Unformatted text preview: first lens can be obtained from the thinlens equation:
1
1
1
=−
do1 f1 di1 (2) Substituting Equation (2) into Equation (1) gives
1
1
1
1 1
1
1
1
1 =
−
=
−
−
−
− =
di2 f 2 d o1 f 2 f1 di1 16.0 cm 12.0 cm 21.0 cm Solving for di2 gives di2 = 37.3 cm .
______________________________________________________________________________ Chapter 26 Problems 1425 123. SSM REASONING AND SOLUTION Usinge the thin–lens equation, to we can find the
first image distance di with respect to the objective:: 1
1
1
1
1
=
−
=
−
di f o do 1.500 m 114.00 m or di = 1.520 m (1) Using tThe magnification equation, we can find the linear magnification m achieved by the
objective is:
d
1.520 m
m=− i =−
= −0.01333
(2)
do
114.00 m We then Now use this "first image" as the object for the second lens (the eyepiece). With
the aid of the thinlens equation, we can determine the distance di′ of the final image with Formatted: Font: Times New Roman, 9 pt,
Lowered by 4 pt
Formatted: Right Formatted: Right Field Code Changed respect to the eyepiece:,
1
1
1
1
1
=
−
=
−
′
di′ f e d o 0.070 m 0.050 m or di' = – 0.18 m (3) Formatted: Right ′
In Equation (3) we have used do = 0.050 m . This follows because we know that the Field Code Changed separation between the objective and the eyepiece is L ≈ f o + f e (see Example 17 in the Field Code Changed text). Therefore, we know that L ≈ 1.500 m + 0.070 m = 1.570 m , and we can determine
that the distance of the “first image” from the second lens (the eyepiece) is
L − 1.520 m ≈ 1.570 m − 1.520 m = 0.050 m .
Using the magnification equation, we can find the linear magnification m′ achieved by the
eyepiece:The magnification in the second case is Field Code Changed Field Code Changed Field Code Changed Formatted: Left d′
−0.18 m
m' = − i = −
= +3.6
′
do
0.050 m (4) Formatted: Right The Using Equations (2) and (4), we find that the total linear magnification achieved by
both the objective and the eyepiece isis therefore,
Ml = M × M ' = (–0.01333)(+3.6) = – 0.048 mTotal = hi′
= m × m′ = ( −0.01333)( +3.6 ) = −0.048
ho Formatted: Right
Field Code Changed (5)
where hi′ is the height of the final image and ho is the height of the initial object. Formatted: Font: Times New Roman, 9 pt,
Lowered by 4 pt
Field Code Changed However, we need the angular magnification, which, according to Equation 26.9, is Formatted: Indent: First line: 0" 1426 THE REFRACTION OF LIGHT: LENSES AND OPTICAL INSTRUMENTS θ′
M= =
θ In Equation (6) θ ′ = − hi′
di′ h′ d + f o + f e = − i o ho
di′ ho do + fo + fe Formatted: Right, Indent: First line: 0" − Field Code Changed (6)so ho
hi′
is the angular size of the final image and θ =
is the
di′
do + fo + fe Field Code Changed
Field Code Changed angular size of the initial object.
Substituting Equations (3) and (5) into Equation (6) and using the given data for do, fo, and
fe, we find that h′ d + f + f 114.00 m + 1.500 m + 0.070 m M = − i o o e = − ( −0.048 ) = −31 ho di′
−0.18 Formatted: Font: Italic
Formatted: Font: Times New Roman, 9 pt,
Lowered by 4 pt
Formatted: Font: Italic
Formatted: Font: Times New Roman, 9 pt,
Lowered by 4 pt
Formatted: Font: Italic θ'=− hi′
di′ and θ= ho
do + f o + f e Formatted: Font: Times New Roman, 9 pt,
Lowered by 4 pt
Formatted: Right
Field Code Changed where we'll use ho = 1 m, hi′ = –0.048 m, do + fo + fe = (114.00 m + 1.500 m + 0.070 m) =
115.57 m, and di' = – 0.18 m. Therefore, θ'=− −0.048 m
= −0.27 rad
−0.18 m and θ= 1.0 m
= 8.65 ×10−3 rad
115.57 m So that, M ang = θ′
−0.27 rad
=
= −31
θ 8.65 ×10−3 rad ______________________________________________________________________________
124. REASONING AND SOLUTION From the drawing in the text we see that do = x + f and
di = x ′ + f. Substituting these two expressions into the thinlens equation, we obtain 1
1
1
1
1
+
=
+
=
do di x + f x ′ + f
f
Combining the terms on the left over a common denominator gives Chapter 26 Problems 1427 x + f + x′ + f
x + x′ + 2 f
1
=
=
x + f )( x′ + f ) ( x + f )( x′ + f ) f
(
Crossmultiplying shows that f ( x + x′ + 2 f ) = ( x + f ) ( x′ + f )
Expanding and simplifying this result, we obtain
f x + f x ′ + 2 f 2 = xx ′ + f x ′ + x f + f 2
or
xx ′ = f 2
______________________________________________________________________________
125. REASONING AND SOLUTION If the near point is 79.0 cm, then di = – 79.0 cm, and
do = 25.0 cm. Using the thinlens equation, we find that the focal length of the correcting
lens is
dd
(25.0 cm)(−79.0 cm)
f= oi =
= +36.6 cm
do + di 25.0 cm + ( −79.0 cm )
′
a. The distance d o to the poster can be obtained as follows: 1
11
1
1
=−=
−
′
do f di′ 36.6 cm −217 cm or ′
do = 31.3 cm b. The image size is d′ −217 cm hi = ho − i = (0.350 m) − = 2.43 m
′ 31.3 cm do
______________________________________________________________________________ CHAPTER 27 INTERFERENCE A...
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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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