Φ21 Fall 2006
1
HW26 Solutions
Problem K22.42
White light (400-700 nm) is incident on a 600 line/mm diraction grating. What is the width of the rstorder rainbow on a screen 2.0 m behind the grating? Give your answer in cm. Do not make small angle
approximations.
Solution:
The two extreme wavelengths of light will diract to dierent angles according to
mλ = d sin θ
The distance between the lines of the grating is d = (1/600) mm = 1.667 × 10−6 m = 1667 nm. Since this is
the rst-order rainbow, m = 1 for both waves. So, the two angles are
θred
= sin−1
θviolet
= sin−1
λ
700 nm
= sin−1
= 24.834◦
d
1667 nm
400 nm
= 13.887◦
1667 nm
The actual distance y from the center of the screen to the diraction line comes from tan θ = Ly , where L is
the distance from the grating to the screen. So, the two distances and the width of the rainbow are:
yred
yviolet
∆y
2
= L tan θred = (2.0 m) tan (24.834◦ ) = 0.925 m
= (2.0 m) tan (13.887◦ ) = 0.494 m
= 0.431 m = 43.1 cm
Problem K22.46
How many lines per millimeter does the grating have? (Do not make
small angle approximations).
Solution:
θ = tan−1
The rst-order line is hitting at y = 0.436 m, so the angle is
= 22.557◦ . Using mλ = d sin θ,
0.436
1.0
d=
mλ
(1) (600 nm)
=
= 1501 nm = 1.501 × 10−3 mm
sin θ
sin (22.557◦ )
Therefore, the density of the grating is 666 lines/mm.
Figure 1: The diraction pattern
of 600 nm light on a screen 1.0 m
behind the diraction grating.
(Notice that unlike the cases where the small angle approximation applies, here the dierent orders of the diraction pattern get further and
further apart.)
3
Problem YF 35.29
What is the thinnest lm of a coating with an index of refraction of n = 1.35 on glass (with an index of
refraction ng > n) for which destructive interference of the red component (a wavelength of 630 nm) of an
incident white light beam in air can take place by reection?
Solution: The thinnest destructive interference comes with a phase dierence of ∆φ = π between the wave
reected o the rst surface and the one reected o the back surface of the lm. Since n > 1 (as is true
for all materials), the wave from the front surface is inverted (given a base phase dierence of π ). But, the
1
wave reected from the back surface is also inverted because n < ng , so in the end, both waves start out in
phase. The only dierence is the pathlength dierence.
The wave from the back surface has to traverse the lm twice, travelling a distance of ∆` = 2y , where y is
the thickness of the lm. The wavelength in the lm is λn = λ/n = 630 nm/1.35 = 466.7 nm (remember
that the frequency is constant for a given wave). Take ∆`/λn (the fraction of a cycle dierence between the
two waves) and multiply it by 2π to get ∆φ, which should be equal to π
(
∆φ = 2π
y
4
∆`
λn
)
= 2π
2y
=π
λ/n
λ
= 116.7 nm
4n
=
Problem YF 36.29
Visible light passes through a diraction grating that has 900 slits/cm and the interference pattern is observed
on a screen that is 2.36 m from the grating. In the rst-order spectrum, maxima for two dierent wavelengths
are separated on the screen by 3.20 mm. What is the dierence in these wavelengths (in meters)? (Here you
need to make small angle approximations.)
Solution:
that
We'll make the small angle approximation that sin θ ≈ tan θ ≈ θ (with θ in radians). That means
y
= tan θ ≈ θ
L
∆θ ≈
and
∆y
0.0032 m
=
= 0.00136 rad
L
2.36 m
The grating spacing is d = (90000 lines/m)−1 = 1.11 × 10−5 m. In the diraction formula, the small-angle
approx. with m = 1 gives:
mλ = d sin θ ≈ dθ
5
→
∆λ ≈ d∆θ = 15.1 nm
Problem K36.33
Plane monochromatic waves with wavelength 540 nm are incident normally on a plane transmission grating
having 350 slits/mm. Part A. Find the angle of deviation in the rst order.
The slit separation is d = (350000 slits/m)−1 = 2.857×10−6 = 2857 nm. First order means m = 1
in the formula mλ = d sin θ.
Solution:
θ1 = sin−1
Part B.
mλ
d
Find the angle of the second order.
θ2 = sin−1
Part C.
(
(
Find the angle of the third order.
θ3 = sin−1
(
)
= sin−1
Solution:
mλ
d
)
(
540
2857
)
= 10.89◦
Second order means m = 2, so
= sin−1
(
2 · 540
2857
)
= 22.21◦
Solution:
mλ
d
)
= sin−1
2
(
3 · 540
2857
)
= 34.54◦