On what:
f (is the focal length of the lens) = - 7.50 cm
p (is the distance from the object to the lens) = ?
p' (is the distance from the image to the spherical lens) = 3.70 cm
Using the Gaussian equation, to know where the object is situated (distance from the point).
\(\frac{1}{f} = \frac{1}{p} + \frac{1}{p'}\)
\(\frac{1}{-7.50} = \frac{1}{p} + \frac{1}{3.70}\)
\(\frac{1}{p} = - \frac{1}{7.50} - \frac{1}{3.70}\)
\(\frac{1}{p} = \frac{-3.70-7.50}{27.75}\)
\(\frac{1}{p} = \frac{- 11.20}{27.75}\)
Product of extremes equals product of means:
\(- 11.20*p = 1*27.75\)
\(- 11.20p = 27.75\:simplify\:by*(-1)\)
\(11.20p = - 27.75\)
\(p = \frac{-27.75}{11.20}\)
\(\boxed{\boxed{p \approx -2.47\:cm}}\end{array}}\qquad\quad\checkmark\)
