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Unformatted text preview: n2 sin θ 2 ) at each
face of the prism. At the first interface where the ray enters the prism, n1 = 1.000 for air and
n2 = ng for glass. Thus, Snell's law gives (1) sin 60.0° = ng sin θ 2 or sin θ 2 = sin 60.0°
ng (1) We will represent the angles of incidence and refraction at the second interface as θ1′ and θ 2′ , respectively. Since the triangle is an equilateral triangle, the angle of incidence at the
′
second interface, where the ray emerges back into air, is θ1 = 60.0° – θ 2 . Therefore, at the
second interface, where n1 = ng and n2 = 1.000, Snell’s law becomes
′
ng sin (60.0° – θ 2 ) = (1) sin θ 2 (2) ′
We can now use Equations (1) and (2) to determine the angles of refraction θ 2 at which the
red and violet rays emerge into the air from the prism. Chapter 26 Problems 1377 SOLUTION
The index of refraction of flint glass at the wavelength of red light is
Red Ray
ng = 1.662. Therefore, using Equation (1), we can find the angle of refraction for the red ray
as it enters the prism: sin θ 2 = sin 60.0°
= 0.521
1.662 θ 2 = sin –1 0.521 = 31.4° or Substituting this value for θ2 into Equation (2), we can find the angle of refraction at which
the red ray emerges from the prism: ′
sin θ 2 = 1.662 sin ( 60.0° – 31.4° ) = 0.796
Violet Ray or ′
θ 2 = sin –1 0.796 = 52.7° For violet light, the index of refraction for glass is ng = 1.698. Again using Equation (1), we find sin θ 2 = sin 60.0°
= 0.510
1.698 θ 2 = sin –1 0.510 = 30.7° or Using Equation (2), we find
′
sin θ 2 = 1.698 sin ( 60.0° – 30.7° ) = 0.831 or ′
θ 2 = sin –1 0.831 = 56.2° ______________________________________________________________________________
48. REASONING At the slanted face, the refracted ray(s)
emerge into the surrounding material, which has an index of
refraction n2. Together, the index of refraction n2 of the
45.00°
surrounding material and the index of refraction n1 of
θ2
crown glass (which varies with the color of the light; see
θ1
Table 26.2) determine the critical angle θc for total internal
n
reflection, as we see from sin θc = 2 (Equation 26.4). To
n1
guarantee that the colors that are to emerge from the prism
at the slanted face do not undergo total internal reflection
there, we must choose n2 such that the angle of incidence θ1 is less than the critical angle θc
for each such color. The incident ray and the totallyinternallyreflected ray are
perpendicular to one another, and the 90.00° angle between them is bisected by the normal
to the slanted surface (see the drawing). Therefore, the angle of incidence is θ1 = 45.00°.
Thus, all colors of light for which the critical angle is greater than 45.00° will not undergo
total internal reflection and will emerge from the slanted face of the prism. 1378 THE REFRACTION OF LIGHT: LENSES AND OPTICAL INSTRUMENTS SOLUTION
a. Solving sin θc = n2
(Equation 26.4) for n2, we obtain
n1
n2 = n1 sin θc (1) Since only red light is to emerge from the slanted face of the prism, we will use n1 = 1.520,
the refractive index for red light in crown glass (see Table 26.2). The indices of refraction
for the remaining colors are all larger than 1.520, so all other colors of light will have
critical angles less than 45.00° and will undergo total internal reflection at the slanted face
of the prism. Thus, we have
n2 = (1.520 ) sin 45.00o = 1.075 b. All colors except violet are to emerge from the slanted face. Therefore, we will use
n1 = 1.531, the index of refraction of blue light in crown glass (see Table 26.2), in
Equation (1):
n2 = (1.531) sin 45.00o = 1.083
______________________________________________________________________________ 49. SSM REASONING The ray diagram is constructed by drawing the paths of two rays
from a point on the object. For convenience, we will choose the top of the object. The ray
that is parallel to the principal axis will be refracted by the lens so that it passes through the
focal point on the right of the lens. The ray that passes through the center of the lens passes
through undeflected. The image is formed at the intersection of these two rays. In this case,
the rays do not intersect on the right of the lens. However, if they are extended backwards
they intersect on the left of the lens, locating a virtual, upright, and enlarged image.
SOLUTION
a. The raydiagram, drawn to scale, is shown below.
Scale:
20 cm 20 cm F
Image F
Object Chapter 26 Problems 1379 From the diagram, we see that the image distance is d i = –75 cm and the magnification
is +2.5 . The negative image distance indicates that the image is virtual. The positive
magnification indicates that the image is larger than the object.
b. From the thinlens equation [Equation 26.6: (1 / do ) + (1 / di ) = (1 / f ) ], we obtain 111
1
1
=–
=
–
di
f d o 50.0 m 30.0 cm or d i = –75.0 cm The magnification equation (Equation 26.7) gives the magnification to be
d
–75.0 cm
m=– i =–
= +2.50
do
30.0 cm
_______________________________________________...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.
 Spring '13
 CHASTAIN
 Physics, The Lottery

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