James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

286 Chapter 4 Applications of Differentiation
rays from sun
rays from sun
42°
138°
The significance of the minimum deviation is that when < 59.48 we have D9sd < 0,
so DDyD < 0. This means that many rays with < 59.48 become deviated by approximately
the same amount. It is the concentration of rays coming from near the direction of
minimum deviation that creates the brightness of the primary rainbow. The figure at the left
shows that the angle of elevation from the observer up to the highest point on the rainbow
is 1808 2 1388 − 428. (This angle is called the rainbow angle.)
observer
laboratory Project
D
å
to
observer
from
sun
∫
∫
å
A
Formation of the secondary rainbow
C
∫ ∫
∫
∫
B
2. Problem 1 explains the location of the primary rainbow, but how do we explain the colors?
Sunlight comprises a range of wavelengths, from the red range through orange, yellow,
green, blue, indigo, and violet. As Newton discovered in his prism experiments of 1666, the
index of refraction is different for each color. (The effect is called dispersion.) For red light
the refractive index is k < 1.3318, whereas for violet light it is k < 1.3435. By repeating
the calculation of Problem 1 for these values of k, show that the rainbow angle is about
42.38 for the red bow and 40.68 for the violet bow. So the rainbow really consists of seven
individual bows corresponding to the seven colors.
The Paradox
Bimportant. 3. Perhaps Figure you have 1 displays seen a data fainter from secondary an observational rainbow above study that the primary clearly depicts bow. That this results trend.
from the part of a ray that enters a raindrop and is refracted at A, reflected twice (at B and
1. C), Draw and the refracted causal as diagram it leaves that the corresponds drop at D (see to the the initial figure expectationt.
the left). This time the deviation
angle Dsd is the total amount of counterclockwise rotation that the ray undergoes in
2. this Suppose. four-stage process. Show that
3. Suppose.
and Dsd has a minimum value when
Dsd − 2 2 6 1 2
cos −Î k 2 2 1
8
Taking k − 4 3 , show that the minimum deviation is about 1298 and so the rainbow angle for
the secondary rainbow is about 518, as shown in the figure at the left.
4. Show that the colors in the secondary rainbow appear in the opposite order from those in
the primary rainbow.
42° 51°
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