$\text{52.0 D}$

(a) $-0\text{.}\text{233 mm}$

(b) The size of the rods and the cones is smaller than the image height, so we can distinguish letters on a page.

(a) $\mathrm{+62.5\; D}$

(b) $\mathrm{–0.250\; mm}$

(c) $\mathrm{–0.0800\; mm}$

2.00 m

(a) $\pm 0\text{.}\text{45 D}$

(b) The person was nearsighted because the patient was myopic and the power was reduced.

0.143 m

1.00 m

20.0 cm

$\mathrm{–5.00\; D}$

25.0 cm

$\mathrm{–0.198\; D}$

30.8 cm

$\mathrm{–0.444\; D}$

(a) 4.00

(b) 1600

(a) 0.501 cm

(b) Eyepiece should be 204 cm behind the objective lens.

(c) No, a distance of 205 cm (which is over 2 m) is unrealistic for the tube of a microscope.

(a) +18.3 cm (on the eyepiece side of the objective lens)

(b) -60.0

(c) -11.3 cm (on the objective side of the eyepiece)

(d) +6.67

(e) -400

$-\text{40}\text{.}0$

$-167$

$+\text{10.0 cm}$

(a) $0\text{.}\text{251}\text{μm}$

(b) Yes, this thickness implies that the shape of the cornea can be very finely controlled, producing normal distant vision in more than 90% of patients.

(a) $\begin{array}{l}\frac{1}{d_i}+\frac{1}{d_0}=\frac{1}{f}\frac{1}{d_i}=\frac{1}{f}-\frac{1}{d_0}=\frac{1}{4.00\text{ mm}}-\frac{1}{6.00\text{ mm}}\\ d_i=12.0\text{ mm}\\ m_0=-\frac{d_i}{d_0}=-\frac{12.0}{6.00}=-2.00\\ \frac{1}{{d^{\prime }}_i}+\frac{1}{{d^{\prime }}_0}=\frac{1}{f}\frac{1}{{d^{\prime }}_i}=\frac{1}{f}-\frac{1}{{d^{\prime }}_0}=\frac{1}{50\text{ mm}}-\frac{1}{13.0\text{ mm}}\\ {d^{\prime }}_i=-17.6\text{ mm}\end{array}$
$$\begin{array}{l}{m}_e=-\frac{-17.6}{13}=1.35\\ m=-2(1.35)=-2.70\\ \frac{1}{d_i}+\frac{1}{d_0}=\frac{1}{f}\frac{1}{d_i}=\frac{1}{f}-\frac{1}{d_0}=\frac{1}{4.00\text{ mm}}-\frac{1}{5.00\text{ mm}}\\ d_i=20.0\text{ mm}\end{array}$$
$$\begin{array}{l}{m}_0=-\frac{d_i}{d_0}=-\frac{20.0}{5.00}=-4.00\\ \frac{1}{{d^{\prime }}_i}+\frac{1}{{d^{\prime }}_0}=\frac{1}{f}\frac{1}{{d^{\prime }}_i}=\frac{1}{f}-\frac{1}{{d^{\prime }}_0}=\frac{1}{50\text{ mm}}-\frac{1}{5.00\text{ mm}}\\ {d^{\prime }}_i=-5.56\text{ mm}\end{array}$$
$\begin{array}{l}{m}_e=-\frac{-5.56}{5}=1.11\\ m=-4(1.11)=-4.44\end{array}$

(b) It increased.

(c) The magnification should decrease.

(d) In both cases the image is inverted.

(a)

(c)

(a)

(b)

(d)

(c)