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).
[tex] \frac{1}{f} = \frac{1}{p} + \frac{1}{p'} [/tex]
[tex]\frac{1}{-7.50} = \frac{1}{p} + \frac{1}{3.70} [/tex]
[tex]\frac{1}{p} = - \frac{1}{7.50} - \frac{1}{3.70}[/tex]
[tex] \frac{1}{p} = \frac{-3.70-7.50}{27.75} [/tex]
[tex]\frac{1}{p} = \frac{- 11.20}{27.75}[/tex]
Product of extremes equals product of means:
[tex]- 11.20*p = 1*27.75[/tex]
[tex]- 11.20p = 27.75\:simplify\:by*(-1)[/tex]
[tex]11.20p = - 27.75[/tex]
[tex]p = \frac{-27.75}{11.20} [/tex]
[tex]\boxed{\boxed{p \approx -2.47\:cm}}\end{array}}\qquad\quad\checkmark[/tex]
