ANSWERS TO CONCEPTUAL QUESTIONS - City …academic.brooklyn.cuny.edu/physics/liu/phys2/hmwk_a… · · 2010-03-27Optical Instruments 363 ANSWERS TO CONCEPTUAL QUESTIONS 2. The objective

Optical Instruments 363

ANSWERS TO CONCEPTUAL QUESTIONS

2. The objective lens of the microscope must form a real image just inside the focal point of the eyepiece lens. In order for this to occur, the object must be located just outside the focal point of the objective lens. Since the focal length of the objective lens is typically quite short ~1 cm( ) , this means that the microscope can focus properly only on objects close to the end of the barrel and will be unable to focus on objects across the room.

4. For a lens to operate as a simple magnifi er, the object should be located just inside the focal point of the lens. If the power of the lens is +20.0 diopters, its focal length is

f = ( ) = ( ) + = =1 00 1 00 20 0 0 050 0 5 00. . . . . m m m cmP

The object should be placed slightly less than 5.00 cm in front of the lens.

6. The aperture of a camera is a close approximation to the iris of the eye. The retina of the eye cor-responds to the fi lm of the camera, and a close approximation to the cornea of the eye is the lens of the camera.

8. You want a real image formed at the location of the paper. To form such an image, the object distance must be greater than the focal length of the lens.

10. Under low ambient light conditions, a photofl ash unit is used to insure that light entering the camera lens will deliver suffi cient energy for a proper exposure to each area of the fi lm. Thus, the most important criterion is the additional energy per unit area (product of intensity and the dura-tion of the fl ash, assuming this duration is less than the shutter speed) provided by the fl ash unit.

12. The angular magnifi cation produced by a simple magnifi er is m f= ( )25 cm . Note that this is proportional to the optical power of a lens, P = 1 f , where the focal length f is expressed in meters. Thus, if the power of the lens is doubled, the angular magnifi cation will also double.

PROBLEM SOLUTIONS

25.1 The f-number (or focal ratio) of a lens is defi ned to be the ratio of focal length of the lens to its diameter. Therefore, the f-number of the given lens is

ff

D-number

cm

4.0 cm= = =28

7 0.

25.2 If a camera has a lens with focal length of 55 mm and can operate at f -numbers that range from f 1 2. to f 22, the aperture diameters for the camera must range from

Df

fmin

max

.= ( ) = =-number

mm

22 mm

552 5

to

D

f

fmax

min

= ( ) = =-number

mm

1.2 mm

5546

56157_25_ch25_p361-381.indd 36356157_25_ch25_p361-381.indd 363 3/20/08 3:36:45 PM3/20/08 3:36:45 PM

364 Chapter 25

25.3 The thin lens equation, 1 1 1

p q f+ = , gives the image distance as

q

pf

p f=

−=

( )( )− ×

=−

100 52 0

100 52 0 1053

m mm

m m

.

.22 0. mm

From the magnitude of the lateral magnifi cation, M h h q p= ′ = − , where the height of the

image is ′ = =h 0 092 0 92 0. . m mm, the height of the object (the building) must be

h hp

q= ′ − = ( ) − =92 0

100

52 0177.

. mm

m

mm m

25.4 Consider rays coming from opposite edges of the object and passing undeviated through the center of the lens as shown at the right. For a very distant object, the image distance equals the focal length of the lens. If the angular width of the object is q, the full image width on the fi lm is

h f= ( )⎡⎣ ⎤⎦ = ( ) °⎛⎝⎜

⎞⎠⎟ =2 2 2 55 0

20

219tan . tanθ mm mmm

so the image easily fi ts within a 23.5 mm by 35.0 mm area.

25.5 The exposure time is being reduced by a factor of

t

t2

1

1 256 1

8= = s

1 32 s

Thus, to maintain correct exposure, the intensity of the light reaching the fi lm should be increased by a factor of 8. This is done by increasing the area of the aperture by a factor of 8, so in terms of

the diameter, π πD D22

124 8 4= ( ) or D D2 18= .

The new f-number will be

ff

D

f

D

f-number

-number( ) = = =( )

= =2

2 1

1

8 8

4 0

81 4

.. or f 1 4.



25.6 (a) The intensity is a measure of the rate at which energy is received by the fi lm per unit area of the image, or I A∞1 image. Consider an object with horizontal and vertical dimensions h hx yand as shown at the right. If the vertical dimension intercepts angle q, the vertical dimension of the image is ′ =h qy θ, or ′ ∞h qy . Similarly for the horizontal dimension,

′ ∞h qx , and the area of the image is A h h qx yimage = ′ ′ ∞ 2. Assuming a very distant object, q f≈ , so A fimage ∞ 2 and we conclude that I f∞1 2.

The intensity of the light reaching the fi lm is also proportional to the cross-sectional area of the lens and hence to the square of the diameter of that lens, or I D∞ 2. Combining this with our earlier conclusion gives

ID

f f D∞ =

( )2

2 2

1 or I

f∞

( )1

2-number

(b) The total light energy hitting the fi lm is proportional to the product of intensity and expo-sure time, It. Thus, to maintain correct exposure, this product must be kept constant, or I t I t2 2 1 1= , giving

tI

It

f

f2

1

21

2

2

1

2=⎛⎝⎜

⎞⎠⎟

=( )( )

⎡

⎣⎢

-number

-number⎢⎢

⎤

⎦⎥⎥

= ⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟ ≈t1

24 0

1 8

1

5001 100

.

. s s

25.7 Since the exposure time is unchanged, the intensity of the light reaching the fi lm must be doubled if the energy delivered is to be doubled. Using the result of Problem 6 (part a), we obtain

fI

If2

2 1

21

2 1

2-number -number( ) =

⎛⎝⎜

⎞⎠⎟

( ) = ⎛⎝⎜

⎞⎠⎟⎟ ( ) =11 61

2, or f2 61 7 8-number = = .

Thus, you should use the f 8 0. setting on the camera.

25.8 The image must always be focused on the fi lm, so the image distance is the distance between the lens and the fi lm. From the thin lens equation, 1 1 1p q f+ = , the object distance is p qf q f= −( ), and the range of object distances this camera can work with is from

pq f

q fminmax

max

=−

=( )( )

−210 175

210 175

mm mm

mm mm mm m= × =1 05 10 1 053. .

to

pq f

q fmaxmin

min

=−

=( )( )

−180 175

180 175

mm mm

mm mm mm m= × =6 30 10 6 303. .


366 Chapter 25

25.9 The corrective lens must form an upright, virtual image at the near point of the eye (i.e., q = − 60 0. cm in this case) for objects located 25.0 cm in front of the eye ( p = +25 0. cm). From the thin lens equation, 1 1 1p q f+ = , the required focal length of the corrective lens is

fpq

p q=

+=

( ) −( )−

=25 0 60 0

25 0 60 0

. .

. .

cm cm

cm cm++ 42 9. cm

and the power (in diopters) of this lens will be

P = =+

= +1 1

0 4292 33

fin meters m diopters

..

25.10 (a) The person is farsighted , able to see distant objects but unable to focus on objects at the normal near point for a human eye.

(b) With the corrective lens 2.00 cm in front of the eye, the object distance for an object 20.0 cm in front of the eye is p = − =20 0 2 00 18 0. . . cm cm cm .

(c) The upright, virtual image formed by the corrective lens will serve as the object for the eye, and this object must be 40.0 cm in front of the eye. With the lens 2.00 cm in front of the eye, the magnitude of the image distance for the lens will be q = − =40 0 2 00 38 0. . . cm cm cm .

(d) The image must be located in front of the corrective lens, so it is a virtual image and the image distance is negative . Thus, q = −38 0. cm.

(e) From the thin lens equation, 1 1 1p q f+ = , the required focal length of the corrective lens is

fpq

p q=

+=

( ) −( )−

=18 0 38 0

18 0 38 0

. .

. .

cm cm

cm cm++ 34 2. cm

(f) The power of the corrective lens is then

P = =+

= +1 1

0 3422 92


..

(g) With a contact lens, the lens to eye distance would be zero, so we would have

p = 20 0. cm , q = − 40 0. cm , giving a required focal length of

fpq

p q=

+=

( ) −( )−

=20 0 40 0

20 0 40 0

. .

. .

cm cm

cm cm++ 40 0. cm

and a power in diopters of

P = =+

= +1 1

0 4002 50


..

25.11 His lens must form an upright, virtual image of a very distant object (p ≈∞) at his far point, 80.0 cm in front of the eye. Therefore, the focal length is f q= = −80 0. cm.

If this lens is to form a virtual image at his near point (q = −18 0. cm), the object distance must be

pqf

q f=

−=

−( ) −( )− − −

18 0 80 0

18 0 80 0

. .

. .

cm cm

cm ccm cm( ) = 23 2.



25.12 (a) When the child clearly sees objects at her far point pmax =( )125 cm , the lens-cornea combination has assumed a focal length suitable for forming the image on the retina q =( )2 00. cm . The thin lens equation gives the optical power under these conditions as

Pfarin meters m m

= = + = + = +1 1 1 1

1 25

1

0 020 05

f p q . .00 8. diopters

When the eye is focused q =( )2 00. cm on objects at her near point pmin .=( )10 0 cm , the optical power of the lens-cornea combination is

Pnearin meters m m

= = + = + =1 1 1 1

0 100

1

0 020 0f p q . .++ 60 0. diopters

(b) If the child is to see very distant objects p → ∞( ) clearly, her eyeglass lens must form an

erect virtual image at the far point of her eye q = −( )125 cm . The optical power of the required lens is

P = = + = +−

= −1 1 10

1

1 250 800

f p qin meters m diopt

.. eers

Since the power, and hence the focal length, of this lens is negative, it is diverging .

25.13 (a) The lens should form an upright, virtual image at the far point q = −( )50 0. cm for very distant objects p ≈( )∞ . Therefore, f q= = − 50 0. cm, and the required power is

P = =−

= −1 1

0 5002 00

f ..

m diopters

(b) If this lens is to form an upright, virtual image at the near point of the unaided eye q = −( )13 0. cm , the object distance should be

pqf

q f=

−=

−( ) −( )− − −

13 0 50 0

13 0 50 0

. .

. .

cm cm

cm ccm cm( ) = 17 6.

25.14 (a) Yes, a single lens can correct the patient’ss vision. The patient needs corrective action in

both the near vision (to allow clear viewing of objects between 45.0 cm and the normal near point of 25 cm) and the distant vision (to allow clear viewing of objects more than 85.0 cm away). A single lens solution is for the patient to wear a bifocal or progressive lens. Alternately, the patient must purchase two pairs of glasses, one for reading, and one for distant vision.

(b) To correct the near vision, the lens must form an upright, virtual image at the patient’s near point (q = − 45 0. cm) when a real object is at the normal near point (p = +25 0. cm). The thin lens equation gives the needed focal length as

fpq

p q=

+=

( ) −( )−

=25 0 45 0

25 0 45 0

. .

. .

cm cm

cm cm++56 3. cm

so the required power in diopters is

P = = = +1 1

0 5631 78


..

continued on next page


368 Chapter 25

(c) To correct the distant vision, the lens must form an upright, virtual image at the patient’s far point (q = − 85 0. cm) for the most distant objects (p → ∞). The thin lens equation gives the needed focal length as f q= = − 85 0. cm, so the needed power is

P = =−

= −1 1

0 8501 18


..

25.15 Considering the image formed by the cornea as a virtual object for the implanted lens, we have p = − +( ) = −2 80 2 53 5 33. . . cm cm cm and q = + 2 80. cm. The thin lens equation then gives

the focal length of the implanted lens as

fpq

p q=

+=

−( )( )− +

5 33 2 80

5 33 2 80

. .

. .

cm cm

cm cm== + 5 90. cm

so the power is P = =+

= +1

0 059 017 0

f

1

m diopters

..

25.16 (a) The upper portion of the lens should form an upright, virtual image of very distant objects p ≈ ∞( ) at the far point of the eye q = −( )1 5. m . The thin lens equation then gives f q= = −1 5. m , so the needed power is

P = =−

= −1 1

1 50 67

f ..

m diopters

(b) The lower part of the lens should form an upright, virtual image at the near point of the eye q = −( )30 cm when the object distance is p = 25 cm. From the thin lens equation,

fpq

p q=

+=

( ) −( )−

= + ×25 30

25 301 5 102 cm cm

cm cm. cm m= +1 5.

Therefore, the power is P = =+

= +1 1

1 50 67

f ..

m diopters

25.17 The corrective lens should form an upright, virtual image at the woman’s far point (q = − 40 0. cm) for a very distant object (p → ∞ ). The thin lens equation gives the required focal length as

f q= = − = −40 0 0 400. . cm m. Since f < 0, it is a diverging lens , and the required power is

P = =−

= −1 1

0 4002 50


..

25.18 (a) f = =−

= − = −1 1

4 000 250 25 0

P .. .

diopters m cm

(b) The corrective lens form virtual images of very distant objects (p → ∞) at

q f= = −25 0. cm. Thus, the person must be very nearsighted , unable to see objects

clearly when they are over (25.0 cm cm+ =2 00 27 0. ) . in front of the eye.

(c) If contact lenses are to be worn, the far point of the eye will be 27.0 cm in front of the lens, so the needed focal length will be f q= = −27 0. cm, and the power is

P = =−

= −1 1

0 2703 70


..



25.19 (a) The simple magnifi er (a converging lens) is to form an upright, virtual image located 25 cm in front of the lens q = −( )25 cm . The thin lens equation then gives

pqf

q f=

−=

−( )( )− −

= +25 7 5

25 7 55 8

cm cm

cm cm

.

.. ccm

so the stamp should be placed 5 8. cm in front of the lens .

(b) When the image is at the near point of the eye, the angular magnifi cation produced by the simple magnifi er is

m mf

= = + = + =max ..1

251

25

7 54 3

cm cm

cm

25.20 (a) The maximum magnifi cation of a simple magnifi er is m fmax (= +1 25 cm) . Thus, if mmax .= +6 0, the focal length of the lens is

f =−

=−

=25

1

25

15 0

cm

m

cm

6.0 cm

max

.

(b) While using a simple magnifi er, the eye is most relaxed if the lens forms the virtual image at infi nity (so parallel rays emerge from the lens) rather than at the near point of the eye. Under these conditions, the magnifi cation produced is

mf

= = = +25 25

5 05 0

cm cm

cm..

25.21 (a) From the thin lens equation,

fpq

p q=

+=

( ) −( )−

=3 25 0

3 25 0

.50 cm cm

.50 cm cm

.

.++ 4 07. cm

(b) With the image at the normal near point, the angular magnifi cation is

m mf

= = + = + = +max

. ..1

25 01

25 07 14

cm cm

4.07 cm

25.22 (a) When the object is at the focal point of the magnifying lens, a virtual image is formed at infi nity and parallel rays emerge from the lens. Under these conditions, the eye is most relaxed and the magnifi cation produced is

mf

= = = +25 25

5 05 0

cm cm

cm..

(b) When the object is positioned so the magnifi er forms a virtual image at the near point of the eye (q = −25 cm), maximum magnifi cation is produced and this is

mfmax .

.= + = + = +125

125

5 06 0

cm cm

cm

(c) From the thin lens equation, the object distance needed to yield the maximum magnifi ca-tion computed in part (b) above is

pq f

q f=

−=

−( )( )− −

=25 5 0

25 5 04 2

cm cm

cm cm c

.

.. mm


370 Chapter 25

25.23 (a) From the thin lens equation, a real inverted image is formed at an image distance of

qpf

p f=

−=

( )( )−

= +7 3

7 39

1.0 cm 9.0 cm

1.0 cm .0 cm886 5. cm

so the lateral magnifi cation produced by the lens is

Mh

h

q

p= ′ = − = − = −86 5

1 22.

. cm

71.0 cm and the magnitude is M = 1 22.

(b) If h is the actual length of the leaf, the small angle approximation gives the angular width of the leaf when viewed by the unaided eye from a distance of d = + =126 71 0 197 cm cm cm. as

θ0 cm≈ =

h

d

h

197

The length of the image formed by the lens is ′ = =h M h h1 22. , and its angular width

when viewed from a distance of ′ = − =d q126 39 5 cm cm. is

θ ≈′′

=h

d

h1 22

3

.

9.5 cm

The angular magnifi cation achieved by viewing the image instead of viewing the leaf directly is

θθ0

9.5 cm

cm

cm

9.5 cm≈ =

( )1 22 3

197

1 22 197

3

. .h

h== 6 08.

25.24 (a) With the image at the normal near point q = −( )25 cm , the angular magnifi cation is

mf

= + = + = +125

125

2 0 cm cm

25 cm.

(b) When the eye is relaxed, parallel rays enter the eye and

mf

= = = +25 251 0

cm cm

25 cm.

25.25 The overall magnifi cation is m M m Mfee

= =⎛⎝⎜

⎞⎠⎟1 1

25 cm

where M1 is the lateral magnifi cation produced by the objective lens. Therefore, the required focal length for the eyepiece is

fM

me =( ) =

−( )( )−

=1 12

1402 1

25 cm 25 cm cm.

25.26 The approximate overall magnifi cation of a compound microscope is given by m L f fo e= −( )( . )25 0 cm , where L is the distance between the objective and eyepiece lenses,

while f fo e and are the focal lengths of the objective and eyepiece lenses, respectively. Thus, the described microscope should have an approximate overall magnifi cation of

mL

f fo e

= −⎛⎝⎜

⎞⎠⎟

= − ⎛⎝⎜

⎞⎠

25.0 cm cm

0.500 cm

20 0.⎟⎟

⎛⎝⎜

⎞⎠⎟ = −25 0

1 70588

.

.

cm

cm



25.27 The magnitude of the magnifi cation of a telescope is m f fo e= , where f fo e and are the focal lengths of the objective element and the eyepiece, respectively. Thus, if m = 45 and fe = 4 0. cm, the focal length of the objective must be f mfo e= = =( )( .45 4 0 180 cm) cm. The overall length of the telescope will therefore be

L f fo e= + = + = =180 4 0 184 1 84 cm cm cm m. .

25.28 It is specifi ed that the fi nal image the microscope forms of the blood cell is 29 0. cm in front of the eye and that the diameter of this image intercepts an angle of θ = 1 43. mrad. The diameter of this fi nal image must then be

h re = = ×( ) ×( ) = ×− − −θ 29 0 10 1 43 10 4 15 102 3. . . m rad 44 m

At this point, it is tempting to use Equation (25.7) from the textbook for the overall magnifi cation of a compound microscope, and compute h h me= as the size of the blood cell serving as the object for the microscope. However, the derivation of that equation is based on several assump-tions, one of which is that the eye is relaxed and viewing a fi nal image located an infi nite distance in front of the eyepiece. This is clearly not true in this case, and the use of Equation (25.7) would introduce considerable error. Instead, we shall return to basics and use the thin lens equation to fi nd the size of the original object.

The image formed by the objective lens is the object for the eyepiece, and we label the size of this image as ′h . The lateral magnifi cation of the objective lens is M h h q p1 1 1= ′ = − and that of the eyepiece is M h h q pe e e e= ′ = − . The overall magnifi cation produced by the microscope is

Mh

h

h

h

h

he e= = ′⎛

⎝⎜⎞⎠⎟ ′

⎛⎝⎜

⎞⎠⎟

which gives the size of the original object as h h Me= .

From the thin lens equation, the required object distance for the eyepiece is

pq f

q fee e

e e

=−

=−( )( )− −

29 0 0

29 0 0

.

.

cm .950 cm

cm ...

9500 920

cm cm=

and the magnifi cation produced by the eyepiece is

Mq

pee

e

= − = −−( )

= +29 0

0 92031 5

.

..

cm

cm

The image distance for the objective lens is then

q L pe1 29 0 0 920 28 1= − = − =. . . cm cm cm

and the object distance for this lens is

pq f

q fo

o1

1

1

28 1 1

2 1 6=

−=

( )( )−

.

.

cm .622 cm

8.1 cm 2221 72

cm cm= .

The magnifi cation by the objective lens is

Mq

p11

1

28 1

116 3= − = −

( ) = −.

. cm

.72 cm

and the overall lateral magnifi cation is M M M e= = −( ) +( ) = −1 16 3 31 5 513. . .

The size of the red blood cell serving as the original object is

hh

Me= = × = × =

−−4 15 10

8 09 10 0 8094

7.. .

m

513 m mμ


372 Chapter 25

25.29 Some of the approximations made in the textbook while deriving the overall magnifi cation of a compound microscope are not valid in this case. Therefore, we start with the eyepiece and work backwards to determine the overall magnifi cation.

If the eye is relaxed, the eyepiece image is at infi nity qe → −( )∞ , so the object distance is p fe e= = 2 50. cm, and the angular magnifi cation by the eyepiece is

mfee

= = =25 0 25 0

2 5010 0

. .

..

cm cm

cm

The image distance for the objective lens is then

q L pe1 15 0 2 50= − = − =. . cm cm 12.5 cm

and the object distance is pq f

q fo

o1

1

1

1 1

1 1 00=

−=

( )( )−

2 .5 cm .00 cm

2.5 cm . cm cm= 1 09.

The magnifi cation by the objective lens is Mq

p11

1

1

111 5= − = −

( )= −

2.5 cm

.09 cm. , and the overall magni-

fi cation of the microscope is

m M me= = −( )( ) = −1 11 5 10 0 115. .

25.30 (a) For a refracting telescope, the overall length is L f fo e= + and the magnifi cation produced is m f fo e= , where f fo e and are the focal lengths of the objective element and the eyepiece, respectively. Thus, we may write f f me o= to obtain

L ff

mf

mf

m

moo

o= + = +⎛⎝⎜

⎞⎠⎟ = +⎛

⎝⎜⎞⎠⎟0 1

1 1

(b) Using the result of part (a), the required change in the length of the telescope will be

ΔL fm

m

m

mo= ′ +′

− +⎛⎝⎜

⎞⎠⎟ = ( ) −1 1

2 00101

100

51 0

5.

. m

00 02 00 10 2 002

.. .⎛

⎝⎜⎞⎠⎟ = − × = −− cm cm

or the telescope must be shortened by moving the eyepiece 2.00 cm forward toward the objective lens.

25.31 The length of the telescope is L f fo e= + = 92 cm

and the angular magnifi cation is mf

fo

e

= = 45

Therefore, f fo e= 45 and L f f f f fo e e e e= + = + = =45 46 92 cm, giving

fe = 2 0. cm and f fo e= −92 cm or fo = 90 cm



25.32 The moon may be considered an infi nitely distant object p → ∞( ) when viewed with this lens, so the image distance will be q fo= = 1 500 cm.

Considering the rays that pass undeviated through the center of this lens as shown in the sketch, observe that the angular widths of the image and the object are equal. Thus, if w is the linear width of an object forming a 1.00 cm wide image, then

θ =×

= =w

fo3 8 10

1 0 1 0

1 5008.

. .

m

cm cm

cm

or w = ×( )⎛⎝⎜

⎞⎠⎟

⎛3 8 10

1 0 18..

m cm

1 500 cm

mi

1 609 m⎝⎝⎜⎞⎠⎟

= ×1 6 102. mi

25.33 (a) From the thin lens equation, qpf

p f=

−, so the lateral magnifi cation by the objective

lens is M h h q p f p f= ′ = − = − −( ). Therefore, the image size will be

′ = = −−

=−

h M hf h

p f

f h

f p

(b) If p f>> , then f p p− ≈ − and ′ ≈ −hf h

p

(c) Suppose the telescope observes the space station at the zenith.

Then, ′ ≈ − = −( )( )

×= − ×h

f h

p

4 00 108 61 07

. ..

m m

407 10 m13 00 m mm3− = −1 07.

25.34 Use the larger focal length (lowest power) lens as the objective element and the shorter focal length (largest power) lens for the eye piece. The focal lengths are

fo =+

= +1

1 200 833

..

diopters m, and fe =

+= +1

9 000 111

..

diopters m

(a) The angular magnifi cation (or magnifying power) of the telescope is then

mf

fo

e

= = ++

=0 833

0 1117 50

.

..

m

m

(b) The length of the telescope is

L f fo e= + = + =0 833 0 111 0 944. . . m m m


374 Chapter 25

25.35 The lens for the left eye forms an upright, virtual image at qL = − 50 0. cm when the object dis-tance is pL = 25 0. cm, so the thin lens equation gives its focal length as

fp q

p qLL L

L L

=+

=( ) −( )

−25 0 50 0

25 0 50

. .

. .

cm cm

cm 0050 0

cm cm= .

Similarly for the other lens, qR = −100 cm when pR = 25 0. cm, and fR = 33 3. cm.

(a) Using the lens for the left eye as the objective,

mf

f

f

fo

e

L

R

= = = =50 0

31 50

..

cm

3.3 cm

(b) Using the lens for the right eye as the eyepiece and, for maximum magnifi cation, requiring that the fi nal image be formed at the normal near point qe = −( )25 0. cm gives

pq f

q fee e

e e

=−

=−( )( )− −

25 0 3

25 0 33

.

.

cm 3.3 cm

cm ...

314 3

cm cm= +

The maximum magnifi cation by the eyepiece is then

mfee

= + = + = +125 0

125 0

33 31 75

. .

..

cm cm

cm

and the image distance for the objective is

q L pe1 10 0 4 3= − = − = −. . cm 14.3 cm cm

The thin lens equation then gives the object distance for the objective as

pq f

q f11 1

1 1

4 3 5

4 3 50 0=

−=

−( )( )− −

.

. .

cm 0.0 cm

cm cm cm= + 4 0.

The magnifi cation by the objective is then

Mq

p11

1

4 3

4 01 1= − = −

−( )= +

.

..

cm

cm

and the overall magnifi cation is m M me= = +( ) +( ) =1 1 1 1 75 1 9. . . .

25.36 Note: We solve part (b) before answering part (a) in this problem.

(b) The objective forms a real, diminished, inverted image of a very distant object at q fo1 = . This image is a virtual object for the eyepiece at p fe e= − , giving

1 1 1 1 1

0q p f f fe e e e e

= − =−

+ =

and qe → ∞




(a) Parallel rays emerge from the eyepiece, so the eye observes a virtual image .

(c) The angular magnifi cation is mf

fo

e

= = 3 00. , giving f fo e= 3 00. . Also, the length of the

telescope is L f f f fo e e e= + = − =3 00 10 0. . cm, giving

f fe e= − = − = −10 05 00

..

cm

2.00 cm and f fo e= =3 00 15 0. . cm

25.37 If just resolved, the angular separation is

θ θ λ= = = ×⎛⎝⎜

⎞⎠⎟

=−

min . ..

1 22 1 22500 10

0 3002

9

D

m

m..03 10 6× − rad

Thus, the altitude is hd= =

×= × =−θ

1 00

2 03 104 92 10 4926

5.

..

m

rad m km

25.38 For a narrow slit, Rayleigh’s criterion gives

θ λmin . .= = ×

×= × =

−

−−

a

500 10

0 101 00 10 1

9

33 m

.500 m000 mrad

25.39 The limit of resolution in air is θ λ μmin . .air

rad= =1 22 0 60D

In oil, the limiting angle of resolution will be

θ λ λ λmin . . .

oiloil oil= =

( )= ⎛

⎝⎜⎞⎠⎟1 22 1 22 1 22

D

n

D D

11

noil

or θθ μ μmin

min ..

oilair

oil

rad

1.5 rad= = =

n

0 600 40

25.40 (a) The wavelength of the light within the eye is λ λn n= . Thus, the limiting angle of resolu-tion for light passing through the pupil (a circular aperture with diameter D = 2 00. mm) is

θ λ λmin . . .

.= = =

×( )( )

−

1 22 1 22 1 22500 10

1 33

9

n

D nD

m

22 00 102 29 10

34

..

×( ) = ×−−

m rad

(b) From s r= θ, the distance from the eye that two points separated by a distance s = 1 00. cm will intercept this minimum angle of resolution is

r = =×

= × =sθ

min

..

1 004 36 10 433cm

2.29 10 radcm

4− ..6 m

25.41 The angular separation of the headlights when viewed from a distance of r = 10 0. km is

θ = =×

= × −s

r

2 00

102 00 103

4..

m

10.0 m rad

If the headlights are to be just resolved, this separation must equal the limiting angle of resolution for the circular aperture, θ λmin .= 1 22 D , so the diameter of the aperture is

D = = =×( )

×

−

−

1 22 1 22 1 22 885 10

2 00 10

9. . .

.min

λθ

λθ

m44

35 40 10 5 40 rad

m mm= × =−. .


376 Chapter 25

25.42 Diffraction occurs when waves pass through an aperture, causing the intensity to go through maxima and minima as one goes from the center of the beam outward as illustrated in the fi gure at the right. The angular separation of the fi rst minimum from the central maximum is a constant determined by the dimension of the aperture, the wavelength of the wave, and the shape of the aperture. For a circular aperture, this angular separation is given by θ λmin .= 1 22 D, where D is the diameter of the aperture. The full angular width of the central maximum is then α θ λ= =2 2 44min . D.

The lateral width of the central maximum, d, increases as the distance r from the aperture increases. When a beam of laser light having wavelength λ = 632 8. nm diffracts through a circular opening of diameter D = 0 200. cm, we estimate the diameter of the beam at distance r = 3 00. km past the opening as equal to the diameter of the central maximum in the diffraction pattern at this location. This gives

d r rD

= = ⎛⎝⎜

⎞⎠⎟ = ×( ) ×

α λ2 443 00 10

2 44 632 8 103.

.. .

m−−

−

( )×

⎡

⎣⎢⎢

⎤

⎦⎥⎥

=9

20 200 102 32

m

m m

..

25.43 If just resolved, the angular separation of the objects is θ θ λ= =min .1 22D

and s r= = ×( ) ×⎛⎝⎜

⎞⎠⎟

−

θ 8 0 10 1 22500 107

9

. . km m

5.00 m

⎡⎡

⎣⎢

⎤

⎦⎥ = 9 8. km

25.44 If just resolved, the angular separation of the objects is θ θ λ= =min .1 22D

and s r= = ×( ) ×⎛⎝⎜

⎞⎠⎟

⎡ −

θ 200 10 1 22550 103

9

m m

0.35 m.

⎣⎣⎢

⎤

⎦⎥ = =0 38 38. m cm

25.45 The grating spacing is d = = × = ×− −1

6 0001 67 10 1 67 104 6 cm

cm m. . , and the highest order of

600 nm light that can be observed is

md

max

m

m= =

×( )( )×

=−

−

sin ..

90 1 67 10 1

600 102

6

9

°

λ778 2 orders→

The total number of slits is N = ( )( ) = ×15 0 6 000 9 00 104. . cm slits cm , and the resolving power of the grating in the second order is

R Nmavailable = = ×( ) = ×9 00 10 2 1 80 104 5. .

The resolving power required to separate the given spectral lines is

Rneeded

nm

0.003 nm= = = ×λ

λΔ600 000

2 0 105..

These lines cannot be separated with this grating.



25.46 The resolving power of a diffraction grating is R Nm= =λλΔ

(a) The number of lines the grating must have to resolve the Hα line in the fi rst order is

NR

m= = ( ) = = ×λ λΔ

1

656 23 6 103..

nm

0.18 nm lines

(b) In the second order m =( )2 , NR= = ( ) = ×2

656 21 8 103.

. nm

2 0.18 nm lines

25.47 A fringe shift occurs when the mirror moves distance λ 4. Thus, if the mirror moves distance ΔL = 0 180. mm, the number of fringe shifts observed is

NL L

shifts

m= =

( ) =×( )

×

−

−

Δ Δλ λ4

4 4 0 180 10

550 10

3

9

.

m fringe shifts= ×1 31 103.

25.48 (a) When the central spot in the interferometer pattern goes through a full cycle from bright to dark and back to bright, two fringe shifts have occurred and the movable mirror has moved a distance of 2 4 2( )λ λ= . Thus, if Ncycles = 1 700 such cycles are observed as the mirror moves distance d = 0 382. mm, it must be true that

d N= ⎛⎝⎜

⎞⎠⎟cycles

λ2

or λ = 2d

Ncycles

and the wavelength of the light illuminating the interferometer is

λ =×( )

= × =−

−2 0 382 10

1 7004 49 10 449

37

..

m m nm

which is in the blue region of the visible spectrum.

(b) Red light has a longer wavelength than blue light, so fewer wavelengths would cover the

given displacement, hence Ncycleswould be smaller .

25.49 A fringe shift occurs when the mirror moves distance λ 4. Thus, the distance moved (length of the bacterium) as 310 shifts occur is

ΔL N= ⎛⎝⎜

⎞⎠⎟ = ×⎛

⎝⎜⎞⎠⎟

=−

shifts

mλ4

310650 10

45 0

9

. 44 10 50 45× =− m m. μ

25.50 A fringe shift occurs when the mirror moves distance λ 4. Thus, the distance the mirror moves as 250 fringe shifts are counted is

ΔL N= ⎛⎝⎜

⎞⎠⎟ = ×⎛

⎝⎜⎞⎠⎟

=−

shifts

mλ4

250632 8 10

43

9... .96 10 39 65× =− m mμ


378 Chapter 25

25.51 When the optical path length that light must travel as it goes down one arm of a Michelson’s interferometer changes by one wavelength, four fringe shifts will occur (one shift for every quarter-wavelength change in path length).

The number of wavelengths (in a vacuum) that fi t in a distance equal to a thickness t is N tvac = λ. The number of wavelengths that fi t in this thickness while traveling through the trans-parent material is N t t n ntn n= = ( ) =λ λ λ. Thus, the change in the number of wavelengths that fi t in the path down this arm of the interferometer is

ΔN N N nt

n= − = −( )vac 1λ

and the number of fringe shifts that will occur as the sheet is inserted will be

# ..

fringe shifts = ( ) = −( ) = −( )4 4 1 4 1 40 115ΔN n

t

λ00 10

406×

×⎛⎝⎜

⎞⎠⎟

=−

−

m

600 10 m9

25.52 A fringe shift will occur each time the effective length of the tube changes by a quarter of a wavelength (that is, for each additional wavelength fi tted into the length of the tube, 4 fringe shifts occur). If L is the length of the tube, the number of fringe shifts observed as the tube is fi lled with gas is

NL L L

n

L L

nshifts

gas

= −⎡

⎣⎢

⎤

⎦⎥ = −

⎡

⎣⎢⎢

⎤

⎦⎥⎥

=4 44

λ λ λ λ λλngas −( )1

Hence, nL

Ngas shifts

m

4 5.00 1= + ⎛

⎝⎜⎞⎠⎟ = + ×

×

−

14

1600 10 9λ

00 m1

2−( )⎡

⎣⎢⎢

⎤

⎦⎥⎥( ) =160 000 5.

25.53 (a) For a refracting telescope, the magnifi cation is m f fo e= , where f fo e and are the focal lengths of the objective lens and the eyepiece, respectively. Thus, when the Yerkes telescope uses an eyepiece with fe = 2 50. cm, the magnifi cation is

mf

fo

e

= =×

= × =−

20 0

108 00 10 8002

2..

m

2.50 m

(b) Standard astronomical telescopes form inverted images. Thus, the observer Martian polar

caps are upside down .

25.54 When viewed from a distance of 50 meters, the angular length of a mouse (assumed to have an actual length of ≈ 10 cm) is

θ = = = × −s

r

0 102 0 10 3.

. m

50 m radians

Thus, the limiting angle of resolution of the eye of the hawk must be

θ θmin .≤ = × −2 0 10 3 rad



25.55 The resolving power of the grating is R Nm= =λ λΔ . Thus, the total number of lines needed on the grating to resolve the wavelengths in order m is

NR

m m= = ( )

λλΔ

(a) For the sodium doublet in the fi rst order,

N = ( )( ) = ×589 30

1 0 590 103.

..

nm

nm1

(b) In the third order, we need N = ( )( ) = ×589 30

3 0 593 3 102.

..

nm

nm

25.56 (a) Since this eye can already focus on objects located at the near point of a normal eye (25 cm), no correction is needed for near objects. To correct the distant vision, a correc-tive lens (located 2.0 cm from the eye) should form virtual images of very distant objects at 23 cm in front of the lens (or at the far point of the eye). Thus, we must require that q = −23 cm when p → ∞. This gives

P = = + = +−

= −1 1 10

1

0 234 3

f p q ..

m diopters

(b) A corrective lens in contact with the cornea should form virtual images of very distant objects at the far point of the eye. Therefore, we require that q = −25 cm when p → ∞, giving

P = = + = +−

= −1 1 10

1

0 254 0

f p q ..

m diopters

When the contact lens f = = −⎛⎝⎜

⎞⎠⎟

125

P cm is in place, the object distance which yields a

virtual image at the near point of the eye (that is, q = −16 cm) is given by

pqf

q f=

−=

−( ) −( )− − −( ) =

16 25

16 2544

cm cm

cm cm ccm

25.57 (a) The lens should form an upright, virtual image at the near point of the eye q = −75 0. cm when the object distance is p = 25 0. cm. The thin lens equation then gives

fpq

p q=

+=

( ) −( )−

=25 0 75 0

25 0 5 0

. .

. .

cm cm

cm 7 cm337 5 0 375. . cm m=

so the needed power is P = = = +1 1

0 3752 67

f ..

m diopters .

(b) If the object distance must be p = 26 0. cm to position the image at q = −75 0. cm, the actual focal length is

fpq

p q=

+=

( ) −( )−

=26 0 75 0

26 0 5 0

. .

. .

cm cm

cm 7 cm00 398. m

and P = = = +1 1

0 3982 51

f ..

m diopters

The error in the power is

ΔP = −( ) =2 67 2 51 0 16. . . diopters diopters too loow


380 Chapter 25

25.58 (a) If q = 2 00. cm when p = =1 00 100. m cm, the thin lens equation gives the focal length as

fpq

p q=

+=

( )( )+

=100 2 00

100 2 001 9

cm cm

cm cm

.

.. 66 cm

(b) The f-number of a lens aperture is the focal length of the lens divided by the diameter of the aperture. Thus, the smallest f-number occurs with the largest diameter of the aperture. For the typical eyeball focused on objects 1.00 m away, this is

ff

D-number

cm

cm( ) = = =

minmax

.

..

1 96

0 6003 27

(c) The largest f-number of the typical eyeball focused on a 1.00-m-distance object is

ff

D-number

cm

cm( ) = = =

maxmin

.

..

1 96

0 2009 80

25.59 (a) The implanted lens should give an image distance of q = 22 4. mm for distant p →( )∞ objects. The thin lens equation then gives the focal length as f q= = 22 4. mm, so the power of the implanted lens should be

Pimplant

1

m diopters= =

×= +−

1

22 4 1044 63f .

.

(b) When the object distance is p = 33 0. cm, the corrective lens should produce parallel rays q →( )∞ . Then the implanted lens will focus the fi nal image on the retina. From the thin

lens equation, the required focal length is f p= = 33 0. cm, and the power of this lens should be

Pcorrective

1

m diopters= = = +1

0 3303 03

f ..

25.60 We use n

p

n

q

n n

R1 2 2 1+ = −

, with p → ∞ and q equal to the cornea to retina distance. Then,

R qn n

n= −⎛

⎝⎜⎞⎠⎟

= ( ) −⎛⎝⎜

⎞2 1

2

2 001 34 1 00

1 34.

. .

. cm ⎠⎠⎟ = =0 507 5 07. . cm mm

25.61 When a converging lens forms a real image of a very distant object, the image distance equals the focal length of the lens. Thus, if the scout started a fi re by focusing sunlight on kindling 5.00 cm from the lens, f q= = 5 00. cm.

(a) When the lens is used as a simple magnifi er, maximum magnifi cation is produced when the upright, virtual image is formed at the near point of the eye (q = −15 cm in this case). The object distance required to form an image at this location is

pqf

q f=

−=

−( )( )− −

=15 5 0

15 5 0

15 cm cm

cm cm

cm.

. 44.0

and the lateral magnifi cation produced is Mq

p= − = − − = +15

4 0 cm

15 cm 4.0.




(b) When the object is viewed directly while positioned at the near point of the eye, its angular size is θ0 15= h cm. When the object is viewed by the relaxed eye while using the lens as a simple magnifi er (with the object at the focal point so parallel rays enter the eye), the angu-lar size of the upright, virtual image is θ = h f . Thus, the angular magnifi cation gained by using the lens is

mh f

h f= = = = =θ

θ0 15

15 153 0

cm

cm cm

5.0 cm.

25.62 The angular magnifi cation is m = θ θo, where θ is the angle subtended by the fi nal image, and θo is the angle subtended by the object as shown in the fi gure. When the telescope is adjusted for minimum eyestrain, the rays entering the eye are parallel. Thus, the objective lens must form its image at the focal point of the eyepiece.

From triangle ABC, θ θo o≈ = ′tan h q1 and from triangle DEF, θ θ≈ = ′tan h fe . The angular

magnifi cation is then mh f

h q

q

fe

e

= = ′′

=θθo 1

1 .

From the thin lens equation, the image distance of the objective lens in this case is

qp f

p f11 1

1 1

300 20 0

300 20 0=

−=

( )( )−

cm cm

cm c

.

. mm cm= 21 4.

With an eyepiece of focal length fe = 2 00. cm, the angular magnifi cation for this telescope is

mq

fe

= = =1 21 410 7

..

cm

2.00 cm


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ANSWERS TO CONCEPTUAL QUESTIONS - City …academic.brooklyn.cuny.edu/physics/liu/phys2/hmwk_a… · · 2010-03-27Optical Instruments 363 ANSWERS TO CONCEPTUAL QUESTIONS 2. The objective