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College Physics 2e

Chapter 26

College Physics 2eChapter 26

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.

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

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.

− 167 − 167

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) 1 d i + 1 d 0 = 1 f 1 d i = 1 f − 1 d 0 = 1 4.00 mm − 1 6.00 mm d i =12.0 mm m 0 =− d i d 0 =− 12.0 6.00 =−2.00 1 d ′ i + 1 d ′ 0 = 1 f 1 d ′ i = 1 f − 1 d ′ 0 = 1 50 mm − 1 13.0 mm d ′ i =−17.6 mm 1 d i + 1 d 0 = 1 f 1 d i = 1 f − 1 d 0 = 1 4.00 mm − 1 6.00 mm d i =12.0 mm m 0 =− d i d 0 =− 12.0 6.00 =−2.00 1 d ′ i + 1 d ′ 0 = 1 f 1 d ′ i = 1 f − 1 d ′ 0 = 1 50 mm − 1 13.0 mm d ′ i =−17.6 mm
m e =− −17.6 13 =1.35 m=−2(1.35)=−2.70 1 d i + 1 d 0 = 1 f 1 d i = 1 f − 1 d 0 = 1 4.00 mm − 1 5.00 mm d i =20.0 mm m e =− −17.6 13 =1.35 m=−2(1.35)=−2.70 1 d i + 1 d 0 = 1 f 1 d i = 1 f − 1 d 0 = 1 4.00 mm − 1 5.00 mm d i =20.0 mm
m 0 =− d i d 0 =− 20.0 5.00 =−4.00 1 d ′ i + 1 d ′ 0 = 1 f 1 d ′ i = 1 f − 1 d ′ 0 = 1 50 mm − 1 5.00 mm d ′ i =−5.56 mm m 0 =− d i d 0 =− 20.0 5.00 =−4.00 1 d ′ i + 1 d ′ 0 = 1 f 1 d ′ i = 1 f − 1 d ′ 0 = 1 50 mm − 1 5.00 mm d ′ i =−5.56 mm
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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