We can solve both parts of the problem by using the lens equation:
[tex] \frac{1}{f} = \frac{1}{d_o}+ \frac{1}{d_i} [/tex]
where
f is the focal length
[tex]d_o[/tex] is the distance of the object from the lens
[tex]d_i [/tex] is the distance of the image from the lens
Let's also keep in mind that for a converging lens, the focal length is taken as positive, so f=+19.0 cm.
(a) In this case, [tex]d_o = 41 cm[/tex]. So, the lens equation becomes
[tex] \frac{1}{d_i}= \frac{1}{f}- \frac{1}{d_o}= \frac{1}{19 cm}- \frac{1}{41 cm} =0.028 cm^{-1} [/tex]
And so the distance of the image from the lens is
[tex]d_i = \frac{1}{0.028 cm^{-1}} =+35.4 cm[/tex]
where the positive sign means the image is real, i.e. is located behind the lens (opposite side of the real object)
(b) In this case, [tex]d_o = 14 cm[/tex]. So, the lens equation becomes
[tex] \frac{1}{d_i}= \frac{1}{f}- \frac{1}{d_o}= \frac{1}{19 cm}- \frac{1}{14 cm} =-0.019 cm^{-1} [/tex]
And so the distance of the image from the lens is
[tex]d_i = \frac{1}{-0.019 cm^{-1}} =-53.2 cm[/tex]
where the negative sign means the image is virtual, i.e. is located in front of the lens (i.e. on the same side of the real object)
