35.41 .. Two flat plates of glass with parallel faces are on a table, one plate

Question

35.41 .. Two flat plates of glass with parallel faces are on a table, one plate on the other. Each plate is 11.0 cm long and has a refrac- tive index of 1.55. A very thin sheet of metal foil is inserted under the end of the upper plate to raise it slightly at that end, in a man- ner similar to that discussed in Example 35.4. When you view the glass plates from above with reflected white light, you observe that, at 1.15 mm from the line where the sheets are in contact, the violet light of wavelength 400.0 nm is enhanced in this reflected light, but no visible light is enhanced closer to the line of contact. (a) How far from the line of contact will green light (of wavelength 550 nm) and orange light (of wavelength 600.0 nm) first be enhanced? (b) How far from the line of contact will the violet, green, and orange light again be enhanced in the reflected light? (c) How thick is the metal foil holding the ends of the plates apart?

Explanation / Answer

Part A)

Apply 2nt = (m + .5)(wavelength)

For the thickness, the tan of the angle between the plates = t/L (L is the distance from the contact point)

tan(angle) = t/L

t = L(tan angle)

In this case, the 2,n, angle, and (m + .5) are consant, so we can set up a ratio

L/wavelength = L/wavelength

1.15/400 = L/550

L = 1.58 mm for Green

1.15/400 = L/600

L = 1.73 mm for Orange

Part B)

For the next violet, m will increase by 1, so the ratio factors changes

L/(m + .5) = L/(m + .5)

1.15/.5 = L/(1.5)

L = 3.45 mm for the Violet

1.58/(.5) = L/(1.5)

L = 4.74 mm for the Green

1.73/.5 = L/1.5

L = 5.18 mm for the Orange

Part C)
The plates are 11 cm long, so we need to determine the maximum m values for one of the colors at 11 cm

We can use the ratio from part B to find the max m value

For violet

1.15/.5 = 110/(m + .5)

m = 47

Then apply 2nt = (m + .5)(wavelength)

2(1)(t) = (47.5)(400 X 10-9)

t = 9.5 X 10-6 m which is 9.5 um

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