On what:
f (is the focal length of the lens) = ?
p (is the distance from the object to the lens) =15.8 cm
p' (is the distance from the image to the spherical lens) = 4.2 cm
Using the Gaussian equation, to know where the object is situated (distance from the point).
\(\frac{1}{f} = \frac{1}{p} + \frac{1}{p'}\)
\(\frac{1}{f} = \frac{1}{15.8} + \frac{1}{4.2}\)
\(\frac{1}{f} = \frac{2.1}{33.18} + \frac{7.9}{33.18}\)
\(\frac{1}{f} = \frac{10}{33.18}\)
Product of extremes equals product of means:
\(10*f = 1*33.18\)
\(10f = 33.18\)
\(f = \frac{33.18}{10}\)
\(\boxed{\boxed{f = 3.318\:cm}}\end{array}}\qquad\quad\checkmark\)
