p35_042

# p35_042 - 42. (a) The ﬁrst image is ﬁgured using Eq....

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Unformatted text preview: 42. (a) The ﬁrst image is ﬁgured using Eq. 35-8, with n1 = 1 (using the rounded-oﬀ 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. 35-8 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 ﬁsh-image is 7.0 cm to the right of the air-wall 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.

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