Farsightedness can be caused by an eyeball that is shorter than normal. Rays from an object inside the near point are brought to a focus behind the retina. Thus a farsighted eye does not converge light enough to focus it on the retina. And this problem can be corrected by “preconverging” the light which is using a converging lens on the eye and a $textbf{convex lens}$ is an example of a converging lens.

$textbf{Given values:}$

$$
begin{align*}
d_{o,1} &= 15 text{ m} \
d_{o,2} &= 5.32 text{ m} \
f &= 45.5 text{ mm}
end{align*}
$$

The distance of the lens that should be moved in order to focus on an object 5.32 meters away is the difference in their image distance from the two object distances. So applying the thin-lens equation to the problem to calculate the image distances :

$$
begin{align*}
dfrac{1}{d_{o,1}} + dfrac{1}{d_{i,1}} &= dfrac{1}{f} \
d_{i,1} &= dfrac{1}{dfrac{1}{f} – dfrac{1}{d_{o,1}}} \
&= dfrac{1}{dfrac{1}{0.0455 text{ m}} – dfrac{1}{15 text{ m}}} \
d_{i,1} &= 45.638 text{ mm}
end{align*}
$$

$$
begin{align*}
\
dfrac{1}{d_{o,2}} + dfrac{1}{d_{i,2}} &= dfrac{1}{f} \
d_{i,2} &= dfrac{1}{dfrac{1}{f} – dfrac{1}{d_{o,2}}} \
&= dfrac{1}{dfrac{1}{0.0455 text{ m}} – dfrac{1}{5.32 text{ m}}} \
d_{i,2} &= 45.893 text{ mm}
end{align*}
$$

Therefore, calculating for the distance of the lens :

$$
{d = d_{i,2} – d_{i,1}}
$$

$$
{d = 45.893 text{ mm} – 45.638 text{ mm}}
$$

$$
{boxed{d = 0.255 text{ mm}}}
$$

$$
{d = 0.255 text{ mm}}
$$

$textbf{Given values:}$

$$
begin{align*}
d_o &= 0.750 text{ m} \
f &= 105 text{ mm}
end{align*}
$$

The distance from a camera lens to the CCD sensor is equal to the image distance from the lens. This can be calculated by applying the thin-lens equation to the problem :

$$
begin{align*}
dfrac{1}{d_o} + dfrac{1}{d_i} &= dfrac{1}{f} \
d_i &= dfrac{1}{dfrac{1}{f} – dfrac{1}{d_o}} \
&= dfrac{1}{dfrac{1}{105 text{ mm}} – dfrac{1}{750 text{ mm}}}
end{align*}
$$

$$
{boxed{d_i = 122.1 text{ mm}}}
$$