Question 1

A thin layer of oil (n = 1.25) is on top of a puddle of water (n = 1.33). If normally incident 500 nm light is strongly reflected, what is the minimum thickness of the oil layer?

Accepted Answer

The condition for strong reflection from a thin film is 2nt = mλ, where m is an integer, n is the index of refraction of the film, t is the thickness of the film, and λ is the wavelength of light. For m = 1 and λ = 500 nm, we have 2nt = 500 nm. Solving for t, we get t = 250 nm.

Question 2

Two pieces of optically flat glass (n = 1.50) touch at one edge and a separated by a thin stretched fiber at the opposite edge. Light of wavelength 500 nm is shone from above and 100 dark fringes are counted. How thick is the fiber?

Accepted Answer

The thickness of the fiber can be calculated using the formula for the thickness of a thin film causing interference: $ t = \frac{m \lambda}{2n} $, where $ m $ is the number of dark fringes, $ \lambda $ is the wavelength of light, and $ n $ is the refractive index. Substituting the given values: $ t = \frac{100 \times 500 \times 10^{-9}}{2 \times 1.5} = 25 \times 10^{-6} $ meters, or 25 μm.

Question 3

A puddle of water has a thin film of gasoline floating on it. A beam of light is shining perpendicular on the film. If the wavelength of light incident on the film is 560 nm and the indices of refraction of gasoline and water are 1.40 and 1.33, respectively, what is the minimum thickness of the film to see a bright reflection?

Accepted Answer

The answer of A puddle of water has a thin...

Question 4

White light incident normal to a thin soap film with n = 1.34 reflects with an interference maximum at 684 nm and an interference minimum at 570 nm with no minima between those two values. What is the thickness of the soap film?

Accepted Answer

The interference maximum occurs when the path difference is an integer multiple of the wavelength, and the minimum occurs when it is a half-integer multiple. The difference in path length for the maximum and minimum is half a wavelength in the film. Using the formula for thin film interference, the thickness can be calculated as t = (mλ_max - (m+0.5)λ_min) / (2n), where m is the order of the maximum. Solving gives t = 638 nm.

Question 5

A coating is being applied to reduce the reflectivity of a pane of glass to light with a wavelength of 522 nm incident near the normal to the pane. If the material has an index of refraction of 1.375, how thick should the coating be?

Accepted Answer

The answer of A coating is being applied to reduce...

Question 6

Light of wavelength 500 nm is incident on a thin sheet of plastic film (n= 1.50) in air. What is the thinnest the sheet can be so that the reflected light from the surfaces undergoes constructive interference?

Accepted Answer

For constructive interference in thin films, the condition is 2nt = mλ, where n is the refractive index, t is the thickness, m is an integer (for the thinnest film, m=1), and λ is the wavelength in the medium. The wavelength in the film is λ/n = 500 nm / 1.5 = 333 nm. Solving 2(1.5)t = 333 nm gives t = 83 nm.

Question 7

Light of wavelength 500 nm illuminates a soap film ( n = 1.33). What is the minimum thickness of film that will give an interference maximum when the light is incident normally on it?

Accepted Answer

The answer of Light of wavelength 500 nm illuminates a...

Question 8

Light of wavelength 500 nm is incident on a thin sheet of plastic film (n= 1.50) in air. What is the thinnest the sheet can be so that the reflected light from the surfaces undergoes destructive interference?

Accepted Answer

For destructive interference in thin films, the condition is 2nt = (m + 0.5)λ, where n is the refractive index, t is the thickness, m is an integer, and λ is the wavelength in the medium. For the thinnest film, m = 0, so t = λ/(4n) = 500 nm / (4 * 1.5) = 83.33 nm. However, this calculation is incorrect, and the correct choice is 167 nm, which corresponds to the next possible thickness for destructive interference.

Question 9

Two flat glass plates, 10.0 cm wide, touch at one end but are separated by a thin wire at the other end, forming a wedge. Light with a wavelength of 450 nm shines almost perpendicularly on the glass and forms fringes which are 1.80 mm apart. What is the diameter of the wire?

Accepted Answer

The answer of Two flat glass plates, 10.0 cm wide,...

Question 10

A very fine thread is placed between two glass plates on one side and the other side is touching to form a wedge. A beam of monochromatic light of wavelength 600 nm illuminates the wedge and 178 bright fringes are observed. What is the thickness of the thread?

Accepted Answer

The number of fringes observed (N) is given by the formula N = 2nt/λ, where t is the thickness of the thread, n is the refractive index of the medium between the plates (assumed to be air here), and λ is the wavelength of light.

Here, λ = 600 nm and N = 178. Let's assume that the refractive index of air is 1 (to account for the fact that the wedge is very thin). Then we have:

178 = 2 × 1 × t/600 nm
t = (178 × 600 nm)/2
t = 53.4 µm

So the thickness of the thread is approximately 53.3 µm. Therefore, the answer is D.

Question 11

If the movable mirror in a Michelson interferometer is moved 120 microns, how many fringes would be counted for 600 nm light?

Accepted Answer

The number of fringes counted is equal to the number of wavelengths that fit into twice the distance the mirror is moved. For a 120 micron movement, the path difference is 240 microns. Dividing 240,000 nm by 600 nm gives 400 fringes.

Question 12

A soap bubble film (n=1.42) of thickness 106 nm results in a constructive interference in the reflected light if this film is illuminated by a beam of light with a wavelength of 601 nm. What are the next three thicknesses of this film that will also result in a constructive interference?

Accepted Answer

The answer of A soap bubble film (n=1.42) of thickness...

Question 13

What is the minimum thickness of a nonreflecting film coating (n = 1.30) on a glass lens (n = 1.50) for wavelength 500 nm?

Accepted Answer

The minimum thickness of a nonreflecting film coating is given by:t = (λ/4n)where λ is the wavelength of light, n is the index of refraction of the film, and t is the thickness of the film.For a wavelength of 500 nm, a film with an index of refraction of 1.30, and a glass lens with an index of refraction of 1.50, the minimum thickness of the film is:t = (500 nm/4 * 1.30) = 96.2 nm

Question 14

A piece of glass has a thin film of gasoline floating on it. A beam of light is shining perpendicular on the film. If the wavelength of light incident on the film is 560 nm and the indices of refraction of gasoline and glass are 1.40 and 1.50, respectively, what is the minimum thickness of the film to see a bright reflection?

Accepted Answer

The answer of A piece of glass has a thin...

Question 15

Light with λ = 420 nm in incident normally onto the flat side of a plano-convex lens which is resting on a glass plate parallel to the flat side of the lens and forms a pattern of concentric luminous and dark rings (Newton's rings). The radius of the second dark ring is 2.0 mm. What is the radius of curvature of the curved side of the lens?

Accepted Answer

The radius of curvature R can be found using the formula for Newton's rings: $ R = \frac{r^2}{m \cdot \lambda} $, where r is the radius of the m-th dark ring, m is the ring number, and λ is the wavelength. For the second dark ring (m=2), $ R = \frac{(2.0 \times 10^{-3})^2}{2 \times 420 \times 10^{-9}} = 4.8 \, \text{m} $.

Question 16

Light with λ = 590 nm enters a Michelson interferometer. How many fringes traverse the field of view if the mirror of one arm of the interferometer is moved 0.44 mm?

Accepted Answer

The answer of Light with λ = 590 nm enters...

Question 17

How far would the movable mirror in a Michelson interferometer need to be moved for 800 fringes to be counted using 633 nm light?

Accepted Answer

The distance moved by the movable mirror for one fringe is given by:$$d = \frac{\lambda}{2} = \frac{633 	ext{ nm}}{2} = 316.5 	ext{ nm}$$For 800 fringes, the distance moved is:$$d = 800 	imes 316.5 	ext{ nm} = 253.2 \mu	ext{m} = 0.253 	ext{ mm}$$

Quiz 34: The Wave Nature of Light; Interference

A thin layer of oil (n = 1.25) is on top of a puddle of water (n = 1.33) . If normally incident 500 nm light is strongly reflected, what is the minimum thickness of the oil layer?