Homework Hint
Assignment (1-95)
Read (27.1, 27.2 and 27.5), Do CONCEPT Q. # (5), Do PROBLEMS # (2, 5, 19) Ch. 27
CQ. # 5
REASONING AND SOLUTION
In Young’s double-slit experiment, is it possible to see interference fringes when the wavelength
( λ ) of the light is greater than the distance (d) between the slits ?
‰ The angle θ for the interference maximum in Young's double-slit experiment is given by the
equation sin θ = mλ / d where m = 0, 1, 2, 3, . . .
‰ When the wavelength ( λ ) of the light is greater than the distance (d) between the slits, the
ratio λ / d is greater than one; however, sin θ CAN NOT be greater than one .
‰ Therefore, it is NOT possible to see interference fringes when the wavelength of the light is
greater than the distance between the slits.
PROB. # 2
A Young’s double-slit experiment is performed using light that has a wavelength of 630 nm. The separation
between the slits is 5.3x10-5 m. Find the angles that locate the first, second and third order bright bands.
REASONING AND SOLUTION
a. We know that θ = sin–1(mλ/d) . Also, m = 1 for the first order bright fringe, so
bgc
L
1 630 × 10 −9 m
M
M
θ = sin M
−5
M
N 5.3 × 10 m
−1
hOPP=
P
P
Q
0.68°
b. For the second-order bright fringe, m = 2 so θ = 1.4° .
c. For the third-order bright fringe, m = 3 so θ = 2 .0 ° .
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5.
In a Young’s double-slit experiment, the angle that locates the second-order bright fringe is 2°. If the slit
separation is 3.8x10-5 m, what is the wavelength of the light ?
REASONING AND SOLUTION
The angular position θ of the bright fringes of a double slit are given by the equation sin θ = mλ / d , with
the order of the fringe specified by m = 0, 1, 2, 3, . . . . Solving for λ , we have
λ =
( 3.8 × 10 –5 m) sin 2.0°
d sin θ
=
= 6.6 × 10 –7 m =
m
2
660 nm
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19. A diffraction pattern forms when light passes through a single slit. The wavelength of the light is 675 nm.
Determine the angle that locates the first dark fringe when the width of the slit is 1.8x10-4 m and 1.8x10-6 m.
REASONING
This problem can be solved by using the equation sin θ = mλ / d for the value of the angle θ when m = 1
for the first dark fringe.
SOLUTION
a. When the slit width is W = 1.8 × 10 –4 m and λ = 675 nm = 675 × 10 –9 m , we find, according to the
small angle approximation,
L(1) 675 × 10
λI
F
= sin M
G
J
HW K N 1.8 × 10
θ = sin −1 m
−1
–9
–4
O
P
Q
m
= 0.21°
m
b. Similarly, when the slit width is W = 1.8 × 10 –6 m and λ = 675 × 10 –9 m , we find
θ = sin
−1
L
675 × 10
( 1)
M
N 1.8 × 10
–9
–6
O
P
Q
m
= 22 °
m