A) 11.1 cm
We can find the focal length of the lens by using the lens equation:
![\frac{1}{f}=\frac{1}{p}+\frac{1}{q}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bf%7D%3D%5Cfrac%7B1%7D%7Bp%7D%2B%5Cfrac%7B1%7D%7Bq%7D)
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:
![\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](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bf%7D%3D%5Cfrac%7B1%7D%7B16.0%20cm%7D%2B%5Cfrac%7B1%7D%7B36.0%20cm%7D%3D0.09%20cm%5E%7B-1%7D%5C%5Cf%3D%5Cfrac%7B1%7D%7B0.09%20cm%5E%7B-1%7D%7D%3D11.1%20cm)
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:
![\frac{h_i}{h_o}=-\frac{q}{p}](https://tex.z-dn.net/?f=%5Cfrac%7Bh_i%7D%7Bh_o%7D%3D-%5Cfrac%7Bq%7D%7Bp%7D)
where
is the heigth of the image
is the height of the object
![q=36.0 cm](https://tex.z-dn.net/?f=q%3D36.0%20cm)
![p=16.0 cm](https://tex.z-dn.net/?f=p%3D16.0%20cm)
Solving the formula for
, we find
![h_i = -h_o \frac{q}{p}=-(8.00 mm)\frac{36.0 cm}{16.0 cm}=-18.0 mm](https://tex.z-dn.net/?f=h_i%20%3D%20-h_o%20%5Cfrac%7Bq%7D%7Bp%7D%3D-%288.00%20mm%29%5Cfrac%7B36.0%20cm%7D%7B16.0%20cm%7D%3D-18.0%20mm)
So the image is 18 mm high.
D) Inverted
From the magnification equation we have that:
- When the sign of
is positive, the image is erect
- When the sign of
is negative, the image is inverted
In this case,
is negative, so the image is inverted.