1) 12.6 cm
2) -6.6 cm
Explanation:
We can solve this problem by using two equations:
1) Lens equation:
[tex]\frac{1}{f}=\frac{1}{p}+\frac{1}{q}[/tex]
where
f is the focal length of the lens
p is the distance of the object from the lens
q is the distance of the image from the lens
2) Magnification equation:
[tex]\frac{y'}{y}=-\frac{q}{p}[/tex]
where
y' is the size of the image
y is the size of the object
In this problem we have:
y = 7.60 cm (size of the turtle)
y'= 4.00 cm (size of the image
So from eq.(2) we get
[tex]q=-p\frac{y'}{y}=-p\frac{4.00}{7.60}=-0.526p[/tex]
Substituting into eq(1), we can find the value of p:
[tex]\frac{1}{f}=\frac{1}{p}-\frac{1}{0.526p}[/tex]
Here we have
f = -14.0 cm is the focal length (negative for a diverging lens)
So we find
[tex]-\frac{1}{14.0}=\frac{1}{p}-\frac{1.901}{p}=\frac{0.901}{p}\\p=\frac{-0.901\cdot 14.0}{1}=12.6 cm[/tex]
2)
Now we want to find how far is the turtle's image from the lens: so, we want to find the value of q.
We can do it by using the magnification equation:
[tex]\frac{y'}{y}=-\frac{q}{p}[/tex]
where here we have:
y' = 4.00 cm is the size of the image
y = 7.60 cm is the size of the object
p = 12.6 cm is the distance of the object from the lens
q is the distance of the image from the lens
Solving for q we find:
[tex]q=-p\frac{y'}{y}=-(12.6)\frac{4.00}{7.60}=-6.6 cm[/tex]
And the negative sign means the image is virtual.
