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Activities 15.1, 15.2
Section 35-2:
Reflection
Problem
1. Through what angle should you rotate a mirror in
order that a reflected ray rotate through 30°?
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11. A light ray propagates in a transparent materid
at 15O to the normal t o the surface. When it
-emerges into the surrounding air, it makes a
24" angle with the normal. What is the refractive
index of the material?
-
Solution
Since el = 0; for specular reflection, (Equation 35-1) a
reflected ray is deviated by 4 = 180° - 2 4 from the
incident direction. If rotating the mirror changes 81 by
hel, then the reflected ray is deviated by A4 =
1-2AQ1( or twice this amount. Thus, if A$ = 30°,
lAO1l = 15".
Solution
Snell's law (Equation 35-3), with air a s medium 1,
gives nz = nlsinBl/sin02 = l x sin24O/sin 15' = 1.57.
-.
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Problem
14. A meter stick lies on the bottom of the
rectangular trough in Fig. 35-34, with its zero
mark a t the left edge of the trough. You look into
the long dimension of the trough at a 45" angle,
with your line of sight just grazing the top edge of
the tank, as shown. What mark on the meter stick
do you see if the trough is (a) empty, (b) half full
of water, and (c) full of water?
Problem 1 Solution.
Problem
2. The mirrors in Fig. 35-33 make a 60" angle; A light
ray enters parallel to the symmetry axis, as shown.
(a) How many reflections does it make? (b) Where
and in what direction does it exit the mirror
system?
FIGURE35-24 Problem 14.
Solution
As shown in the diagram, the mark seen on the meter
stick is at position x = x1 + xz = (e - h) h tane2,
where 82 = sin-'(sin 45011.333). Thus, x =
40 cm - 0.374h. For h = 0 (empty), 20 cm (half full),
and 40 cm (full), x = 40 cm, 32.5 cm, and 25.0 cm,
respectively.
+
FIGURE35-33 Problem 2 Solution.
.
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Solution
The first reflected ray leaves the upper mirror at a
grazing angle of 30°, and therefore strikes the lower
mirror normally. It is then reflected twice more in
retracing its path in the opposite direction.
Sec3on 35-3:
Refraction
Problem
8. In which substance in Table 35-1 does the speed of
. light have the value 2 . 2 9 2 ~
lo8 m/s?
Solution
Since the speed of light in a medium is v = cln, n =
3x108/2.292x lo8 = 1.309. This matches ice in
Table 35-1.
Problem 14 Solution.
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Problem
20. The prism in Fig. 35-36 has n = 1.52, cr = 60°,
and-is surrounded by air. A light beam is incident
at B1 = 37". Find the angle b through which the
beam is deflected.
FIGURE 35-36 Problem 20.
Problem
33. A scuba diver sets off a camera flash a distance h
below the surface of water with refractive index n.
Show that light emerges from the water surface
through a circle of diameter 2 h / d G .
Solution
Light from the flash will strike the water surface at the
critical angle for a distance r = h tan 8, from a point
directly over the flash. Therefore, the diameter of the
circle through which light emerges is 2r = 2h tan8,.
But sine, = 1l.n (Equation 33-5 at the water-air
interface), and tan2 Oc = (csc2OC- 1)-I (a
trigonometric identity), so we can write 2r = 2h +
d m .
Solution
A more general case of refraction through the same
type of prism is treated in Problem 55. For the data in
this problem, the other angles and the deflection are:
62 = sin-'(sin 37011.52) = 23.3", 4 2 = 60" - 6% =
36.P, = sin-'(1.52 sin 36.70) = 65.2") and 6 =
41 - 23" = 42.2".
Section 35-4:
Total Internal Reflection
Problem
21. Find the critical angle for total internal reflection
in (a) ice, (b) polystyrene, and (c) rutile. Assume
the surrounding medium is air.
Solution
For nai, k: 1, the critical angle for total internal
reflection in a medium of refractive index n is OC =
sin-'(lln). (Air is medium-2 in Equation 35-5.) ~romTable 35-1, n = 1.309 (ice), 1.49 (polystyrene), and
2.62 (rutile), so 8, = sin-'(1/1.309) = 49.8", 42.2' and
22.4", respectively, for these media.
Problem
27. What is the minimum refractive index for which
total internal reflection will occur as shown in
Fig. 35-15a? Assume the surrounding medium is
air and that the prism is an isosceles right triangle.
Solution
Figure 35-15a shows a 45" right-triangle prism with
critical angle less than 45". Thus, 8, = sin-'(;) <
45", or n > l/sin45O = 4.(We used nz = 1 for air,
and nl = n for the prism, in Equation 35-5.)
Problem 33 Solution.
Problem
36. White light propagating in air is incident a t 45O
on the equilateral prism of Fig. 3538. Find the
angular dispersion 7 of the outgoing beam, if the
prism has refractive indices nred = 1.582, nviolet =
1.633.
FIGURE 35-38 Problem 36 Solution.
Solution
This geometry for refraction through a prism L
treated in Problem 55. The difference in the directions
of the exiting red and violet light is the dispersion. For
the given values of cr = 60°, O1 = 45", rr, = 1.582, and
n, = 1.633, we find:
= 26.5", &,+= 33.5",
=
60.7", and 6, = 45.P, while B2,, = 25.6", 42,, = 34.3",
4l,, = 67.1°, and 6, = 52.1". The angular dispersion is
7 = 6, - 6, (or 41,, - 41,,) = 6.41'. (Note: We did not
need t o calculate the deflection, but did so because
this is the customary parameter in prism optics.)
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Problem
37. Two of the prominent spectral lines-discrete
wavelengths of light--emitted by glowing
hydrogen are hydrogen-a at 656.3 nm and
hydrogen-P at 486.1 nm. Light from glowing
hydrogen passes through a prism like that of
Fig. 35-22, then falls on a screen 1.0 m from the
prism. How far apart will these two spectral lines
be? Use Fig. 35-20 for the refractive index.
Solution
The angular dispersion of Ha and HD light in the
prism of Fig. 35-22 can be found from the analysis in
Example 35-6. For normal incidence on the prism, rays
emerge with refraction angles of sin-' (nsin 40'). From
Fig. 35-20, we estimate that na = 1.517 and no =
1.528, so the angular dispersion is 7 = 79.2' - 77.2' =
1.98O. We can assume that the size of the prism is
small compared to the distance, r, to the screen, so the
separation on the screen corresponding to 7 is Ax =
7.7-= (1.98')(1r/180°)(1 m) = 3.45 cm.
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Section 35-6:
--
Reflection and Polarization
Problem
39. Find the polarizing angle for diamond when light
is incident from air.
Solution
Equation 35-6 gives the polarizing angle, for light in
air reflected from diamond; 0, = tan-'(2.419/1) =
67.5'.