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Appendix C C-73<br />
vertically downward to the point at the center of a coin lying at the bottom of a<br />
cup of water. The solid red rays emerge from this point and enter the eye after<br />
bending at the water’s surface. The ray emerging from the center of the coin to<br />
the center of the eye is called the chief ray. On the right side of the figure, one<br />
ray emerging from the center of the coin is shown refracted at the surface of the<br />
water and entering the pupil of the eye. The dashed line extends the part of this<br />
ray in air back through the water intersecting the chief ray. We could perform<br />
this same construction for any other non-chief ray in Figure D. Our goal is to<br />
show that all the dashed extensions would intersect the chief ray at approximately<br />
the same point.<br />
In Figure E we lay out the geometry for a typical ray in great detail.<br />
Air: nª=1<br />
P<br />
˙ª<br />
1<br />
y<br />
O<br />
yª<br />
A<br />
˙<br />
I B<br />
4<br />
Water: n=<br />
3<br />
Figure D<br />
Figure E<br />
Exercise 1<br />
(a) Show that a ray emerging from the coin making an angle f with the vertical<br />
intersects the air-water boundary with angle of incidence f. Then show<br />
that angle BPI is equal to the angle of refraction f.<br />
(b) Show that the length of line segment IB is sin f, which by Snell’s law is<br />
n sin f, then show that the length of PB is 21 n 2 sin 2 f. Caution: Pay<br />
attention to the direction of the rays in Figure E.<br />
(c) Show that the length of line segment OI is y tan f.<br />
(d) Show that triangles IOA and IBP are similar.<br />
y tan f21 n 2 sin 2 f<br />
(e) Use parts (a) through (d) to show that y¿ <br />
and<br />
n sin f<br />
simplify to show<br />
y¿ 21 n2 sin 2 f<br />
cos f<br />
1<br />
n y<br />
The last formula allows us to compute the point at which the center of the<br />
coin would appear to be located if we just consider the chief ray and a particular<br />
ray emerging from the center point at an angle f to the vertical. To complete