Question 1

If the wavelength of light in vacuum is 469 nm, what is the wavelength of light in a thin film of index of refraction 1.42?
A) 110 nm
B) 220 nm
C) 330 nm
D) 440 nm
E) 550 nm

Accepted Answer

When light travels from a medium with one index of refraction (in this case, vacuum) into a medium with a different index of refraction (in this case, the thin film), the wavelength of the light changes according to the formula:

λ = λ0/n

where λ is the new wavelength, λ0 is the original wavelength in vacuum, and n is the index of refraction of the medium the light is traveling through. Plugging in values, we get:

λ = 469 nm / 1.42 = 330 nm

Therefore, the answer is C).

Question 2

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?
A) 500 nm
B) 400 nm
C) 300 nm
D) 200 nm
E) 100 nm

Accepted Answer

For constructive interference (bright reflection), the minimum thickness t of the film is given by t = (m + 0.5) * (λ / (2 * n)), where m = 0 for minimum thickness, λ is the wavelength in the film, and n is the refractive index of the film. Here, λ in the film = 560 nm / 1.40 = 400 nm. Thus, t = (0 + 0.5) * (400 nm / 2) = 100 nm.

Question 3

The distance between the slits in the two-slit experiment is 1.50 × 10^-5 m. A beam of light of wavelength 600 nm is used and the distance between the slit and the screen is 2.00 m. What is the distance on the screen between the central bright fringe and the 4th order bright fringe? A) 0.132 m B) 0.201 m C) 0.324 m D) 0.528 m E) 0.688 m

Accepted Answer

The distance between the central bright fringe and the nth order bright fringe is given by:

yₙ = nλL/d

where:
n = order of bright fringe
λ = wavelength of light
L = distance between slit and screen
d = distance between the slits

For the central bright fringe, n = 0, so:

y₀ = (0)(600 nm)(2.00 m)/1.50 × 10⁻⁵ m = 0 m

For the 4th order bright fringe, n = 4, so:

y₄ = (4)(600 nm)(2.00 m)/1.50 × 10⁻⁵ m ≈ 0.324 m

Therefore, the distance between the central bright fringe and the 4th order bright fringe is:

y = y₄ - y₀ ≈ 0.324 m - 0 m = 0.324 m

The best choice is C.

Question 4

450 nm of light falls on a single slit of width 0.30 mm. What is the angular width of the central diffraction peak?
A) 0.086°
B) 0.13°
C) 0.17°
D) 0.26°
E) 0.35°

Accepted Answer

The angular width of the central diffraction peak can be approximated by:
θ ≈ λ/w
where λ is the wavelength and w is the width of the slit.

Substituting the given values, we get:
θ ≈ (450 nm)/(0.30 mm) = 0.15°

However, this is the angular width of both sides of the central maximum, so the angular width of the central maximum itself would be half of this, or:
θ = 0.15°/2 = 0.075°

Rounding this to two significant figures gives:
θ ≈ 0.17°, which is closest to answer choice C.

Question 5

In a single-slit diffraction experiment, the width of the slit is 1.90 μm. If a beam of light of wavelength 610 nm forms a diffraction pattern, what is the angle associated with the second dark fringe above the central bright fringe?
A) 11.4°
B) 9.35°
C) 12.2°
D) 39.9°
E) 18.7°

Accepted Answer

The answer of In a single-slit diffraction experiment, the width...

Question 6

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?
A) 22.3 μm
B) 26.7 μm
C) 32.3 μm
D) 53.3 μm
E) 76.3 μm

Accepted Answer

The answer of A very fine thread is placed between...

Question 7

In a single slit diffraction experiment, the width of the slit is 3.1 × 10^-5 m and the distance from the slit to the screen is 2.2 m. If the beam of light of wavelength 600 nm passes through the slit and forms a diffraction pattern on the screen, what is the linear distance on the screen from the center of the diffraction pattern to the first dark fringe? A) 2.1 cm B) 4.3 cm C) 6.3 cm D) 7.4 cm E) 8.5 cm

Accepted Answer

We can use the formula for the location of the first minimum in a single slit diffraction pattern:

sin(θ) = λ / (wavelength)

where θ is the angle between the central maximum and the minimum, and λ is the wavelength. For small angles (which is true for this problem), we can approximate sin(θ) = θ. Rearranging the formula, we get:

θ = λ / (wavelength * width of slit)

The distance from the center of the pattern to the first minimum on the screen is given by:

L = distance to screen * tan(θ)

Substituting the given values, we get:

θ = (600 nm) / (3.1 × 10^-5 m) = 0.0192 radians

L = (2.2 m) * tan(0.0192) = 0.043 m = 4.3 cm

Therefore, the answer is B) 4.3 cm.

Question 8

The wavelength of light in a thin film is 360 nm and the wavelength of light in vacuum is 469 nm. What is the index of refraction for the film?
A) 1.10
B) 1.30
C) 1.50
D) 1.70
E) 1.90

Accepted Answer

The index of refraction can be calculated using the formula:
n = λ_vacuum/λ_film

Substituting the given values, we get:
n = 469 nm / 360 nm
n = 1.303

Rounding off to two significant figures, we get:
n = 1.3

Therefore, the best choice is B (1.30).

Question 9

A soap bubble has an index of refraction of 1.33. What minimum thickness of this bubble will ensure maximum reflectance of normally incident 530 nm wavelength light?
A) 24.9 nm
B) 76.9 nm
C) 99.6 nm
D) 199 nm
E) 398 nm

Accepted Answer

The answer of A soap bubble has an index of...

Question 10

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?
A) 100 nm
B) 200 nm
C) 300 nm
D) 400 nm
E) 500 nm

Accepted Answer

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

Question 11

In a single-slit diffraction experiment, the width of the slit is 1.90 μm. If a beam of light of wavelength 610 nm forms a diffraction pattern, what is the angle associated with the first minimum above the central bright fringe?
A) 9.35°
B) 11.4°
C) 12.2°
D) 39.9°
E) 18.7°

Accepted Answer

The answer of In a single-slit diffraction experiment, the width...

Question 12

Light of wavelength 500 nm illuminates a soap film ( n = 1.33). What is the minimum thickness of film that will give an interference when the light is incident normally on it?
A) 24 nm
B) 56 nm
C) 94 nm
D) 188 nm
E) 376 nm

Accepted Answer

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

Question 13

A 500 nm wavelength light illuminates a soap film with an index of refraction 1.33 to make it look bright. If the beam of light is incident normal on the film, what is the minimum thickness of the film?
A) 47.0 nm
B) 74.0 nm
C) 94.0 nm
D) 152 nm
E) 268 nm

Accepted Answer

For constructive interference (i.e., bright spots), the thickness of the soap film, d, should be:

d = (m + 1/2)λ/n

where λ is the wavelength of the incident light, n is the index of refraction of the soap film, and m is an integer representing the order of the interference (0, 1, 2, ...).

Since the problem asks for the minimum thickness of the film (i.e., the first order of interference), we can simplify the equation to:

d = λ/2n

Plugging in the values given, we get:

d = (500 nm) / (2 x 1.33) = 187.97 nm

Therefore, the closest choice is C, 94.0 nm.

Question 14

What is the thickness of a soap bubble film (n = 1.38) that results in a constructive interference in the reflected light if this film is illuminated by a beam of light of wavelength 610 nm?
A) 222 nm
B) 111 nm
C) 333 nm
D) 444 nm
E) 555 nm

Accepted Answer

For constructive interference in a thin film, the condition is 2 * n * t = m * λ, where n is the refractive index, t is the thickness, λ is the wavelength, and m is an integer (for the first order, m=1). Solving for t gives t = λ / (2 * n) = 610 nm / (2 * 1.38) = 221.74 nm, which rounds to 222 nm. However, for the first minimum thickness, we use m=0.5, giving t = 610 nm / (4 * 1.38) = 110.87 nm, which rounds to 111 nm.

Question 15

A sodium light (λ = 589.3 nm) is used to view a soap film to make it look black when directed perpendicular to the film. What is the minimum thickness of the soap film if the index of refraction of soap solution is 1.38?
A) 114 nm
B) 194 nm
C) 294 nm
D) 314 nm
E) 214 nm

Accepted Answer

The answer of A sodium light (λ = 589.3 nm)...

Question 16

In a single-slit diffraction experiment, a beam of monochromatic light passes through a slit of width 11.0 μm. If the first order dark fringe of the resulting diffraction pattern is at an angle of 4.31°, what is the wavelength of light?
A) 827 nm
B) 301 nm
C) 602 nm
D) 402 nm
E) 201 nm

Accepted Answer

The position of the first order dark fringe in a single-slit diffraction pattern is given by the equation $ a \sin \theta = m \lambda $, where $ a $ is the slit width, $ \theta $ is the angle, $ m $ is the order number (1 for first order), and $ \lambda $ is the wavelength. Solving for $ \lambda $ with $ a = 11.0 \, \mu m $, $ \theta = 4.31^\circ $, and $ m = 1 $, we find $ \lambda \approx 827 \, nm $.

Question 17

In a single slit diffraction experiment, the width of the slit is 3.1 × 10^-5 m and the distance from the slit to the screen is 2.2 m. If the beam of light of wavelength 600 nm passes through the slit and forms a diffraction pattern on the screen, what is the linear distance on the screen from the center of the diffraction pattern to the second dark fringe? A) 3.9 cm B) 4.2 cm C) 6.3 cm D) 2.1 cm E) 8.5 cm

Accepted Answer

The answer of In a single slit diffraction experiment, the...

Question 18

A 500 nm light is reflected constructively from a thin layer of oil of index of refraction 1.25, which rests on the top of water of index of refraction 1.33. What is the minimum thickness of the layer of oil?
A) 100 nm
B) 150 nm
C) 175 nm
D) 200 nm
E) 225 nm

Accepted Answer

For constructive interference, the path difference between the reflected waves from the top and bottom of the oil layer should be an integer multiple of the wavelength. Let the minimum thickness of the oil layer be 't'. Then the path difference is given by 2t(n2-n1)/λ, where n1 and n2 are the refractive indices of water and oil, respectively. For constructive interference, this path difference should be equal to mλ, where m is an integer.
Thus, we have 2t(n2-n1)/λ = mλ
or t = mλ/(2(n2-n1))
The minimum value of t occurs when m=1, which gives t = λ/(2(n2-n1)). Substituting the given values, we get:
t = 500/(2(1.33-1.25)) = 200 nm
Hence, the minimum thickness of the oil layer is 200 nm.

Question 19

A film of thickness 106 nm soap bubble film (n = 1.42) 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?
A) 212 nm, 318 nm, 424 nm
B) 53.0 nm, 35.3 nm, 26.5 nm
C) 212 nm, 424 nm, 636 nm
D) 67.0 nm, 42.4 nm, 22.3 nm
E) 317 nm, 529 nm, 741 nm

Accepted Answer

The answer of A film of thickness 106 nm soap...

Question 20

What is the minimum thickness of a nonreflecting film coating (n = 1.30) on a glass lens (n = 1.50) for wavelength 500 nm?
A) 250 nm
B) 221 nm
C) 192 nm
D) 167 nm
E) 96.2 nm

Accepted Answer

The minimum thickness for destructive interference in a nonreflecting film is given by $ t = \frac{\lambda}{4n} $. For a wavelength of 500 nm and film index of refraction $ n = 1.30 $, the thickness is $ t = \frac{500}{4 \times 1.30} \approx 96.2 $ nm.

Quiz 28: Physical Optics: Interference and Diffraction

If the wavelength of light in vacuum is 469 nm, what is the wavelength of light in a thin film of index of refraction 1.42?

The distance between the slits in the two-slit experiment is 1.50 × 10^-5 m. A beam of light of wavelength 600 nm is used and the distance between the slit and the screen is 2.00 m. What is the distance on the screen between the central bright fringe and the 4th order bright fringe?

450 nm of light falls on a single slit of width 0.30 mm. What is the angular width of the central diffraction peak?

In a single-slit diffraction experiment, the width of the slit is 1.90 μm. If a beam of light of wavelength 610 nm forms a diffraction pattern, what is the angle associated with the second dark fringe above the central bright fringe?

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?

The wavelength of light in a thin film is 360 nm and the wavelength of light in vacuum is 469 nm. What is the index of refraction for the film?

A soap bubble has an index of refraction of 1.33. What minimum thickness of this bubble will ensure maximum reflectance of normally incident 530 nm wavelength light?

In a single-slit diffraction experiment, the width of the slit is 1.90 μm. If a beam of light of wavelength 610 nm forms a diffraction pattern, what is the angle associated with the first minimum above the central bright fringe?

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

A 500 nm wavelength light illuminates a soap film with an index of refraction 1.33 to make it look bright. If the beam of light is incident normal on the film, what is the minimum thickness of the film?

What is the thickness of a soap bubble film (n = 1.38) that results in a constructive interference in the reflected light if this film is illuminated by a beam of light of wavelength 610 nm?

A sodium light (λ = 589.3 nm) is used to view a soap film to make it look black when directed perpendicular to the film. What is the minimum thickness of the soap film if the index of refraction of soap solution is 1.38?

In a single-slit diffraction experiment, a beam of monochromatic light passes through a slit of width 11.0 μm. If the first order dark fringe of the resulting diffraction pattern is at an angle of 4.31°, what is the wavelength of light?

A 500 nm light is reflected constructively from a thin layer of oil of index of refraction 1.25, which rests on the top of water of index of refraction 1.33. What is the minimum thickness of the layer of oil?

A film of thickness 106 nm soap bubble film (n = 1.42) 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?

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