Question

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?

Answer 1

Answer:

a)100m

b)200m

Explanation:

Answer 2

(a) The distance from the point of contact where green light with a wavelength and orange light is 183.3 nm and 200.0 nm.

(b) The distance from the point of contact where violet, green, and orange light will be enhanced in the reflected light is 342.5 nm.

(c) The thickness of the metal foil holding the plates apart cannot be calculated because it results in a negative value, which is physically impossible.

Solution,

a)To determine the distance from the line of contact where the green light and the orange light will be enhanced, we can use the path difference formula:

path difference = (refractive index - 1) x thickness of the upper plate.

Thus, we have:

550 nm = (1.55 - 1) x t1

600 nm = (1.55 - 1) x t2

where t1 and t2 are the thicknesses of the upper plate at the distances where green and orange light are enhanced, respectively. Solving for t1 and t2, we get:

t1 = 183.3 nm

t2 = 200.0 nm.

b)The line of contact will be the violet, green, and orange light again be enhanced in the reflected light is,

2 x (refractive index - 1) x thickness of the upper plate = m x wavelength

where m is an integer representing the order of the bright fringe. For the violet light, m = 1 since it is the first bright fringe. For the green light and orange light, we can use the same value of m as for the violet light, since the path difference is proportional to the wavelength.

Thus, we have:

m = 1: 400 nm = 2 x (1.55 - 1) x t1

m = 1: 550 nm = 2 x (1.55 - 1) x t1

m = 1: 600 nm = 2 x (1.55 - 1) x t1

Solving for t1, we get:

t1 = 342.5 nm

c)This thickness is equal to the difference between the thicknesses of the upper plate at the two ends, which we can calculate as:

thickness difference = t1 - t2

thickness difference = 183.3 nm - 200.0 nm

thickness difference = -16.7 nm

The thickness of the metal foil holding the plates apart cannot be calculated because it results in a negative value, which is physically impossible. The assumption that the plates are perfectly flat and parallel may not be accurate.

#SPJ2