Jarak antara lensa cembung dan cekung 120 cm. Lensa cembung mempunyai panjang fokus 50 cm dan lensa .... Question from @Chromium

Asruul $\bold{Diketahui}\\
l ~~= 120~cm\\
f_{1} = 50~cm\\
f_{2} =-100~cm\\
s_1=100~cm\\
\bold{Ditanya}\\
a)~jarak~bayangan..?\\
b)~perbesaran~bayangan...?\\
\bold{Jawab}$
$$\begin{align}\Rightarrow &~Jarak~bayangan~lensa~cembung\\
 &~~\frac{1}{s_1'} = \frac{1}{f_1} ~-~ \frac{1}{s_1}\\
&~~ \frac{1}{s_1'} = \frac{1}{50}~-~ \frac{1}{100}\\
 &~~ \frac{1}{s_1'} = \frac{2-1}{100}\\
&~~ s_1' ~= \frac{ \frac{1}{100} }{1}\\
&~~s_1'~= 100~cm\\\
 \end{align}
$
$$\begin{align} \Rightarrow &~Jarak~bayangan~lensa~cekung\\
& ~~s_2 = 120-100=20\\
 &~~\frac{1}{s_2'} = \frac{1}{f_2} ~-~ \frac{1}{s_2}\\
&~~ \frac{1}{s_2'} =- \frac{1}{10}~-~ \frac{1}{20}\\
 &~~ \frac{1}{s_2'} = \frac{-1-5}{100}\\
&~~ \frac{1}{s_2'} = \frac{-6}{100} 
\\ &~~ s_2' ~= \frac{-100}{6}\\
&~~ s_2' ~=16,7~cm \end{align}$

$$\begin{align}\Rightarrow &~M~lensa~cembung\\
&~~M_1 = \frac{s_1'}{s_1} = \frac{-100}{100} = -1~kali\\
&~~M_2= \frac{s_2'}{s_2} = -\frac{-16,7}{20}=0,8~kali\\\\
&~~M_{total} = \frac{M_1}{M_2} = \frac{-1}{0,8} = -0,8~kali \end{align}$

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