A) 11.0 cm
The focal length of the lens can be find by using the lens equation:
[tex]\frac{1}{f}=\frac{1}{p}+\frac{1}{q}[/tex]
where
p = 16.0 cm is the distance of the object from the lens
q = 36.0 cm is the distance of the image from the lens (positive, because it is on the opposite side to the lens with respect to the object)
Substituting into the equation, we find
[tex]\frac{1}{f}=\frac{1}{16.0 cm}+\frac{1}{36.0 cm}=0.091 cm^{-1}[/tex]
[tex]f=\frac{1}{0.091 cm^{-1}}=11.0 cm[/tex]
B) Converging
The sign convention for the lenses states that:
- For a converging lens, the focal length is positive
- For a diverging lens, the focal length is negative
In this case, the focal length is positive (+11.0 cm), so the lenses must be a converging lens.
C) -18.0 mm
We can solve this part of the problem by using the magnification equation:
[tex]\frac{h_i}{h_o}=-\frac{q}{p}[/tex]
where
[tex]h_i[/tex] is the heigth of the image
[tex]h_o=8.00 mm[/tex] is the height of the object
[tex]q=36.0 cm[/tex] is the distance of the image from the lens
[tex]p=16.0 cm[/tex] is the distance of the object from the lens
Solving the equation for [tex]h_i[/tex], we find
[tex]h_i = -h_o \frac{q}{p}=-(8.00 mm)\frac{36.0 cm}{16.0 cm}=-18.0 mm[/tex]
D) Inverted
For the sign convention, we have:
- if [tex]h_i[/tex] is positive, then the image is erect
- if [tex]h_i[/tex] is negative, then the image is upside-down (inverted)
In this case, [tex]h_i[/tex] is negative, therefore the image is inverted.
