This preview shows page 1. Sign up to view the full content.
Unformatted text preview: possible values for m, and we wish to determine what the
specific value is. At point A the value is m = 0. As you walk from point A toward either
empty corner, the maximal loudness that indicates constructive interference does not occur
again until you arrive at the corner. Thus, the next possibility for m applies at the corner; in
other words, m = 1.
SOLUTION Consider the constructive interference that occurs at either empty corner.
Using L to denote the length of a side of the square and taking advantage of the Pythagorean
theorem, we have l1 = L and l2 = L2 + L2 = 2 L The specific condition for the constructive interference at the empty corner is (using m = 1)
l 2 − l1 = 2 L − L = (1) λ
Solving for the wavelength of the waves gives λ=L 5. 4
d2 − 1i = b.6 mg 2 − 1i =
d REASONING According to Equation 27.2,
Double
the wavelength λ of the light is related to
slit
the angle θ between the central bright fringe
and the seventh dark fringe according to λ= d sin θ
m+ 1
2 1.9 m Seventh
dark fringe
y θ
L Central
bright
fringe where d is the separation between the slits, and m = 0, 1, 2, 3, … The first dark fringe occurs
when m = 0, so the seventh dark fringe occurs when m = 6. The distance d is given, and we 1434 INTERFERENCE AND THE WAVE NATURE OF LIGHT can determine the angle θ by using the inverse tangent function, θ = tan −1 ( y / L ) , since both
y and L are known (see the drawing).
SOLUTION We will first compute the angle between the central bright fringe and the
seventh dark fringe using the geometry shown in the drawing: y
L 0.025 m = 1.3° 1.1 m θ = tan −1 = tan −1 The wavelength of the light is λ= 6. d sin θ (1.4 × 10−4 m ) sin1.3°
=
= 4.9 × 10 −7 m
1
1
m+ 2
6+ 2 REASONING The drawing shows the double slit
and the screen, The variable y denotes one half of the
screen width. Each bright fringe is located by an
angle θ that is determined by sin θ = mλ/d (Equation
27.1), where λ is the wavelength, d is the separation
between the slits, and m = 0, 1, 2, 3, …. The
maximum number m of bright fringes that can fall on
the screen (on one side of the central bright fringe)
occurs when tan θ = y/L. y
θ L SOLUTION For version A of the setup, the drawing
indicates that
tan θ = y
L or θ = tan −1 y
0
F I = tan F.10 m I = 16°
GJ G m J
HK H K
L
0.35
−1 But Equation 27.1 indicates that sin θ = mλ/d. Solving this expression for m and using
θ = 16°, we find m= d sin θ λ = 1
c.4 × 10 −5 h m sin 16 ° 625 × 10 −9 m = 6. 2 Thus, for version A, there are 6 bright fringes on the screen (on one side of the central
bright fringe), not counting the central bright fringe (which corresponds to m = 0). For
version B, we find Chapter 27 Problems 1435 y
0
F I = tan F.10 m I = 11°
GJ G m J
HK H K
L
0.50
1
sin
d sin θ c.4 × 10 m h 11°
m=
=
= 4 .3 θ = tan −1 −1 −5 λ 625 × 10 −9 m For this version, there are 4 bright fringes on the screen (on one side of the central bright
fringe), again not counting the central bright fringe. 7. REASONING AND SOLUTION The position of the third dark fringe (m = 2) is given by λ
λ
2
c h = c+ h .
d
d sin θ = m +
sin θ = m λ′
d 1
2 =4 1
2 λ′
d The position of the fourth bright fringe (m = 4) is given by . Equating these two expressions and solving for λ' yields 2
2
λ
b+ g = b+ g645 nmg=
b
λ′ =
1
2 1
2 4 8. 4 403 nm REASONING AND SOLUTION The angular position of a third order fringe is given by
θ = sin–1 3λ/d, and the position of the fringe on the screen is y = L tan θ. For the thirdorder red fringe, θ = sin −1 L F × 10 −9 m I O
3 665
MH
P
K
M
M .158 × 10 −3 m P= 0.724 °
P
M
P
N0
Q and y = (2.24 m) tan 0.724° = 2.83 × 10–2 m
For the thirdorder green fringe, θ = 0.615° and y = 2.40 × 10 m.
–2 The distance between the fringes is 2.83 × 10 m − 2.40 × 10 m = 4.3 × 10 −3 m .
–2 9. –2 REASONING The drawing shows a top
Secondorder bright
view of the double slit and the screen, as Double
fringe (m = 2)
slit
well as the position of the central bright
y
fringe (m = 0) and the secondorder
θ
bright fringe (m = 2). The vertical
Central bright
fringe (m = 0)
distance y in the drawing can be obtained
L
from
the
tangent
function
as
y = L tan θ. According to Equation 27.1, the angle θ is related to the wavelength λ of the 1436 INTERFERENCE AND THE WAVE NATURE OF LIGHT light and the separation d between the slits by sin θ = mλ / d , where m = 0, 1, 2, 3, … If the
angle θ is small, then tan θ ≈ sin θ , so that mλ y = L tan θ ≈ L sin θ = L d (1) We will use this relation to find the value of y when λ = 585 nm.
SOLUTION When λ = 425 nm, we know that y = 0.0180 m, so m ( 425 nm ) 0.0180 m = L d (2) Dividing Equation (1) by Equation (2) and algebraically eliminating the common factors of
L, m, and d, we find that mλ L λ
y
d
=
=
0.0180 m m ( 425 nm ) 425 nm
L d When λ = 585 nm, the distance to the secondorder bright fringe is 585 nm y = ( 0.0180 m ) = 0.0248 m 425 nm 10. REASONING AND SOLUTION The firstorder orange fringes occur farther out from the
center than do the firstord...
View
Full
Document
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

Click to edit the document details