The Correct Option is B
Solution and Explanation
Using lens maker’s formula,
$\frac{1}{f}=\left(\mu-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)$
$\frac{1}{f_{1}}=\left(\frac{1.5}{1}-1\right)\left(\frac{1}{\infty}-\frac{1}{-20}\right)$
$\Rightarrow\,f_{1}=40$ cm
$\frac{1}{f_{2}}=\left(\frac{1.7}{1}-1\right)\left(\frac{1}{-20}-\frac{1}{+20}\right)$
$\Rightarrow\quad f_{2}=-\frac{100}{7} cm$
and $\frac{1}{f_{3}}=\left(\frac{1.5}{1}-1\right)\left(\frac{1}{\infty}-\frac{1}{-20}\right)$
$\Rightarrow \,f_{3}=40$ cm
$\frac{1}{f_{eq}}=\frac{1}{f_{1}}+\frac{1}{f_{2}}+\frac{1}{f_{3}}$
$\Rightarrow\, \frac{1}{f_{eq}}=\frac{1}{40}+\frac{1}{100\backslash7}+\frac{1}{40}$
$\therefore\quad f_{eq}=-50\,$ cm
Therefore, the focal length of the combination is - 50 cm.