College Physics 2e

# Chapter 26

### Problems & Exercises

1.

$52.0 D 52.0 D$

3.

(a) $− 0 . 233 mm − 0 . 233 mm$

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

5.

(a) $+62.5 D+62.5 D$

(b) $–0.250 mm–0.250 mm$

(c) $–0.0800 mm–0.0800 mm$

6.

2.00 m

8.

(a) $±0.45 D±0.45 D$

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

10.

0.143 m

12.

1.00 m

14.

20.0 cm

16.

$–5.00 D –5.00 D$

18.

25.0 cm

20.

$–0.198 D –0.198 D$

22.

30.8 cm

24.

$–0.444 D –0.444 D$

26.

(a) 4.00

(b) 1600

28.

(a) 0.501 cm

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

30.

(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

33.

$− 40 . 0 − 40 . 0$

35.

$− 1 . 67 − 1 . 67$

37.

$+ 10.0 cm + 10.0 cm$

39.

(a) $0.251 μm0.251 μm$

(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.

40.

(a)

$m e =− −5.56 5 =1.11 m=−4(1.11)=−4.44 m e =− −5.56 5 =1.11 m=−4(1.11)=−4.44$

(b) It increased.

(c) The magnification should decrease.

(d) In both cases the image is inverted.

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