(a) -15.2 cm
We can solve the problem by using the lens equation:
[tex]\frac{1}{f}=\frac{1}{p}+\frac{1}{q}[/tex]
where
f = -22.8 cm is the focal length of the lens (negative because it is a diverging lens)
p = 45.6 cm is the distance of the object from the lens
q is the distance of the image from the lens
Solving the equation for q, we find the position of the image:
[tex]\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{-22.8 cm}-\frac{1}{45.6 cm}=-0.066 cm^{-1}[/tex]
[tex]q=\frac{1}{-0.066 cm^{-1}}=-15.2 cm[/tex]
and the negative sign means that the image is virtual.
(b) -11.4 cm
In this case, the distance of the object from the lens is
p = 22.8 cm
Substituting into the lens equation, we can find the new image distance, q:
[tex]\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{-22.8 cm}-\frac{1}{22.8 cm}=-\frac{2}{22.8 cm}[/tex]
[tex]q=\frac{-22.8 cm}{2}=-11.4 cm[/tex]
and the negative sign means that the image is virtual.
(c) -7.6 cm
In this case, the distance of the object from the lens is
p = 11.4 cm
Substituting into the lens equation, we can find the new image distance, q:
[tex]\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{-22.8 cm}-\frac{1}{11.4 cm}=-0.132 cm^{-1}[/tex]
[tex]q=\frac{1}{0.132 cm^{-1}}=-7.6 cm[/tex]
and again, the negative sign means that the image is virtual.
