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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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. ...
View Full Document

This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

Ask a homework question - tutors are online