Question 1

Three experiments involving a thin film (in air) are shown. If t denotes the film thickness and $\lambda$ denotes the wavelength of the light in the film, which experiments will produce constructive interference as seen by the observer?

Accepted Answer

In thin film interference, the condition for constructive interference is given by:

2t = mλ/(n_eff)

where 
t = thickness of the film 
m = an integer (1, 2, 3, ...) 
λ = wavelength of light in vacuum 
n_eff = effective refractive index of the film (depends on the refractive indices of the film and the surrounding medium)

Let's analyze each experiment:

Experiment I: The film has a thickness of 1/4λ in air. The light travels through air (n = 1) and then through the film, which has a refractive index greater than 1. Therefore, the effective refractive index is greater than 1.

Thus, for constructive interference, we need:

2t = mλ/(n_eff)

2(1/4λ) = mλ/n_eff

m = 2

Therefore, only the second order maxima (m = 2) will be observed, and the answer is "I only".

Experiment II: The film has a thickness of λ/2 in air. Following the same logic as above, we get:

2t = mλ/(n_eff)

2(λ/2) = mλ/n_eff

m = 2

Therefore, only the second order maxima (m = 2) will be observed, and the answer is "II only".

Experiment III: The film has a thickness of λ/4 in air. The light first travels through air, then through the film, which has a refractive index less than 1. Therefore, the effective refractive index is less than 1.

Thus, for constructive interference, we need:

2t = mλ/(n_eff)

2(λ/4) = mλ/n_eff

m = 2, 4, 6, ...

Therefore, the second, fourth, sixth, etc. order maxima will be observed, and the answer is "I and III only".

Overall, only experiments I and III will produce constructive interference, so the best answer is D.

Question 2

If two light waves are coherent:

Accepted Answer

Coherence refers to the relationship between the phase difference of two waves. If the phase difference between two waves is constant, then they are considered coherent. The amplitudes, frequencies, and wavelengths of the waves do not have to be the same for them to be coherent.

Question 3

A glass (n = 1.6) lens is coated with a thin film (n = 1.3) to reduce reflection of certain incident light. If $\lambda$is the wavelength of the light in the film, the least film thickness is:

Accepted Answer

To reduce reflection of certain incident light, the thin film must be of such thickness that the light reflected from the top surface and the bottom surface of the film combines destructively. This means that the path difference between the two reflected beams should be λ/2, where λ is the wavelength in the film.

Let t be the thickness of the film. Then, using the formula for the optical path difference due to a medium of refractive index n,
2nt = (m+1/2)λ

where m is an integer. Since the film is thin, we can take m = 0. Also, the wavelength of the light in the glass is given by λ/n, where n is the refractive index of the glass. Therefore, we have
2nt/n = λ/2
or t = λ/4n

Substituting the given values, we get
t = F1F1F1S1F1F1F10/4

Therefore, the least film thickness is F1F1F1S1F1F1F10/4. Answer choice B is correct.

Question 4

One of the two slits in a Young's experiment is painted over so that it transmits only one-half the intensity of the other slit. As a result:

Accepted Answer

The intensity of light from one slit has been reduced to half of that from the other slit. Hence, the pattern will not be symmetrical, and the bright fringes will occur where the intensity of light from both slits is significant while the dark fringes will occur where the intensity of light differs greatly between the two slits. Therefore, the dark fringes get brighter, and the bright ones get darker.

Question 5

In a thin film experiment, a wedge of air is used between two glass plates. If the wavelength of the incident light in air is 480 nm, how much thicker is the air wedge at the 16th dark fringe than it is at the 6th?

Accepted Answer

The thickness difference between two dark fringes in a thin film interference pattern is given by the formula: $ \Delta t = \frac{m \lambda}{2} $, where $ m $ is the fringe order difference. For the 16th and 6th dark fringes, $ m = 16 - 6 = 10 $. Thus, $ \Delta t = \frac{10 \times 480 \, \text{nm}}{2} = 2400 \, \text{nm} $.

Question 6

An air wedge is formed from two glass plates which are in contact at their left edges. There are ten dark bands when viewed by reflection using monochromatic light. The left edge of the top plate is now slowly lifted until the plates are parallel. During this process:

Accepted Answer

The answer of An air wedge is formed from two...

Question 7

A soap film is illuminated by white light normal to its surface. The index of refraction of the film is 1.50. Wavelengths of 480 nm and 800 nm and no wavelengths between are be intensefied in the reflected beam. The thickness of the film is:

Accepted Answer

The answer of A soap film is illuminated by white...

Question 8

A light wave with an electric field amplitude of 2E₀ and a phase constant of zero is to be combined with one of the following waves. Which of these combinations produces the least intensity?

Accepted Answer

The answer of A light wave with an electric field...

Question 9

In a Young's experiment, it is essential that the two beams:

Accepted Answer

The two beams in a Young's experiment need to come originally from the same source in order to create interference patterns. If they are from different sources or created differently, they will not produce the desired interference pattern. The intensity, parallelism, and equal distances of the beams can affect the resulting pattern, but they are not essential for the experiment to work. The wavelength and frequency of the light source are also important factors.

Question 10

Monochromatic light, at normal incidence, strikes a thin film in air. If $\lambda$ denotes the wavelength in the film, what is the thinnest film in which the reflected light will be a maximum?

Accepted Answer

When the reflected waves from both surfaces of the thin film interfere constructively, the light is strongest. This happens when the path difference between the reflected waves is an integer multiple of the wavelength in the film, given by 2nt, where n is the refractive index of the film and t is the thickness of the film. The thinnest film for maximum reflection occurs when the path difference is one wavelength in the film, i.e. 2nt = λ. Rearranging, we get t = λ/2n. According to the question, the wavelength in the film is F1F1F1S1F1F1F10. Therefore, the thinnest film is (F1F1F1S1F1F1F10)/2n. As we are dealing with normal incidence in air, the refractive index of the film is simply its own refractive index, so n = 1. Therefore, the thinnest film is F1F1F1S1F1F1F10/2. Answer choice B is the closest to this value.

Question 11

In a Young's double-slit experiment, a thin sheet of mica is placed over one of the two slits. As a result, the center of the fringe pattern (on the screen) shifts by an amount corresponding to 30 dark bands. The wavelength of the light in this experiment is 480 nm and the index of the mica is 1.60. The mica thickness is:

Accepted Answer

The shift in the fringe pattern is due to the additional optical path length introduced by the mica. The optical path difference is given by (n - 1)t, where n is the refractive index and t is the thickness. This path difference corresponds to 30 dark bands, or 30 wavelengths. Therefore, (1.60 - 1)t = 30 * 480 nm. Solving for t gives t = 0.024 mm.

Question 12

Yellow light is viewed by reflection from a thin vertical soap film. Let $\lambda$ be the wavelength of the light within the film. Why is there a large dark space at the top of the film?

Accepted Answer

The large dark space, known as a dark fringe, occurs due to destructive interference of the reflected light waves. At the top of the film, the reflected light waves from both the top and bottom surfaces of the film are out of phase with each other. This means that the light waves are canceling each other out, creating a dark fringe. The phase change of 180 degrees occurs when the reflected wave undergoes a change in direction due to reflection at the top surface of the film, resulting in a path difference of half a wavelength between the two reflected waves. This choice is the best because it provides a direct and accurate explanation for the occurrence of the dark fringe.

Question 13

Two point sources, vibrating in phase, produce an interference pattern in a ripple tank. If the frequency is increased by 20%, the number of nodal lines:

Accepted Answer

The answer of Two point sources, vibrating in phase, produce...

Question 14

A light wave with an electric field amplitude of E₀ and a phase constant of zero is to be combined with one of the following waves. Which of these combinations produces the greatest intensity?

Accepted Answer

The greatest intensity is produced when the waves are in phase and have the largest amplitude. Wave C has an amplitude of 2E₀ and a phase constant of zero, meaning it is in phase with the original wave, resulting in constructive interference and maximum intensity.

Question 15

An air wedge is formed using two glass plates that are in contact along their left edge. When viewed by highly monochromatic light, there are exactly 4001 dark bands in the reflected light. The air is now evacuated (with the glass plates remaining rigidly fixed) and the number of dark bands decreases to exactly 4000. The index of refraction of the air is:

Accepted Answer

The number of dark fringes in the reflected light, $N$, is given by:

$$N = \frac{2et}{\lambda}$$

Where $e$ is the thickness of the air wedge at its thickest point (which is the right edge when viewed from above), $t$ is the distance between the two glass plates, and $\lambda$ is the wavelength of the light.

When the air is evacuated, the thickness of the air wedge reduces to zero, hence the new number of fringes, $N'$, is given by:

$$N'=\frac{2t}{\lambda}$$

We can now use these two equations to find the index of refraction of the air. Let $n$ be the index of refraction of the air. Then we have:

$$N' = N\frac{n+1}{2n}$$

Substituting the given values:

$$4000 = 4001\frac{n+1}{2n}$$

Solving for $n$ gives:

$$n = \frac{1}{(4001/4000)-1} = 1.00025$$

Therefore, the answer is C.

Question 16

Binoculars and microscopes are frequently made with coated optics by adding a thin layer of transparent material to the lens surface as shown. One wants:

Accepted Answer

The coated optics in binoculars and microscopes work by creating constructive interference between the waves reflected off the thin layer of transparent material and the waves transmitted through the lens. This results in a brighter and clearer image for the viewer. Option A is incorrect because it mentions constructive interference between waves 1 and 2, which are not relevant in this context. Option B is incorrect because destructive interference would result in a dimmer and less clear image, which is not desirable for binoculars or microscopes. Option D is incorrect because the coating is meant to reduce reflections and increase light transmission, not to be more transparent than the lens. Option E is also incorrect because the speed of light in the coating should be greater than or equal to that in the lens to avoid any bending or distortion of the light.

Question 17

A lens with a refractive index of 1.5 is coated with a material of refractive index 1.2 in order to minimize reflection. If $\lambda$denotes the wavelength of the incident light in air, what is the thinnest possible such coating?

Accepted Answer

The answer of A lens with a refractive index of...

Question 18

A liquid of refractive index n = 4/3 replaces the air between a fixed wedge formed from two glass plates as shown. As a result, the spacing between adjacent dark bands in the interference pattern:

Accepted Answer

The answer of A liquid of refractive index n =...

Question 19

Red light is viewed through a thin vertical soap film. At the third dark area shown, the thickness of the film, in terms of the wavelength within the film, is:

Accepted Answer

The answer of Red light is viewed through a thin...

Question 20

To obtain an observable double-slit fringe pattern:

Accepted Answer

For a double-slit interference pattern to be visible, the light must be coherent, which means that the waves are in phase with each other. Incoherent light produces a random distribution of photons that does not interfere with itself constructively or destructively. The other choices (A, B, C, and E) are not necessary for an observable interference pattern but may affect the quality or intensity of the pattern.

Quiz 35: Interference

If two light waves are coherent:

A glass (n = 1.6) lens is coated with a thin film (n = 1.3) to reduce reflection of certain incident light. If $\lambda$ is the wavelength of the light in the film, the least film thickness is:

One of the two slits in a Young's experiment is painted over so that it transmits only one-half the intensity of the other slit. As a result:

In a thin film experiment, a wedge of air is used between two glass plates. If the wavelength of the incident light in air is 480 nm, how much thicker is the air wedge at the 16th dark fringe than it is at the 6th?

An air wedge is formed from two glass plates which are in contact at their left edges. There are ten dark bands when viewed by reflection using monochromatic light. The left edge of the top plate is now slowly lifted until the plates are parallel. During this process:

A soap film is illuminated by white light normal to its surface. The index of refraction of the film is 1.50. Wavelengths of 480 nm and 800 nm and no wavelengths between are be intensefied in the reflected beam. The thickness of the film is:

A light wave with an electric field amplitude of 2E₀ and a phase constant of zero is to be combined with one of the following waves. Which of these combinations produces the least intensity?

In a Young's experiment, it is essential that the two beams:

Monochromatic light, at normal incidence, strikes a thin film in air. If $\lambda$ denotes the wavelength in the film, what is the thinnest film in which the reflected light will be a maximum?

Two point sources, vibrating in phase, produce an interference pattern in a ripple tank. If the frequency is increased by 20%, the number of nodal lines:

A light wave with an electric field amplitude of E₀ and a phase constant of zero is to be combined with one of the following waves. Which of these combinations produces the greatest intensity?

Binoculars and microscopes are frequently made with coated optics by adding a thin layer of transparent material to the lens surface as shown. One wants:

To obtain an observable double-slit fringe pattern:

Quiz 35: Interference

Three experiments involving a thin film (in air) are shown. If t denotes the film thickness and λ\lambdaλ denotes the wavelength of the light in the film, which experiments will produce constructive interference as seen by the observer?

If two light waves are coherent:

A glass (n = 1.6) lens is coated with a thin film (n = 1.3) to reduce reflection of certain incident light. If λ\lambdaλ is the wavelength of the light in the film, the least film thickness is:

One of the two slits in a Young's experiment is painted over so that it transmits only one-half the intensity of the other slit. As a result:

In a thin film experiment, a wedge of air is used between two glass plates. If the wavelength of the incident light in air is 480 nm, how much thicker is the air wedge at the 16th dark fringe than it is at the 6th?

An air wedge is formed from two glass plates which are in contact at their left edges. There are ten dark bands when viewed by reflection using monochromatic light. The left edge of the top plate is now slowly lifted until the plates are parallel. During this process:

A soap film is illuminated by white light normal to its surface. The index of refraction of the film is 1.50. Wavelengths of 480 nm and 800 nm and no wavelengths between are be intensefied in the reflected beam. The thickness of the film is:

A light wave with an electric field amplitude of 2E0 and a phase constant of zero is to be combined with one of the following waves. Which of these combinations produces the least intensity?

In a Young's experiment, it is essential that the two beams:

Monochromatic light, at normal incidence, strikes a thin film in air. If λ\lambdaλ denotes the wavelength in the film, what is the thinnest film in which the reflected light will be a maximum?

Yellow light is viewed by reflection from a thin vertical soap film. Let λ\lambdaλ be the wavelength of the light within the film. Why is there a large dark space at the top of the film?

Two point sources, vibrating in phase, produce an interference pattern in a ripple tank. If the frequency is increased by 20%, the number of nodal lines:

A light wave with an electric field amplitude of E0 and a phase constant of zero is to be combined with one of the following waves. Which of these combinations produces the greatest intensity?

Binoculars and microscopes are frequently made with coated optics by adding a thin layer of transparent material to the lens surface as shown. One wants:

A lens with a refractive index of 1.5 is coated with a material of refractive index 1.2 in order to minimize reflection. If λ\lambdaλ denotes the wavelength of the incident light in air, what is the thinnest possible such coating?

A liquid of refractive index n = 4/3 replaces the air between a fixed wedge formed from two glass plates as shown. As a result, the spacing between adjacent dark bands in the interference pattern:

Red light is viewed through a thin vertical soap film. At the third dark area shown, the thickness of the film, in terms of the wavelength within the film, is:

To obtain an observable double-slit fringe pattern:

A glass (n = 1.6) lens is coated with a thin film (n = 1.3) to reduce reflection of certain incident light. If $\lambda$ is the wavelength of the light in the film, the least film thickness is:

A light wave with an electric field amplitude of 2E₀ and a phase constant of zero is to be combined with one of the following waves. Which of these combinations produces the least intensity?

Monochromatic light, at normal incidence, strikes a thin film in air. If $\lambda$ denotes the wavelength in the film, what is the thinnest film in which the reflected light will be a maximum?

A light wave with an electric field amplitude of E₀ and a phase constant of zero is to be combined with one of the following waves. Which of these combinations produces the greatest intensity?