f}\
d_{i}&= frac{8.0 hspace{0.5mm} mathrm{cm} cdot 20.0 hspace{0.5mm} mathrm{cm}}{20.0 hspace{0.5mm} mathrm{cm} – 8.0 hspace{0.5mm} mathrm{cm}}\
d_{i}&= 13.3 hspace{0.5mm} mathrm{cm}
end{align*}
$$
Step 2
2 of 4
Now, when we know the positions of the original and the image, we can determinate the height of the image as [m=- frac{d_{i}}{d_{o}}= frac{h_{i}}{h_{o}} ] where the height of the original is $h_{o}=3.0 hspace{0.5mm} mathrm{cm}$, and the $m$ is a magnification. Now, we write
begin{align*}
frac{h_{i}}{h_{o}} &= – frac{d_{i}}{d_{o}}\
h_{i} &= -h_{o} frac{d_{i}}{d_{o}}\
h_{i} &= -3.0 hspace{0.5mm} mathrm{cm} frac{13.3 hspace{0.5mm} mathrm{cm}}{20.0 hspace{0.5mm} mathrm{cm}}\
h_{i} &= -1.995 hspace{0.5mm} mathrm{cm} \
h_{i} &= -2.0 hspace{0.5mm} mathrm{cm}
end{align*}
So, the position of the image is [ framebox[1.1width]{$ therefore d_{i}= 13.3 hspace{0.5mm} mathrm{cm} $}] and the height of the image is [ framebox[1.1width]{$ therefore h_{i} = -2.0 hspace{0.5mm} mathrm{cm} $}]
begin{align*}
frac{h_{i}}{h_{o}} &= – frac{d_{i}}{d_{o}}\
h_{i} &= -h_{o} frac{d_{i}}{d_{o}}\
h_{i} &= -3.0 hspace{0.5mm} mathrm{cm} frac{13.3 hspace{0.5mm} mathrm{cm}}{20.0 hspace{0.5mm} mathrm{cm}}\
h_{i} &= -1.995 hspace{0.5mm} mathrm{cm} \
h_{i} &= -2.0 hspace{0.5mm} mathrm{cm}
end{align*}
So, the position of the image is [ framebox[1.1width]{$ therefore d_{i}= 13.3 hspace{0.5mm} mathrm{cm} $}] and the height of the image is [ framebox[1.1width]{$ therefore h_{i} = -2.0 hspace{0.5mm} mathrm{cm} $}]
Step 3
3 of 4
Result
4 of 4
$d_{i}= 13.3 hspace{0.5mm} mathrm{cm}$
$$
h_{i} = -2.0 hspace{0.5mm} mathrm{cm}
$$
Exercise 15
Solution 1
Solution 2
Step 1
1 of 5
We need to determine the height of an object that is $2.4,,rm{cm}$ tall and is $16,,rm{cm}$ away from the mirror that has a focal length of $7,,rm{cm}$
Step 2
2 of 5
In order to calculate the distance of the image we will use the next equation:
$$frac{1}{d_o}+frac{1}{d_i}=frac{1}{f}$$
$$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}$$
Step 3
3 of 5
inserting values we get.
$$d_i=frac{0.07cdot 0.16}{0.16-0.07}$$
$$boxed{d_i=12.44,,rm{cm}}$$
$$d_i=frac{0.07cdot 0.16}{0.16-0.07}$$
$$boxed{d_i=12.44,,rm{cm}}$$
Step 4
4 of 5
We can find the height of the image using the next equation:
$$-frac{d_i}{d_o}=frac{h_i}{h_o}$$
From this we can get:
$$h_i=-h_o frac{d_i}{d_o}$$
Inserting values:
$$h_i=- 0.024cdot frac{0.1244}{0.16}$$
$$boxed{h_i=-1.867,,rm{cm}}$$
Where negative value means that the image is inverted.
$$-frac{d_i}{d_o}=frac{h_i}{h_o}$$
From this we can get:
$$h_i=-h_o frac{d_i}{d_o}$$
Inserting values:
$$h_i=- 0.024cdot frac{0.1244}{0.16}$$
$$boxed{h_i=-1.867,,rm{cm}}$$
Where negative value means that the image is inverted.
Result
5 of 5
$$d_i=12.44,,rm{cm}$$
$$h_i=-1.867,,rm{cm}$$
$$h_i=-1.867,,rm{cm}$$
Step 1
1 of 3
{We will use : $text{color{#c34632}$$dfrac{1}{f} = dfrac{1}{d_o}+dfrac{1}{d_i}$$}$ $Given that$d_o = 16$and$f=7 $Note that the sign of$f$is positive because it is a concave mirror$ $dfrac{1}{7} = dfrac{1}{16}+dfrac{1}{d_i}$ $dfrac{1}{7} – dfrac{1}{16}=dfrac{1}{d_i}$ $dfrac{1}{7}times dfrac{16}{16} – dfrac{1}{16}timesdfrac{7}{7}=dfrac{1}{d_i}$ $dfrac{16}{112} -dfrac{7}{112}=dfrac{1}{d_i}$ $dfrac{9}{112}=dfrac{1}{d_i}$ $Take reciprocal of both sides$ $dfrac{112}{9}=d_i$ $d_iapprox 12.44$ $
The positive sign means that the Image is formed in front of the mirror }$
Step 2
2 of 3
To find the image height we need to find magnification first
$$
text{color{#c34632}Recall that : $$m = -dfrac{d_i}{d_o}$$}
$$
$$
m = -dfrac{12.44}{16} = -0.778
$$
The negative sign means that the Image will be Inverted
The height of the image will be $mtimestext{Object Height} =-0.778times2.4 = -1.867$
Result
3 of 3
$$
text{color{#4257b2}The Image is formed 12.44 cm in front of the mirror
text{color{#4257b2}The Image is formed 12.44 cm in front of the mirror
$ $
The height of the image is $-1.867$ cm
$ $
The negative sign means that the Image is Inverted }
$$
Exercise 16
Solution 1
Solution 2
Step 1
1 of 5
Mirror has a focal length of $f=7,,rm{cm}$. We need to determine the position and height of an object if the image is $h_i=3,,rm{cm}$ tall, inverted and $d_i=16,,rm{cm}$ away from the mirror.
Step 2
2 of 5
The equation that we will use is:
$$frac{1}{d_o}+frac{1}{d_i}=frac{1}{f}$$
$$frac{1}{d_o}+frac{1}{d_i}=frac{1}{f}$$
From the previous equation we can express $d_o$:
$$frac{1}{d_o}=frac{1}{f}-frac{1}{d_i}$$
$$frac{q}{d_o}=frac{d_i-f}{fd_i}$$
And finally:
$$d_o=frac{fd_i}{d_i-f}$$
Step 3
3 of 5
Inserting values we get:
$$d_o=frac{0.10cdot 0.16}{0.16-0.10}$$
$$boxed{d_o=26.67,,rm{cm}}$$
$$d_o=frac{0.10cdot 0.16}{0.16-0.10}$$
$$boxed{d_o=26.67,,rm{cm}}$$
Step 4
4 of 5
In order to find height of an object we can use an equation for magnification:
$$-frac{d_o}{d_i}=frac{h_o}{h_i}$$
From this we can get:
$$h_o=-h_ifrac{d_o}{d_i}$$
inserting values we get.
$$h_o=-(-3)cdot frac{26.67}{16}$$
$$boxed{h_o=5,,rm{cm}}$$
$$-frac{d_o}{d_i}=frac{h_o}{h_i}$$
From this we can get:
$$h_o=-h_ifrac{d_o}{d_i}$$
inserting values we get.
$$h_o=-(-3)cdot frac{26.67}{16}$$
$$boxed{h_o=5,,rm{cm}}$$
Result
5 of 5
$$d_o=26.67,q,rm{cm}$$
$$h_0=5,,rm{cm}$$
$$h_0=5,,rm{cm}$$
Step 1
1 of 3
{We will use : $text{color{#c34632}$$dfrac{1}{f} = dfrac{1}{d_o}+dfrac{1}{d_i}$$}$ $f = +10$ $Note that the sign of$f$is positive because it is a concave mirror$ $d_i = +16$ $Note that :$d_i$is positive because the Image is formed in front of the mirror.$ $But how do I know that the Image is formed in front of the mirror?$ $Because it is given that the Image is Inverted. This means that$m<0$.$ $According to the formula fbox{$m = -dfrac{d_i}{d_o}$},$d_i$must be positive$ $Substitute the values of$f$and$d_i $dfrac{1}{10} = dfrac{1}{d_o}+dfrac{1}{16}$ $dfrac{1}{10}-dfrac{1}{16} = dfrac{1}{d_o}$ $dfrac{8}{80}-dfrac{5}{80} = dfrac{1}{d_o}$ $
}$
}$
Step 2
2 of 3
$$
dfrac{3}{80} = dfrac{1}{d_o}
$$
dfrac{3}{80} = dfrac{1}{d_o}
$$
Take Reciprocal of both sides
$$
dfrac{80}{3} = d_o
$$
$$
d_oapprox 26.67
$$
To find the object height we need to find magnification first
$$
text{color{#c34632}Recall that : $$m = -dfrac{d_i}{d_o}$$}
$$
$$
m = -dfrac{16}{26.67} = -0.6
$$
$$
text{color{#4257b2}Recall that : $m=dfrac{h_i}{h_o}$}
$$
$$
-0.6=dfrac{-3}{h_o}
$$
$$
h_o=dfrac{-3}{-0.6} = +5
$$
Result
3 of 3
$$
text{color{#4257b2}Height of the Object is 5 cm
text{color{#4257b2}Height of the Object is 5 cm
$ $
It is placed 26.67 cm in front of the mirror }
$$
Exercise 17
Step 1
1 of 4
In the diagram below, we solved the problem in the scale $1:4$. But firstly, we are going to solve the problem by using the mirror equation [ frac{1}{f}= frac{1}{d_{0}}+ frac{1}{d_{i}}] where $f=-15.0 hspace{0.5mm} mathrm{cm}$ is the focal length, $d_{o}= 0.0 hspace{0.5mm} mathrm{cm}$ is the position of the object, and $d_{i}$ is the position of the image. So, we can write
begin{align*}
frac{1}{d_{o}}+ frac{1}{d_{i}}&= frac{1}{f}\
frac{1}{d_{i}}&= frac{1}{f} – frac{1}{d_{o}}\
frac{1}{d_{i}}&= frac{d_{o} – f}{f hspace{0.5mm} d_{o}} \
d_{i}&= frac{f hspace{0.5mm} d_{o}}{d_{o} – f}\
d_{i}&= frac{-15.0 hspace{0.5mm} mathrm{cm} cdot 20.0 hspace{0.5mm} mathrm{cm}}{20.0 hspace{0.5mm} mathrm{cm} + 15.0 hspace{0.5mm} mathrm{cm}}\
d_{i}&= -8.6 hspace{0.5mm} mathrm{cm}
end{align*}
So, the distance of the image is [ framebox[1.1width]{$ therefore d_{i} = -8.6 hspace{0.5mm} mathrm{cm} $}]
begin{align*}
frac{1}{d_{o}}+ frac{1}{d_{i}}&= frac{1}{f}\
frac{1}{d_{i}}&= frac{1}{f} – frac{1}{d_{o}}\
frac{1}{d_{i}}&= frac{d_{o} – f}{f hspace{0.5mm} d_{o}} \
d_{i}&= frac{f hspace{0.5mm} d_{o}}{d_{o} – f}\
d_{i}&= frac{-15.0 hspace{0.5mm} mathrm{cm} cdot 20.0 hspace{0.5mm} mathrm{cm}}{20.0 hspace{0.5mm} mathrm{cm} + 15.0 hspace{0.5mm} mathrm{cm}}\
d_{i}&= -8.6 hspace{0.5mm} mathrm{cm}
end{align*}
So, the distance of the image is [ framebox[1.1width]{$ therefore d_{i} = -8.6 hspace{0.5mm} mathrm{cm} $}]
Step 2
2 of 4
Now, we are going to look at the diagram. We use two rays from one point (the top of the arrow), to get one point of the image. We use one ray that goes on the center of the convex mirror, and the other ray is one parallel to the optical axis. Our image is behind the convex mirror. Also, we can see the image is about $8 hspace{0.5mm} mathrm{cm}$ behind the mirror. In the diagram, big square has side of $1.0 hspace{0.5mm} mathrm{cm}$.
Step 3
3 of 4
Result
4 of 4
$$
d_{i} = -8.6 hspace{0.5mm} mathrm{cm}
$$
d_{i} = -8.6 hspace{0.5mm} mathrm{cm}
$$
Exercise 18
Solution 1
Solution 2
Step 1
1 of 6
We need to find position and diameter of an image of a lightbulb. Lightbulb has a diamter of $_o=6,,rm{cm}$ and is placed $d_o=60,,rm{cm}$ away from the mirror that has a focal length of $f=-13,,rm{cm}$.
Step 2
2 of 6
We know that this is convex mirror because the focal length is negative.
We can use the next equation in order to calculate the distance of an image:
$$frac{1}{d_o}+frac{1}{d_i}=frac{1}{f}$$
We can use the next equation in order to calculate the distance of an image:
$$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}$$
Step 3
3 of 6
Inserting values we get.
$$d_i=frac{-13cdot 60}{60+13}$$
$$boxed{d_i=-10.7,,rm{cm}}$$
$$d_i=frac{-13cdot 60}{60+13}$$
$$boxed{d_i=-10.7,,rm{cm}}$$
Step 4
4 of 6
We can calculate the height of the image using the next equation:
$$frac{h_i}{h_o}=-frac{d_i}{d_o}$$
From this we can get:
$$h_i=h_ofrac{d_i}{d_o}$$
$$frac{h_i}{h_o}=-frac{d_i}{d_o}$$
From this we can get:
$$h_i=h_ofrac{d_i}{d_o}$$
Step 5
5 of 6
Inserting values we get.
$$h_i=-6cdot frac{-10.7}{60}$$
$$boxed{h_i=1.07,,rm{cm}}$$
$$h_i=-6cdot frac{-10.7}{60}$$
$$boxed{h_i=1.07,,rm{cm}}$$
Result
6 of 6
$$d_i=-10.7,,rm{cm}$$
$$h_i=1.07,,rm{cm}$$
$$h_i=1.07,,rm{cm}$$
Step 1
1 of 2
For this convex mirror, the focal length is $f= -13.0 hspace{0.5mm} mathrm{cm}$. Our object has a diameter of $h_{o}=6.0 hspace{0.5mm} mathrm{cm}$, and it is at the position of $d_{o}= 60.0 hspace{0.5mm} mathrm{cm}$. To calculate the diameter of image, we can observe the height og the object, and determinate the height of the image. Firstly, we are going to calculate the position of theimage by using the mirror equation
begin{align*}
frac{1}{d_{o}}+ frac{1}{d_{i}}&= frac{1}{f}\
frac{1}{d_{i}}&= frac{1}{f} – frac{1}{d_{o}}\
frac{1}{d_{i}}&= frac{d_{o} – f}{f hspace{0.5mm} d_{o}} \
d_{i}&= frac{f hspace{0.5mm} d_{o}}{d_{o} – f}\
d_{i}&= frac{-13.0 hspace{0.5mm} mathrm{cm} cdot 60.0 hspace{0.5mm} mathrm{cm}}{60.0 hspace{0.5mm} mathrm{cm} + 13.0 hspace{0.5mm} mathrm{cm}}\
d_{i}&= -10.7 hspace{0.5mm} mathrm{cm}
end{align*}
So, the distance of the image is [ framebox[1.1width]{$ therefore d_{i} = -10.7 hspace{0.5mm} mathrm{cm} $}]
To calculate the height of the image, we use the equation [ frac{h_{i}}{h_{o}} =- frac{d_{i}}{d_{o}} ] where $h_{i}$ is the image of the diameter, so we can write
begin{align*}
frac{h_{i}}{h_{o}} &=- frac{d_{i}}{d_{o}} \
h_{i} &=-h_{o} frac{d_{i}}{d_{o}} \
h_{i} &=-6.0 hspace{0.5mm} mathrm{cm} frac{-10.7 hspace{0.5mm} mathrm{cm}}{60.0 hspace{0.5mm} mathrm{cm}} \
h_{i} &=-6.0 hspace{0.5mm} mathrm{cm} cdot (-0.18) \
h_{i} &=1.07 hspace{0.5mm} mathrm{cm}
end{align*}
The diameter of the image is [ framebox[1.1width]{$ therefore h_{i} =1.07 hspace{0.5mm} mathrm{cm} $}]
begin{align*}
frac{1}{d_{o}}+ frac{1}{d_{i}}&= frac{1}{f}\
frac{1}{d_{i}}&= frac{1}{f} – frac{1}{d_{o}}\
frac{1}{d_{i}}&= frac{d_{o} – f}{f hspace{0.5mm} d_{o}} \
d_{i}&= frac{f hspace{0.5mm} d_{o}}{d_{o} – f}\
d_{i}&= frac{-13.0 hspace{0.5mm} mathrm{cm} cdot 60.0 hspace{0.5mm} mathrm{cm}}{60.0 hspace{0.5mm} mathrm{cm} + 13.0 hspace{0.5mm} mathrm{cm}}\
d_{i}&= -10.7 hspace{0.5mm} mathrm{cm}
end{align*}
So, the distance of the image is [ framebox[1.1width]{$ therefore d_{i} = -10.7 hspace{0.5mm} mathrm{cm} $}]
To calculate the height of the image, we use the equation [ frac{h_{i}}{h_{o}} =- frac{d_{i}}{d_{o}} ] where $h_{i}$ is the image of the diameter, so we can write
begin{align*}
frac{h_{i}}{h_{o}} &=- frac{d_{i}}{d_{o}} \
h_{i} &=-h_{o} frac{d_{i}}{d_{o}} \
h_{i} &=-6.0 hspace{0.5mm} mathrm{cm} frac{-10.7 hspace{0.5mm} mathrm{cm}}{60.0 hspace{0.5mm} mathrm{cm}} \
h_{i} &=-6.0 hspace{0.5mm} mathrm{cm} cdot (-0.18) \
h_{i} &=1.07 hspace{0.5mm} mathrm{cm}
end{align*}
The diameter of the image is [ framebox[1.1width]{$ therefore h_{i} =1.07 hspace{0.5mm} mathrm{cm} $}]
Result
2 of 2
$d_{i} = -10.7 hspace{0.5mm} mathrm{cm}$
$$
h_{i} =1.07 hspace{0.5mm} mathrm{cm}
$$
Exercise 19
Solution 1
Solution 2
Step 1
1 of 5
We need to determine the focal length of a convex mirror that has magnification of $M=0.75$ and creates an image at $d_i=24,,rm{cm}$ behind mirror.
Step 2
2 of 5
Magnification is calculated as:
$$M=-frac{d_i}{d_o}$$
From this we can get.
$$d_o=-frac{24}{0.75}$$
$$d_o=32,,rm{cm}$$
$$M=-frac{d_i}{d_o}$$
From this we can get.
$$d_o=-frac{24}{0.75}$$
$$d_o=32,,rm{cm}$$
Step 3
3 of 5
In order to find a focal length we can use the equation for convex mirrors:
$$frac{1}{d_o}+frac{1}{d_i}=frac{1}{f}$$
From this we can extract $f$:
$$frac{1}{f}=frac{d_o+d_i}{d_od_i}$$
$$f=frac{d_od_i}{d_o+d_i}$$
$$frac{1}{d_o}+frac{1}{d_i}=frac{1}{f}$$
From this we can extract $f$:
$$frac{1}{f}=frac{d_o+d_i}{d_od_i}$$
$$f=frac{d_od_i}{d_o+d_i}$$
Step 4
4 of 5
Now we can insert values:
$$f=frac{32cdot (-24)}{32-24}$$
$$boxed{f=-96,,rm{cm}}$$
$$f=frac{32cdot (-24)}{32-24}$$
$$boxed{f=-96,,rm{cm}}$$
Result
5 of 5
$$f=-96,,rm{cm}$$
Step 1
1 of 2
To find the position of the object, we are going to use the equation for a magnification [ M=- frac{d_{i}}{d_{o}}= frac{h_{i}}{h_{o}}] We know that the magnification is $M= frac{3}{4}$, and the position of the image is $d_{i}= -24 hspace{0.5mm} mathrm{cm}$, so we can write
begin{align*}
M&=- frac{d_{i}}{d_{o}}\
d_{o}&= – frac{d_{i}}{M}\
d_{o}&= – frac{-24 hspace{0.5mm} mathrm{cm}}{ frac{3}{4}}\
d_{o}&= frac{4 cdot 24 hspace{0.5mm} mathrm{cm}}{ 3}\
d_{o}&= 32 hspace{0.5mm} mathrm{cm}
end{align*}
To find the focal length, we will use the mirror equation [ frac{1}{d_{o}}+ frac{1}{d_{i}}= frac{1}{f}] so we are going to write
begin{align*}
frac{1}{f} &= frac{1}{d_{o}}+ frac{1}{d_{i}}\
frac{1}{f} &= frac{d_{o}+d_{i}}{ d_{o} hspace{0.5mm} d_{i}}\
frac{1}{f} &= frac{d_{o}+d_{i}}{ d_{o} hspace{0.5mm} d_{i}}\
f &= frac{ d_{o} hspace{0.5mm} d_{i}}{ d_{o}+d_{i}} \
f &= frac{ 32 hspace{0.5mm} mathrm{cm} hspace{0.5mm}(-24 hspace{0.5mm} mathrm{cm})}{ 32 hspace{0.5mm} mathrm{cm}-24 hspace{0.5mm} mathrm{cm}} \
f&= -96 hspace{0.5mm} mathrm{cm}
end{align*}
The focal length is [ framebox[1.1width]{$ therefore f = -96 hspace{0.5mm} mathrm{cm} $}]
begin{align*}
M&=- frac{d_{i}}{d_{o}}\
d_{o}&= – frac{d_{i}}{M}\
d_{o}&= – frac{-24 hspace{0.5mm} mathrm{cm}}{ frac{3}{4}}\
d_{o}&= frac{4 cdot 24 hspace{0.5mm} mathrm{cm}}{ 3}\
d_{o}&= 32 hspace{0.5mm} mathrm{cm}
end{align*}
To find the focal length, we will use the mirror equation [ frac{1}{d_{o}}+ frac{1}{d_{i}}= frac{1}{f}] so we are going to write
begin{align*}
frac{1}{f} &= frac{1}{d_{o}}+ frac{1}{d_{i}}\
frac{1}{f} &= frac{d_{o}+d_{i}}{ d_{o} hspace{0.5mm} d_{i}}\
frac{1}{f} &= frac{d_{o}+d_{i}}{ d_{o} hspace{0.5mm} d_{i}}\
f &= frac{ d_{o} hspace{0.5mm} d_{i}}{ d_{o}+d_{i}} \
f &= frac{ 32 hspace{0.5mm} mathrm{cm} hspace{0.5mm}(-24 hspace{0.5mm} mathrm{cm})}{ 32 hspace{0.5mm} mathrm{cm}-24 hspace{0.5mm} mathrm{cm}} \
f&= -96 hspace{0.5mm} mathrm{cm}
end{align*}
The focal length is [ framebox[1.1width]{$ therefore f = -96 hspace{0.5mm} mathrm{cm} $}]
Result
2 of 2
$$
f = -96 hspace{0.5mm} mathrm{cm}
$$
f = -96 hspace{0.5mm} mathrm{cm}
$$
Exercise 20
Solution 1
Solution 2
Step 1
1 of 7
We need to determine the position and diameter of the image if the object has diameter of $h_o=7.6,,rm{cm}$ and is located at a distance of $d_o=22,,rm{cm}$ away from the convex mirror of a radius of $60,,rm{cm}$
Step 2
2 of 7
First, we need to determine a focal length:
$$f=frac{r}{2}$$
$$f=frac{60}{2}$$
$$f=30,,rm{cm}$$
$$f=frac{r}{2}$$
$$f=frac{60}{2}$$
$$f=30,,rm{cm}$$
Step 3
3 of 7
Now we can calculate the distance of an image using the next equation:
$$frac{1}{d_o}+frac{1}{d_i}=frac{1}{f}$$
$$frac{1}{d_o}+frac{1}{d_i}=frac{1}{f}$$
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:
$$d_i=frac{fd_o}{d_o-f}$$
Step 4
4 of 7
Inserting values we get:
$$d_i=-frac{30cdot 22}{30+22}$$
$$boxed{d_i=12.7,,rm{cm}}$$
$$d_i=-frac{30cdot 22}{30+22}$$
$$boxed{d_i=12.7,,rm{cm}}$$
Step 5
5 of 7
Now we can calculate its diameter:
$$frac{h_i}{h_o}=-frac{d_i}{d_o}$$
$$h_i=h_ofrac{d_i}{d_o}$$
$$frac{h_i}{h_o}=-frac{d_i}{d_o}$$
$$h_i=h_ofrac{d_i}{d_o}$$
Step 6
6 of 7
Inserting values:
$$h_i=7.6cdot frac{12.7}{22}$$
$$boxed{h_i=4.4,,rm{cm}}$$
$$h_i=7.6cdot frac{12.7}{22}$$
$$boxed{h_i=4.4,,rm{cm}}$$
Result
7 of 7
$$d_i=12.7,,rm{cm}$$
$$h_i=4.4,,rm{cm}$$
$$h_i=4.4,,rm{cm}$$
Step 1
1 of 3
$$
text{color{#4257b2}Recall that : Focal length = $dfrac{text{Radius of Curvature}}{2} color{Black}= dfrac{60}{2}$}
$$
text{color{#4257b2}Recall that : Focal length = $dfrac{text{Radius of Curvature}}{2} color{Black}= dfrac{60}{2}$}
$$
Therefore $f = -30$
Note that $f$ is negative because the mirror is convex
$$
text{color{#c34632}Recall that : $$dfrac{1}{f} = dfrac{1}{d_o}+dfrac{1}{d_i}$$}
$$
Substitute $f = -30$ and $d_o = 22$
$$
dfrac{1}{-30} = dfrac{1}{22}+dfrac{1}{d_i}
$$
$$
-dfrac{1}{30} – dfrac{1}{22}=dfrac{1}{d_i}
$$
$$
-dfrac{1}{30}times dfrac{11}{11} – dfrac{1}{22}times dfrac{15}{15}=dfrac{1}{d_i}
$$
$$
-dfrac{11}{330} – dfrac{15}{330}=dfrac{1}{d_i}
$$
$$
-dfrac{26}{330}=dfrac{1}{d_i}
$$
Take Reciprocal of both sides
$$
-dfrac{330}{26}=d_i
$$
$$
d_i = -dfrac{330}{26}approx -12.69
$$
The image is formed 12.69 cm behind the mirror
Step 2
2 of 3
$$
text{color{#4257b2}Recall that : $$m = -dfrac{q}{p}= dfrac{text{Image Height}}{text{Object Height}}$$}
$$
text{color{#4257b2}Recall that : $$m = -dfrac{q}{p}= dfrac{text{Image Height}}{text{Object Height}}$$}
$$
Substitute $q = -12.69$ and $p = 22$ and Object Height = 7.6
$$
-dfrac{-12.69}{22}= dfrac{text{Image Height}}{7.6}
$$
Multiply both sides by 7.6
$$
dfrac{12.69times7.6}{22}= text{Image Height}
$$
$$
text{Image Height} = dfrac{12.69times7.6}{22}approx 4.38
$$
Note that diameter is same as the height
Result
3 of 3
$$
text{color{#4257b2}The image is formed 12.69 cm behind the mirror
text{color{#4257b2}The image is formed 12.69 cm behind the mirror
$ $
Diameter of the image is 4.38 cm }
$$
Exercise 21
Solution 1
Solution 2
Step 1
1 of 3
$$
text{color{#c34632}Recall that : $$text{Magnification} = dfrac{text{Image Height}}{text{Object Height}} = -dfrac{d_i}{d_o}$$}
$$
text{color{#c34632}Recall that : $$text{Magnification} = dfrac{text{Image Height}}{text{Object Height}} = -dfrac{d_i}{d_o}$$}
$$
Substitute : Object Height = 1.8 m and Image Height = 0.36 m
And $d_o = 2.4$
$$
dfrac{0.36}{1.8} = -dfrac{d_i}{2.4}
$$
$$
– dfrac{0.36times 2.4}{1.8} = d_i
$$
$$
d_i= – dfrac{0.36times 2.4}{1.8} approx -0.48
$$
Step 2
2 of 3
$$
text{color{#4257b2}Recall that : $$dfrac{1}{f} = dfrac{1}{d_o}+dfrac{1}{d_i}$$}
$$
text{color{#4257b2}Recall that : $$dfrac{1}{f} = dfrac{1}{d_o}+dfrac{1}{d_i}$$}
$$
Substitute $d_o = 2.4$ and $d_i = -0.48$
$$
dfrac{1}{f} = dfrac{1}{2.4}+dfrac{1}{-0.48}
$$
$$
dfrac{1}{f} = dfrac{1}{2.4}-dfrac{1}{0.48}times dfrac{5}{5}
$$
$$
dfrac{1}{f} = dfrac{1}{2.4}-dfrac{5}{2.4}
$$
$$
dfrac{1}{f} = -dfrac{4}{2.4}
$$
Take Reciprocal of both sides
$$
f = -dfrac{2.4}{4} = -0.6
$$
Result
3 of 3
$$
text{color{#4257b2}$f = -0.6$ m
text{color{#4257b2}$f = -0.6$ m
$ $
The negative sign means that the mirror is convex }
$$
Step 1
1 of 6
We need to determine the focal length of a mirror if an object that is standing at a distance of $d_o=2.4,,rm{m}$ with a height of $h_o=1.8,,rm{m}$ creates an image of a height of $h_i=0.36,,rm{m}$
Step 2
2 of 6
In order to calculate the distance of an image we can use the next equation:
$$frac{h_i}{h_o}=-frac{d_i}{d_o}$$
$$d_i=-d_ofrac{h_i}{h_o}$$
$$frac{h_i}{h_o}=-frac{d_i}{d_o}$$
$$d_i=-d_ofrac{h_i}{h_o}$$
Step 3
3 of 6
Inserting values we get.
$$d_i=-2.4cdot frac{0.36}{1.8}$$
$$d_i=-0.48,,rm{m}$$
$$d_i=-2.4cdot frac{0.36}{1.8}$$
$$d_i=-0.48,,rm{m}$$
Step 4
4 of 6
Now we can calculate the focal length using the equation for mirrors:
$$frac{1}{d_o}+frac{1}{d_i}=frac{1}{f}$$
$$frac{1}{f}=frac{d_i+d_o}{d_id_o}$$
From that we can get:
$$f=frac{d_od_i}{d_o+d_i}$$
$$frac{1}{d_o}+frac{1}{d_i}=frac{1}{f}$$
$$frac{1}{f}=frac{d_i+d_o}{d_id_o}$$
From that we can get:
$$f=frac{d_od_i}{d_o+d_i}$$
Step 5
5 of 6
Inserting values:
$$f=frac{2.4cdot (-0.48)}{2.4-0.48}$$
$$boxed{f=-0.6,,rm{m}}$$
$$f=frac{2.4cdot (-0.48)}{2.4-0.48}$$
$$boxed{f=-0.6,,rm{m}}$$
Result
6 of 6
$$f=-0.6,,rm{m}$$
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