Page 36 - Modern Optical Engineering The Design of Optical Systems

Optics Overview 19

ANSWER: The optical path ( n d) is 1.333 0.003 in 0.004 in, or 200 wave-
lengths. At one fringe per half wavelength, there will be 400 fringes.
6 The convex surface of a lens is in contact with a flat plate of glass. If the
radius of the lens surface is 20 in, at what diameter will the first dark inter-
ference band/ring be seen? The second? The third? What are the ring diameters
if the radius is 200 in?
ANSWER: There will be a dark spot at the center contact point (“Newton’s
black spot”) because of the reflection phase change at the lower surface. The
first dark ring will occur where the airspace is one-half wavelength, or
0.000010 in. This is of course simply the sagittal height of the lens surface. Per
Eq. 1.8 this is:
2
2 1/2
2 1/2
SH R – (R – Y ) 0.000010 in 20 – (400 – Y )
2
2
Squaring, we get: Y 400 – 19.99999 0.0004, and Y 0.02 in; the ring
diameter is 0.040 in.
The second ring is located where the airspace equals one wavelength
(0.000020 in); its diameter is 0.05657 in. The third ring is at 1.5 wavelength
spacing, and its diameter is 0.06928 in. Note that the diameters are related by
approximately the square root of the ring number; ring #2 diameter equals
ring #1 diameter times √2, and ring #3 diameter equals #1 diameter times √3.
When the radius is 200 in, the diameters are 0.1265, 0.1789, and 0.2191 in.
These are less than for the 20-in radius surface by a factor of about 3.162,
which is about √10, or √200/20. This ratio is exact only for small diameters.