Physics
Physics
1st Edition
Walker
ISBN: 9780133256925
Textbook solutions

All Solutions

Page 582: Practice Problems

Exercise 19
Step 1
1 of 2
$tt{According to the mirror equation $frac{1}{d_o}+frac{1}{d_i}=frac{1}{f}$}$

$$
frac{1}{d_i}=frac{1}{f}-frac{1}{d_o}Rightarrow frac{1}{d_i}=frac{d_0-f}{fd_0}
$$

$$
d_i=frac{fd_0}{d_0-f}
$$

the mirror is concave therefore $f>0$

(a)
$$
d_i=frac{fd_0}{d_0-f}=frac{5*9}{9-5}=11.25cm
$$

(a)
$$
d_i=frac{fd_0}{d_0-f}=frac{5*2}{2-5}=-3.33cm
$$

Result
2 of 2
$$
tt{(a)11.25cm,(b)-3.33cm}
$$
Exercise 20
Step 1
1 of 2
$tt{The mirror equation states: $frac{1}{f}=frac{1}{d_i}+frac{1}{d_o}$}$

The image is behind the mirror (Virtual image) $Rightarrow d_i<0$

$$
frac{1}{f}=frac{d_i+d_o}{d_od_i} Rightarrow f=frac{d_od_i}{d_i+d_o}=frac{1.8*(-4.1)}{-4.1+1.8}=boxed{3.2cm}
$$

Result
2 of 2
$$
tt{$f=3.2cm$}
$$
Exercise 21
Step 1
1 of 2
$tt{The mirror equation states: $frac{1}{f}=frac{1}{d_i}+frac{1}{d_o}$}$

The mirror is concave $Rightarrow f>0$

The image is in front of the mirror (Real image) $Rightarrow d_i>0$

$$
frac{1}{d_o}=frac{1}{f}-frac{1}{d_i}=frac{d_i-f}{d_if}Rightarrow d_o=frac{d_if}{d_i-f}=frac{4.8*8.9}{8.9-4.8}=boxed{10.4cm}
$$

Result
2 of 2
$$
d_o=10.4cm
$$
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