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).
[tex] \frac{1}{f} = \frac{1}{p} + \frac{1}{p'} [/tex]
[tex]\frac{1}{f} = \frac{1}{15.8} + \frac{1}{4.2} [/tex]
[tex]\frac{1}{f} = \frac{2.1}{33.18} + \frac{7.9}{33.18} [/tex]
[tex] \frac{1}{f} = \frac{10}{33.18} [/tex]
Product of extremes equals product of means:
[tex]10*f = 1*33.18[/tex]
[tex]10f = 33.18[/tex]
[tex]f = \frac{33.18}{10} [/tex]
[tex]\boxed{\boxed{f = 3.318\:cm}}\end{array}}\qquad\quad\checkmark[/tex]
