EXAMPLE 24–10 Thin film of air, wedge-shaped. A very fine wire 7.35 × 10-3 mm in diameter is placed between two flat glass plates as in Fig. 24–33a. Light whose wavelength in air is 600 nm falls (and is viewed) per- pendicular to the plates and a series of bright and dark bands is seen, Fig. 24–33b. How many light and dark bands will there be in this case? Will the area next to the wire be bright or dark? APPROACH We need to consider two effects: (1) path differences for rays reflecting from the two close surfaces (thin wedge of air between the two glass plates), and (2) the -cycle phase shift at the lower surface (point E in Fig. 24–33a), where rays in air can enter glass (or be reflected). Because there is a phase shift only at the lower surface, there will be a dark band (no reflection) when the path difference is 0, 1, 21, 3A, and so on. Since the light rays are perpendicular to the plates, the extra path length (DEF) equals 21, where t is the thickness of the air gap at any point. SOLUTION Dark bands will occur where m = 0, 1, 2, ·.. Bright bands occur when 2t = (m + })a, where m is an integer. At the position of the wire, t = 7.35 × 10- m. At this point there will be 2t/A = (2)(7.35 × 10-6 m)/(6.00 × 10-7 m) = 24.5 wavelengths. This is a “half integer," so the area next to the wire will be bright. There will be a total of 25 dark lines along the plates, corresponding to path lengths DEF of 0A, 12, 21, 32, -·, 24A, including the one at the point of contact A (m = 0). Between them, there will be 24 bright lines plus the one at the end, or 25. 21 = ma, NOTE The bright and dark bands will be straight only if the glass plates are extremely flat. If they are not, the pattern is uneven, as in Fig. 24–33c. Thus we see a very precise way of testing a glass surface for flatness. Spherical lens surfaces can be tested for precision by placing the lens on a flat glass surface and observing Newton's rings (Fig. 24–31b) for perfect circularity.

EXAMPLE 24–10 Thin film of air, wedge-shaped. A very fine wire 7.35 × 10-3 mm in diameter is placed between two flat glass plates as in Fig. 24–33a. Light whose wavelength in air is 600 nm falls (and is viewed) per- pendicular to the plates and a series of bright and dark bands is seen, Fig. 24–33b. How many light and dark bands will there be in this case? Will the area next to the wire be bright or dark? APPROACH We need to consider two effects: (1) path differences for rays reflecting from the two close surfaces (thin wedge of air between the two glass plates), and (2) the -cycle phase shift at the lower surface (point E in Fig. 24–33a), where rays in air can enter glass (or be reflected). Because there is a phase shift only at the lower surface, there will be a dark band (no reflection) when the path difference is 0, 1, 21, 3A, and so on. Since the light rays are perpendicular to the plates, the extra path length (DEF) equals 21, where t is the thickness of the air gap at any point. SOLUTION Dark bands will occur where m = 0, 1, 2, ·.. Bright bands occur when 2t = (m + })a, where m is an integer. At the position of the wire, t = 7.35 × 10- m. At this point there will be 2t/A = (2)(7.35 × 10-6 m)/(6.00 × 10-7 m) = 24.5 wavelengths. This is a “half integer," so the area next to the wire will be bright. There will be a total of 25 dark lines along the plates, corresponding to path lengths DEF of 0A, 12, 21, 32, -·, 24A, including the one at the point of contact A (m = 0). Between them, there will be 24 bright lines plus the one at the end, or 25. 21 = ma, NOTE The bright and dark bands will be straight only if the glass plates are extremely flat. If they are not, the pattern is uneven, as in Fig. 24–33c. Thus we see a very precise way of testing a glass surface for flatness. Spherical lens surfaces can be tested for precision by placing the lens on a flat glass surface and observing Newton's rings (Fig. 24–31b) for perfect circularity.