Unformatted text preview: to the distance L between the
lenses minus the focal length fe of the eyepiece, or di1 ≈ L – fe. Thus, the object distance is
given by
1
1
1
1
1
=
−
≈
−
do1 f o di1 f o L − fe
The magnification due to the objective is given by Equation 26.7 as
mobjective = –di1/do1. Since both di1 and do1 are now known, the magnification can be
evaluated.
SOLUTION
a. According to Equation 26.11, the angular magnification of the compound microscope is 1410 THE REFRACTION OF LIGHT: LENSES AND OPTICAL INSTRUMENTS M ≈− ( L − fe ) N = − ( 26.0 cm − 6.50 cm ) ( 35.0 cm ) =
( 3.50 cm )( 6.50 cm ) fo fe −30.0 b. Using the thinlens equation, we can determine the object distance from the objective as
follows:
1
1
1
1
1
=
−
=
−
= 0.234 cm −1
do1
f o L − f e 3.50 cm 26.0 cm − 6.50 cm or d o1 = 4.27 cm .
c. The magnification m of the objective is given by Equation 26.7 as di1
26.0 cm − 6.50 cm
=−
= −4.57
do1
4.27 cm
______________________________________________________________________________
mobjective = − 96. REASONING The angular magnification M of an astronomical telescope is given by
f
M ≈ − o (Equation 26.12), where fo is the focal length of the objective lens, and fe is the
fe
focal length of the eyepiece. The length of the barrel must be adjusted so that the image
formed by the objective appears very close to the focal point of the eyepiece. The result is
that the length L of the barrel is approximately equal to the sum of the focal lengths of the
telescope’s two lenses: L ≈ fo + fe . Therefore, if the barrel can only be shortened by
0.50 cm, then the replacement eyepiece can have a focal length fe2 that is only 0.50 cm
shorter than the focal length fe1 of the current eyepiece.
SOLUTION The focal length of the replacement eyepiece is f e 2 = f e1 − 0.50 cm = 1.20 cm − 0.50 cm = 0.70 cm
Therefore, from Equation 26.12, the angular magnification with the second eyepiece in
place is
fo 180 cm
=−
= −260
fe
0.70 cm
______________________________________________________________________________
M ≈− Chapter 26 Problems 97. 1411 SSM REASONING Knowing the angles subtended at the unaided eye and with the
telescope will allow us to determine the angular magnification of the telescope. Then, since
the angular magnification is related to the focal lengths of the eyepiece and the objective,
we will use the known focal length of the eyepiece to determine the focal length of the
objective.
SOLUTION From Equation 26.12, we have M= f
θ′
=– o
θ
fe where θ is the angle subtended by the unaided eye and θ′ is the angle subtended when the
telescope is used. We note that θ′ is negative, since the telescope produces an inverted
image. Thus, using Equation 26.12, we find fo = – fe θ ′ = 1.1 m
8.0 × 10–5 rad
______________________________________________________________________________ θ =– ( 0.032 m ) ( –2.8 × 10–3 rad ) 98. REASONING The refractive power of the eyepiece (in diopters) is the reciprocal of its focal
length fe (in meters), according to the definition in Equation 26.8: ( Refractive power )e = 1
fe (1) To obtain the focal length of the eyepiece, we consider the angular magnification of the
telescope. According to Equation 26.12, the angular magnification M is
M ≈− fo (26.12) fe To use this expression to determine fe, however, we need a value for fo, which is the focal
length of the objective. Although a value for fo is not given directly, the value of the
refractive power of the objective is given, and it can be used in Equation 26.8 to obtain fo: ( Refractive power )o = 1
fo (2) SOLUTION Substituting fe and fo from Equations (1) and (2) into Equation 26.12, we find 1412 THE REFRACTION OF LIGHT: LENSES AND OPTICAL INSTRUMENTS M ≈− fo
fe =− 1/ ( Refractive power )o
1/ ( Refractive power )e =− ( Refractive power )e
( Refractive power )o Solving for the refractive power of the eyepiece gives ( Refractive power )e ≈ − M ( Refractive power )o = − ( −132 ) (1.50 diopters ) = 198 diopters
99. SSM REASONING AND SOLUTION The angular magnification of an astronomical
telescope, is given by Equation 26.12 as M ≈ – f o / f e . Solving for the focal length of the
eyepiece, we find
fo 48.0 cm
= 0.261 cm
M
( –184 )
______________________________________________________________________________
fe ≈ – =– 100. REASONING AND SOLUTION
a. The lens with the largest focal length should be used for the objective of the telescope.
Since the refractive power is the reciprocal of the focal length (in meters), the lens with the
smallest refractive power is chosen as the objective, namely, the 1.3  diopter lens .
b. According to Equation 26.8, the refractive power is related to the focal length f by
Refractive power (in diopters) =1/[f (in meters)] . Since we know the refractive powers of
the two lenses, we can solve Equation 26.8 for the focal lengths of the objective and the
eyepiece. We find that fo = 1/(1.3 diopters) = 0.77 m . Similarly, for the eyepiece,
fe =...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.
 Spring '13
 CHASTAIN
 Physics, The Lottery

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