Babinet’s principle. A monochromatic beam of parallel light is incident on a “collimating” hole of diameter x ≫ λ. Point P lies in the geometrical shadow region on a distant screen (Fig. 36-39 a ). Two diffracting objects, shown in Fig. 36-39 b, are placed in turn over the collimating hole. Object A is an opaque circle with a hole in it, and B is the “photographic negative” of A. Using superposition concepts, show that the intensity at P is identical for the two diffracting objects A and B. Figure 36-39 Problem 16.
Babinet’s principle. A monochromatic beam of parallel light is incident on a “collimating” hole of diameter x ≫ λ. Point P lies in the geometrical shadow region on a distant screen (Fig. 36-39 a ). Two diffracting objects, shown in Fig. 36-39 b, are placed in turn over the collimating hole. Object A is an opaque circle with a hole in it, and B is the “photographic negative” of A. Using superposition concepts, show that the intensity at P is identical for the two diffracting objects A and B. Figure 36-39 Problem 16.
Babinet’s principle. A monochromatic beam of parallel light is incident on a “collimating” hole of diameter x ≫ λ. Point P lies in the geometrical shadow region on a distant screen (Fig. 36-39a). Two diffracting objects, shown in Fig. 36-39b, are placed in turn over the collimating hole. Object A is an opaque circle with a hole in it, and B is the “photographic negative” of A. Using superposition concepts, show that the intensity at P is identical for the two diffracting objects A and B.
10. A light ray of given wavelength, initially in air, strikes a 90°
prism at P (see Fig. 39-53) and is refracted there and at Q to
such an extent that it just grazes the right-hand prism surface
at Q. (a) Determine the index of retraction of the prism for
this wavelength in terms of the angle of incidence , that
gives rise to this situation. (b) Give a numerical upper bound
for the index of refraction of the prism. Show, by ray dia-
grams, what happens if the angle of incidence at P is
(c) slightly greater or (d) slightly less than 0₁.
90
FIGURE 39-53. Problem 10.
109 In Fig. 34-54, a fish watcher at
point P watches a fish through a
glass wall of a fish tank. The watcher
is level with the fish; the index of re-
fraction of the glass is 8/5, and that Watcher
of the water is 4/3. The distances are
di = 8.0 cm, dz = 3.0 cm, and dz =
6.8 cm. (a) To the fish, how far away
does the watcher appear to be?
(Hint: The watcher is the object.
Light from that object passes
through the wall's outside surface, which acts as a refracting sur-
face. Find the image produced by that surface. Then treat that im-
age as an object whose light passes through the wall's inside sur-
face, which acts as another refracting surface.) (b) To the watcher,
how far away does the fish appear to be?
de
D
Wall
Figure 34-54
Problem 109.
(1), A light wave is incident upon an air/glass (ZF13, n=1.78 @ 632.8 nm) interface at an
angle of incidence 0-60°. The incident electric field can be described by:
8-(18+,+,)
√√√3
e,+e, cos(at-k), where k = --
2
The x, y and z directions are depicted in the Figure
Ꮎ
Air
Glass
(a) What fractions of the input power are in the TE and TM components, respectively,
of the input wave?
(b) What fraction of the incident wave power is transmitted?
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