Question 1

A 360-nm thick oil film floats on the surface of a pool of water. The indices of refraction of the oil and the water are 1.50 and 1.33, respectively. When the surface of the oil is illuminated from above at normal incidence with white light, what is the wavelength of light between 400 nm to 800 nm wavelength that is most weakly reflected?
A) 520 nm
B) 540 nm
C) 560 nm
D) 580 nm
E) 600 nm

Accepted Answer

The wavelength of light that is most weakly reflected is the one for which the reflection from the top surface of the oil film and the bottom surface of the oil film are completely out of phase, leading to destructive interference. We can use the formula for the thickness of a film producing destructive interference for a given wavelength:

t = (m + 1/2) * lambda / (2 * n)

where t is the thickness of the film, m is an integer representing the order of interference, lambda is the wavelength of light, and n is the index of refraction of the film.

For white light, we can assume that the wavelength of the most weakly reflected light is somewhere in the middle of the visible spectrum, around 600 nm. Plugging in t = 360 nm, n = 1.50, and lambda = 600 nm, we get:

360 nm = (m + 1/2) * 600 nm / (2 * 1.50)

Solving for m gives us:

m = 2

This means that the most weakly reflected wavelength is the one for which the path difference between the reflected waves from the top and bottom surfaces of the oil film is equal to two wavelengths. Using the formula for the path difference:

delta = 2 * n * t

where delta is the path difference, n is the index of refraction of the film, and t is the thickness of the film, we can find the most weakly reflected wavelength to be:

delta = 2 * 1.50 * 360 nm = 1080 nm

This path difference corresponds to a wavelength of:

lambda = delta / 2 = 540 nm

Therefore, the most weakly reflected wavelength is 540 nm, which corresponds to choice B.

Question 2

A 500-nm-wavelength beam of light illuminates a soap film with air on both sides and an index of refraction 1.33. If the beam of light is incident normally on the film, what is the minimum thickness of the film to make the reflected light enhanced in brightness?
A) 47.0 nm
B) 74.0 nm
C) 94.0 nm
D) 152 nm
E) 268 nm

Accepted Answer

For a soap film with air on both sides and an index of refraction 1.33, the minimum thickness for constructive interference (i.e. maximum brightness of reflected light) can be found using the formula: 
2nt = mλ, where n is the refractive index of the soap film (1.33), t is the thickness of the film, m is an integer representing the order of interference, and λ is the wavelength of the incident light (500 nm). Since the problem asks for the minimum thickness to achieve maximum brightness, we can use the first order of interference (m = 1). 
Thus, we have 2(1.33)t = (1)500 nm, which gives us t = 94.0 nm. Therefore, the correct answer is choice C.

Question 3

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

Accepted Answer

The formula for the minimum thickness of a soap bubble for maximum reflectance of normally-incident light is:

t = λ/(2n)

where t is the thickness of the bubble, λ is the wavelength of the incident light, and n is the index of refraction of the soap bubble.

Substituting the given values, we get:

t = (530 nm)/(2 x 1.33) = 99.6 nm

Therefore, the minimum thickness of the soap bubble for maximum reflectance of normally-incident light at a wavelength of 530 nm is 99.6 nm, which is closest to option C, 99.6 nm.

Question 4

A 360-nm thick oil film floats on the surface of a pool of water water. The indices of refraction of the oil and the water are 1.5 and 1.33, respectively. When the surface of the oil is illuminated from above at normal incidence with white light, what are the two wavelengths of light between 400 nm to 800 nm wavelength that are most strongly reflected?
A) 410 nm and 700 nm
B) 430 nm and 720 nm
C) 450 nm and 740 nm
D) 470 nm and 760 nm
E) 490 nm and 780 nm

Accepted Answer

For constructive interference to occur in reflected light, the path length difference between two reflected rays should be an integer multiple of the wavelength of light. This condition can be written as:

2ntcosθ = mλ

where n is the refractive index of the medium, t is the thickness of the film, θ is the angle of incidence, m is the order of the interference fringe, and λ is the wavelength of light.

At normal incidence, θ = 0 and cosθ = 1, so the equation becomes:

2nt = mλ

For white light, we need to find the wavelengths that reflect most strongly at different points in the film. We can use the equation above to calculate the wavelengths that correspond to different orders of interference. The first few orders are:

m = 1: λ1 = 2nt
m = 2: λ2 = nt
m = 3: λ3 = 2nt/3
m = 4: λ4 = nt/2
m = 5: λ5 = 2nt/5
m = 6: λ6 = nt/3

For a film of 360 nm thickness, we have:

λ1 = 2 × 1.5 × 360 nm = 1080 nm
λ2 = 1.5 × 360 nm = 540 nm
λ3 = 2 × 1.5 × 360 nm / 3 = 720 nm
λ4 = 1.5 × 360 nm / 2 = 270 nm
λ5 = 2 × 1.5 × 360 nm / 5 = 432 nm
λ6 = 1.5 × 360 nm / 3 = 180 nm

Therefore, the two wavelengths of light that are most strongly reflected lie between λ2 and λ3, i.e. between 540 nm and 720 nm. The closest choices are B (430 nm and 720 nm) and C (450 nm and 740 nm). Out of these, the correct answer is B because it includes one wavelength from the lower end (between 400 nm and 540 nm) and one wavelength from the upper end (between 720 nm and 800 nm) of the visible spectrum.

Question 5

An oil film with refractive index 1.48 and thickness 290 nm is floating on water and illuminated with white light at normal incidence. What is the wavelength of the dominant color in the reflected light?
A) green (541 nm)
B) blue-green (493 nm)
C) violet (404 nm)
D) yellow (572 nm)
E) blue (470 nm)

Accepted Answer

The answer of An oil film with refractive index 1.48...

Question 6

Light in air of wavelength 500 nm illuminates a soap film that has an index of refraction of 1.3 and air on both sides. What is the minimum thickness of this film that will produce cancellation in the reflected light when the light is incident normally on the film?

Accepted Answer

The answer of Light in air of wavelength 500 nm...

Question 7

A glass plate that is 2.5 cm long is separated from another glass plate at one end by a strand of someone's hair that is 0.010 mm in diameter. How far apart are the adjacent interference bands when viewed with light of wavelength 600 nm?

Accepted Answer

The answer of A glass plate that is 2.5 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=fu/f, 
where n is the index of refraction, f is the frequency of light in vacuum, and fu is the frequency of light in the medium (in this case, thin film).
Using the given wavelengths, we can find the frequency: 
f=c/λ=3.00x10^8/469x10^-9=6.40x10^14 Hz
fu=c/λu=3.00x10^8/360x10^-9=8.33x10^14 Hz
Substituting these values into the formula, we get:
n=fu/f=8.33x10^14 Hz / 6.40x10^14 Hz = 1.30
Therefore, the index of refraction for the film is 1.30. 
The best choice is B.

Question 9

Light is incident perpendicularly from air onto a liquid film that is on a glass plate. The liquid film is $199 \mathrm {~nm}$ thick, and the liquid has index of refraction 1.60. The glass has index of refraction 1.50. Calculate the longest visible wavelength (as measured in air) of the light for which there will be totally destructive interference between the rays reflected from the top and bottom surfaces of the film. Assume that the visible spectrum lies between 400 nm and 700 nm.

Accepted Answer

The answer of Light is incident perpendicularly from air onto...

Question 10

Light has wavelength 600 nm in a vacuum. It passes into glass, which has an index of refraction of 1.5. What is the frequency of the light inside the glass? (c = 3.0 × 10⁸ m/s) A) 3.3 × 10¹⁴ Hz B) 5.0 × 10¹⁴ Hz C) 3.3 × 10⁵ Hz D) 5.0 × 10⁵ Hz

Accepted Answer

The frequency of light does not change when it passes from one medium to another. The frequency $f$ can be calculated using the formula $f = \frac{c}{\lambda}$, where $c$ is the speed of light in vacuum and $\lambda$ is the wavelength in vacuum. Given $c = 3.0 \times 10^{8} \, \text{m/s}$ and $\lambda = 600 \, \text{nm} = 600 \times 10^{-9} \, \text{m}$, $f = \frac{3.0 \times 10^{8}}{600 \times 10^{-9}} = 5.0 \times 10^{14} \, \text{Hz}$.

Question 11

A soap bubble, when illuminated at normal incidence with light of 463 nm, appears to be especially reflective. If the index of refraction of the film is 1.35, what is the minimum thickness the soap film can be if it is surrounded by air?

Accepted Answer

The answer of A soap bubble, when illuminated at normal...

Question 12

A light beam with wavelength 500 nm is reflected constructively from a thin layer of oil having index of refraction 1.25. The oil floats 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 to occur, the path difference between the two reflected rays must be an integer multiple of the wavelength. In this case, we have:

2nt = mλ

where n is the refractive index of the oil, t is the thickness of the oil layer, m is an integer, and λ is the wavelength of the light. We want to find the minimum value of t that satisfies this condition.

Using Snell's law, we can find the angle of incidence at the air/oil interface:

sin θ1 = n2 sin θ2
sin θ1 = (1.33/1.25) sin 90° = 1.064
θ1 = 44.1°

Since the oil layer is thin, we can assume that the angle of reflection at the oil/water interface is approximately equal to the angle of incidence at the air/oil interface. Thus, we have:

2t = (m + 1/2)λ/ sin θ1

For the minimum value of t, we take m = 0. Thus,

2t = λ/2 sin θ1

t = λ/4 sin θ1

Plugging in the values, we get:

t = (500 nm)/(4 sin 44.1°) ≈ 201 nm

Therefore, the minimum thickness of the layer of oil is 200 nm, which corresponds to choice D.

Question 13

A sodium light, having wavelength 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, having wavelength 589.3 nm,...

Question 14

A soap bubble with air on both sides has an index of refraction of 1.33. What minimum thickness of the wall of this bubble will ensure maximum reflectance of normally incident light having a wavelength of 530 nm?

Accepted Answer

The answer of A soap bubble with air on both...

Question 15

What is the minimum thickness of a soap bubble film with refractive index 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

The condition for constructive interference in reflected light from soap film is given by 2nt = mλ, where t is the thickness of the film, n is the refractive index of the film, m is an integer, and λ is the wavelength of the incident light.

Solving for t, we get t = mλ/2n.

We are given n = 1.38, λ = 610 nm, and we want m = 1 for constructive interference.

So, t = (1 x 610 nm)/(2 x 1.38) = 111 nm.

Therefore, the minimum thickness of the soap bubble film is 111 nm. The answer is B.

Question 16

A metallic sheet has a large number of parallel slits, 5.0 mm wide and 20 cm apart. This sheet is used as a diffraction grating for microwaves. A wide parallel beam of microwaves is incident normally on the sheet. What is the smallest microwave frequency for which only the central maximum will occur? (c = 3.00 × 10⁸ m/s) A) 0.50 GHz B) 0.70 GHz C) 1.0 GHz D) 1.5 GHz E) 2.0 GHz

Accepted Answer

The answer of A metallic sheet has a large number...

Question 17

A metallic sheet has a large number of parallel slits, 5.0 mm wide and 20 cm apart. This sheet is used as a diffraction grating for microwaves. A wide parallel beam of microwaves is incident normally on the sheet. If the intensity maxima occur 2.0° apart in the central region, what is the wavelength of the microwaves?
A) 5.0 mm
B) 6.0 mm
C) 7.0 mm
D) 8.0 mm
E) 9.0 mm

Accepted Answer

The distance between the slits, d = 20 cm = 0.2 m, and the width of each slit, a = 5.0 mm = 5.0 x 10^-3 m. The angle between the intensity maxima, θ = 2.0° = 2.0 x π/180 radians. The distance between the central maximum and the first-order maximum is given by:
sinθ = λ/d
λ = d sinθ
λ = 0.2 x sin(2.0° x π/180)
λ = 7.02 x 10^-3 m
λ = 7.0 mm (to two significant figures)

Therefore, the wavelength of the microwaves is 7.0 mm and the best choice is C.

Question 18

A metallic sheet has a large number of parallel slits, 5.0 mm wide and 20 cm apart. This sheet is used as a diffraction grating for microwaves. A wide parallel beam of microwaves is incident normally on the sheet. If the microwave wavelength is 6.0 cm, what is the largest angle from the normal, at which an intensity maximum occurs?
A) 64°
B) 69°
C) 74°
D) 79°
E) 84°

Accepted Answer

The answer of A metallic sheet has a large number...

Question 19

A diffraction grating has 450 lines per millimeter. What is the highest order m that contains the entire visible spectrum from 400 nm to 700 nm?
A) m = 2
B) m = 3
C) m = 4
D) m = 5
E) m = 6

Accepted Answer

The answer of A diffraction grating has 450 lines per...

Question 20

Light has a wavelength of 600 nm in a vacuum. It passes into glass, which has an index of refraction of 1.50. What is the wavelength of the light in the glass?
A) 600 nm
B) 500 nm
C) 400 nm
D) 300 nm
E) 900 nm

Accepted Answer

The relationship between the wavelength of light in vacuum and the wavelength of light in a medium is given by:

n = λ_vacuum / λ_medium

where n is the refractive index of the medium, λ_vacuum is the wavelength of light in vacuum and λ_medium is the wavelength of light in the medium.

Rearranging the equation to solve for λ_medium, we get:

λ_medium = λ_vacuum / n

Substituting the values given, we get:

λ_medium = 600 nm / 1.5 = 400 nm

Therefore, the wavelength of the light in the glass is 400 nm.

Quiz 17: Wave Optics

A 500-nm-wavelength beam of light illuminates a soap film with air on both sides and an index of refraction 1.33. If the beam of light is incident normally on the film, what is the minimum thickness of the film to make the reflected light enhanced in brightness?

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

An oil film with refractive index 1.48 and thickness 290 nm is floating on water and illuminated with white light at normal incidence. What is the wavelength of the dominant color in the reflected light?

Light in air of wavelength 500 nm illuminates a soap film that has an index of refraction of 1.3 and air on both sides. What is the minimum thickness of this film that will produce cancellation in the reflected light when the light is incident normally on the film?

A glass plate that is 2.5 cm long is separated from another glass plate at one end by a strand of someone's hair that is 0.010 mm in diameter. How far apart are the adjacent interference bands when viewed with light of wavelength 600 nm?

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?

Light has wavelength 600 nm in a vacuum. It passes into glass, which has an index of refraction of 1.5. What is the frequency of the light inside the glass? (c = 3.0 × 108 m/s)

A soap bubble, when illuminated at normal incidence with light of 463 nm, appears to be especially reflective. If the index of refraction of the film is 1.35, what is the minimum thickness the soap film can be if it is surrounded by air?

A light beam with wavelength 500 nm is reflected constructively from a thin layer of oil having index of refraction 1.25. The oil floats on the top of water of index of refraction 1.33. What is the minimum thickness of the layer of oil?

A sodium light, having wavelength 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 soap bubble with air on both sides has an index of refraction of 1.33. What minimum thickness of the wall of this bubble will ensure maximum reflectance of normally incident light having a wavelength of 530 nm?

What is the minimum thickness of a soap bubble film with refractive index 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 diffraction grating has 450 lines per millimeter. What is the highest order m that contains the entire visible spectrum from 400 nm to 700 nm?

Light has a wavelength of 600 nm in a vacuum. It passes into glass, which has an index of refraction of 1.50. What is the wavelength of the light in the glass?

Light has wavelength 600 nm in a vacuum. It passes into glass, which has an index of refraction of 1.5. What is the frequency of the light inside the glass? (c = 3.0 × 10⁸ m/s)