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Unformatted text preview: 42. (a) The ﬁrst image is ﬁgured using Eq. 358, with n1 = 1 (using the roundedoﬀ value for air) and
n2 = 8/5.
1
8
1.6 − 1
+
=
p 5i
r
For a “ﬂat lens” r = ∞, so we obtain i = −8p/5 = −64/5 (with the unit cm understood) for that
object at p = 10 cm. Relative to the second surface, this image is at a distance of 3 + 64/5 = 79/5.
This serves as an object in order to ﬁnd the ﬁnal image, using Eq. 358 again (and r = ∞) but
with n1 = 8/5 and n2 = 4/3.
4
8
+
=0
5p
3i
which produces (for p = 79/5) i = −5p/6 = −79/6 ≈ −13.2. This means the observer appears
13.2 + 6.8 = 20 cm from the ﬁsh.
(b) It is straightforward to “reverse” the above reasoning, the result being that the ﬁnal ﬁshimage is
7.0 cm to the right of the airwall interface, and thus 15 cm from the observer. ...
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.
 Fall '08
 SPRUNGER
 Physics

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