A) 11.1 cm
We can find the focal length of the lens by using the lens equation:
[tex]\frac{1}{f}=\frac{1}{p}+\frac{1}{q}[/tex]
where
f is the focal length
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 (taken with positive sign since it is on the opposide side to the image, so it is a real image)
Solving the equation for f:
[tex]\frac{1}{f}=\frac{1}{16.0 cm}+\frac{1}{36.0 cm}=0.09 cm^{-1}\\f=\frac{1}{0.09 cm^{-1}}=11.1 cm[/tex]
B) Converging
The focal length is:
- Positive for a converging lens
- Negative for a diverging lens
In this case, the focal length is positive, so it is a converging lens.
C) 18.0 mm
The magnification equation states that:
[tex]\frac{h_i}{h_o}=-\frac{q}{p}[/tex]
where
[tex]h_i[/tex] is the heigth of the image
[tex]h_o[/tex] is the height of the object
[tex]q=36.0 cm[/tex]
[tex]p=16.0 cm[/tex]
Solving the formula 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]
So the image is 18 mm high.
D) Inverted
From the magnification equation we have that:
- When the sign of [tex]h_i[/tex] is positive, the image is erect
- When the sign of [tex]h_i[/tex] is negative, the image is inverted
In this case, [tex]h_i[/tex] is negative, so the image is inverted.
