00:01
Okay, so we're going to do chapter 32, problem 82.
00:05
So we have this figure here which shows a cylindrical rod whose end has a radius of curvature r of 20, oh, so not 20.
00:15
That is 2 .0 centimeters.
00:18
And the rod is immersed in water with an index of refraction 1 .33.
00:26
Okay, and the rod itself has an index of refraction of 1 .53.
00:31
We want to find a location in height of the image of an object 2 .0 millimeters high.
00:39
So the object height is 2 millimeters and its object distance is 23 centimeters.
00:51
Okay, so now from this we can use the spherical index of refraction equation 32 -8 to calculate the image location.
01:00
So that's n1 over the object distance plus n2 over the image distance equals n2 minus n1 over r.
01:13
Rearranging for the image distance, we see this as n2, n2 minus n1 over r minus n1 over d...