Question 1

The pattern of light and dark fringes formed in Young's double-slit experiment is due to
A) interference and dispersion.
B) diffraction and refraction.
C) refraction and interference.
D) refraction and dispersion.
E) interference and diffraction.

Accepted Answer

The pattern of light and dark fringes formed in Young's double-slit experiment is due to interference, which is the result of the superposition of waves from the two slits, and diffraction, which is the bending of light around the edges of the slits. Therefore, option E, which includes both interference and diffraction, is the best choice. The other options involve refraction and/or dispersion, which are not relevant to this experiment.

Question 2

The graphs are plots of relative intensities of various diffraction patterns versus the sine of the angle from the central maximum. The graph that represents the diffraction pattern from the widest single slit is
A) 1
B) 2
C) 3
D) 4
E) 5

Accepted Answer

The answer of  The graphs are plots of relative...

Question 3

Five coherent sources are used to produce an interference pattern. The phasor diagram shown could be used to calculate the intensity of the
A) first minimum in the interference pattern.
B) second maximum in an interference pattern.
C) first maximum in an interference pattern.
D) second minimum in an interference pattern.
E) None of these is correct.

Accepted Answer

The phasor diagram shows that the five sources are in phase and have equal amplitudes. This means that the interference between them will result in a maximum intensity at the point where all the waves add up constructively, which occurs at the first maximum in the interference pattern. Therefore, choice C is correct. The other options refer to points in the interference pattern where the waves interfere destructively, which would not be the case for five coherent sources in phase.

Question 4

The interference and diffraction envelopes of a double slit are shown separately but to the same scale in the figure. For this arrangement, the number of fringes in the second diffraction maximum is
A) 3
B) 4
C) 5
D) 7
E) 0

Accepted Answer

In the second diffraction maximum, the central maximum of the diffraction envelope coincides with the first minimum of the interference envelope. Thus, the distance from the center of the pattern to the first minimum is equal to the distance from the center of the pattern to the second maximum. Counting the fringes from the first maximum, we see that there are 4 fringes between the first minimum and the central maximum, and 3 fringes between the central maximum and the second maximum. Therefore, the total number of fringes in the second diffraction maximum is 4+3=7, which corresponds to answer choice D.

Question 5

Which of the phasor diagrams shows the first minimum for five equally spaced in-phase sources?
A) 1
B) 2
C) 3
D) 4
E) 5

Accepted Answer

The phasor diagram in option C shows the first minimum for five equally spaced in-phase sources. The sources would be cancelling each other out when their respective phasors are at 180 degrees to each other, which is shown in option C. In options A, B and E, the phasors are not at the required angle to show complete cancellation, while in option D, there are only four phasors instead of five.

Question 6

Two slits of width a = 0.020 mm are separated by a distance d = 0.05 mm and illuminated by light of wavelength $\lambda$ = 500 nm. The number of bright fringes seen in the central maximum is
A) 2.5
B) 3
C) 4
D) 5
E) 6

Accepted Answer

The number of bright fringes in the central maximum is given by the formula (2a/d)*(λ/D), where D is the distance between the slits and the screen. In this case, D is not given but is assumed to be large enough to consider it as infinity. Therefore, the formula simplifies to (2a/d)*(λ), which equals (2*0.020 mm/0.05 mm)*(500 nm) = 4. Therefore, the answer is (C) 4 bright fringes.

Question 7

In dealing with the diffraction pattern of a single slit, we are usually interested in the location of the first minimum in light intensity because
A) it is the only one that can be determined accurately.
B) it is the only one for which the small-angle approximation holds.
C) nearly all the light energy is contained in the central maximum.
D) it is the only one for which the location is directly proportional to the wavelength of the light.
E) it is the only one that can be described by Fraunhofer diffraction.

Accepted Answer

The central maximum contains nearly all of the light energy and thus is the most important feature of the diffraction pattern. The location of the first minimum can be determined accurately and is related to the width of the slit, but it is not the only significant feature of the pattern. The small-angle approximation is used to determine the location of the minima and maxima in the diffraction pattern, but it applies to all of the features, not just the first minimum. The location of the minima is proportional to the wavelength of light, but again, this is true for all of the minima, not just the first one. Fraunhofer diffraction is a type of diffraction pattern that occurs when the source is far from the diffracting aperture, but it is not relevant to the importance of the first minimum in a single slit diffraction pattern.

Question 8

The bending of light around an obstacle such as the edge of a slit is called
A) diffraction
B) dispersion
C) reflection
D) refraction
E) polarization

Accepted Answer

Diffraction is the bending of light around an obstacle or through a narrow aperture such as the edge of a slit. Dispersion refers to the separation of light into its constituent colors. Reflection occurs when light bounces off a surface. Refraction is the bending of light as it passes through a material with a different refractive index. Polarization refers to the orientation of the electric field vector of light waves.

Question 9

Suppose you observe fifteen bright fringes in the central maximum of a double-slit interference pattern. If you know that the separation of the slits d = 0.01 mm, you can conclude that the width of each slit is
A) 0.63 mm
B) 1.25 mm
C) 1.68 mm
D) 1.88 mm
E) 1.99 mm

Accepted Answer

In a double-slit interference pattern, the location of bright fringes is given by the equation:
dsinθ = mλ, where d is the separation of the slits, θ is the angle between the line from the slits to the bright fringe and the central maximum, m is the order of the bright fringe (m = 0 for the central maximum), and λ is the wavelength of light.

For the central maximum (m=0), sinθ = 0, so we have:
dsinθ = 0 = mλ
or λ = 0 (which is not possible)
Thus, for the central maximum, we have the condition:
dsinθ = 0

If we count 15 bright fringes in the central maximum, it means that there are 15 bright fringes on each side of the central maximum (m = ±1, ±2, ..., ±15). Thus, the total number of bright fringes in the double-slit interference pattern is 31 (including the central maximum).

The condition for the first bright fringe on one side of the central maximum (m = 1) is:
dsinθ = λ
The condition for the first bright fringe on the other side of the central maximum (m = -1) is:
dsinθ = -λ

Adding these two equations, we get:
2dsinθ = 0.02 mm

Since sinθ ≈ θ in radians for small angles, we can use:
sinθ ≈ θ = y/L,
where y is the distance from the central maximum to the first bright fringe, and L is the distance from the double-slit to the screen.

Assuming L is much larger than y, we can use the approximation y ≈ Lθ. Substituting this into the previous equation, we get:
2d/L ≈ y/L ≈ θ ≈ sinθ ≈ 0.02/15
or d ≈ 0.00027 m

Now, we can use the condition for the central maximum to find the wavelength:
dsinθ = 0
or λ = 0 (which is not possible)

Thus, we use the condition for the first bright fringe on one side of the central maximum (m = 1):
dsinθ = λ
or d = λ/θ
Substituting θ ≈ sinθ ≈ 0.02/15, we get:
d ≈ λ/θ ≈ 0.01/0.0027 m
or λ ≈ 0.0027d m

Using λ = hc/E, where h is Planck's constant, c is the speed of light, and E is the energy of a photon, we get:
E = hc/λ ≈ 4.63 eV

For visible light, the energy of a photon is typically between 1.65 eV (red light) and 3.26 eV (violet light). Thus, we can assume that the double-slit was illuminated by a laser with a wavelength of about 532 nm (green light).

Finally, we can use the condition for the width of each slit:
d ≈ λ/θ
or w ≈ λL/d
Substituting θ ≈ sinθ ≈ 0.02/15 and L = 1 m, we get:
w ≈ λL/d ≈ 532*10^-9 m * 1 m / 0.01 mm
or w ≈ 1.25 mm

Question 10

Light of wavelength 650 nm is incident on a slit of width 25.0 µm. At what angle is the second diffraction minimum observed?
A) 0.052º
B) 1.5º
C) 2.2º
D) 3.0º
E) 3.7º

Accepted Answer

The answer of Light of wavelength 650 nm is incident...

Question 11

The fringes are the result of
A) diffraction from a single slit.
B) interference from a double slit in addition to diffraction from the two slits.
C) interference from three slits.
D) diffraction from three slits.
E) None of these is correct.

Accepted Answer

The presence of fringes indicates interference, which is a result of the double slit setup. The diffraction from the two slits is also a contributing factor to the overall pattern observed. Option A is incorrect because diffraction from a single slit would produce a different pattern, with a central maximum and alternating dark and bright fringes on either side. Option C is incorrect because interference from three slits would produce a different pattern as well, with more bright fringes but also more dark fringes. Option D is incorrect because diffraction from three slits would also produce a different pattern with more fringes than observed. Therefore, option B is the best choice.

Question 12

Suppose you observe nineteen bright fringes in the central maximum of a double-slit interference pattern. If you know that the separation of the slits d = 0.06 mm, you can conclude that the width of each slit is
A) 0.048 mm
B) 0.024 mm
C) 0.003 mm
D) 0.006 mm
E) 0.012 mm

Accepted Answer

The bright fringes are given by the equation 
y = mλL/d
where y is the distance from the central maximum, m is the order of the fringe, λ is the wavelength of the light, L is the distance from the slits to the screen, and d is the separation of the slits.

In the central maximum, m = 0, so 
y = 0.

Thus, we have 
0 = mλL/d
or 
L = mdλd/2.

Since L is much larger than d, we can assume that 
L ≈ mdλd/2.

For m = 19, we have 
L = 19λd/2.

Now, we can solve for the wavelength of the light using the known value of L and the known value of d: 
L = 19λd/2
0.06 m = 19λ(0.06 mm)/2
λ ≈ 633 nm.

Finally, we can find the width of each slit using the equation 
d sinθ ≈ mλ,
where θ is the angle from the central maximum to the fringe and m is the order of the fringe.

For the first bright fringe, m = 1 and sinθ ≈ θ ≈ y/L ≈ (0.024 mm)/(19λd/2) ≈ 0.0033.

Thus, we have 
d ≈ mλ/sinθ ≈ 1(633 nm)/0.0033 ≈ 0.006 mm.

Therefore, the width of each slit is D) 0.006mm.

Question 13

Which of the following statements concerning the interference pattern described by the phasor diagram is true?
A) There are exactly seven coherent sources.
B) There is a relative minimum at the point in the pattern corresponding to the phasor diagram.
C) The amplitude of the resultant wave at the point in the pattern corresponding to the phasor diagram is less than the amplitudes of the individual waves.
D) The diagram corresponds to the third bright spot in the interference pattern.
E) None of these is correct.

Accepted Answer

The phasor diagram shows that the waves have different amplitudes and are out of phase by π/2. Therefore, at the point in the interference pattern corresponding to the phasor diagram, there will be destructive interference, resulting in a lower amplitude of the resultant wave than the individual waves. Choice A is incorrect as there are only four phasors shown in the diagram. Choice B is incorrect as there is a relative maximum at the point in the pattern corresponding to the phasor diagram. Choice D is incorrect as the phasor diagram does not correspond to any specific bright spot in the pattern.

Question 14

The diffraction pattern of a single slit is shown in the figure. The point at which the path difference of the extreme rays is two wavelengths is
A) 1
B) 2
C) 3
D) 4
E) 5

Accepted Answer

The answer of  The diffraction pattern of a single...

Question 15

Two slits of width a = 0.0016 mm are separated by a distance d = 0.04 mm and illuminated by light of wavelength $\lambda$= 450 nm. The number of bright fringes seen in the central maximum is
A) 8
B) 16
C) 25
D) 37
E) 49

Accepted Answer

The answer of Two slits of width a = 0.0016...

Question 16

A single-slit diffraction pattern is displayed on a screen 0.900 m away from the slit. If the wavelength of the light is 600 nm and the slit is 1.50 * 10^-4 m wide, the distance from the first minimum on the right to the first minimum on the left is A) 0.164 mm B) 1.83 mm C) 3.99 mm D) 7.20 mm E) 7.98 mm

Accepted Answer

The answer of A single-slit diffraction pattern is displayed on...

Question 17

As the width of the slit producing a single-slit diffraction pattern is slowly and steadily reduced (always remaining larger than the wavelength of the light), the diffraction pattern
A) slowly and steadily gets wider.
B) slowly and steadily gets brighter.
C) does not change because the wavelength of the light does not change.
D) slowly and steadily gets narrower.
E) None of these is correct.

Accepted Answer

As the width of the slit decreases, the diffraction pattern becomes wider due to the increased spreading of light waves. This is a fundamental principle of wave optics, where the diffraction pattern's width is inversely proportional to the slit width.

Question 18

When a parallel beam of light is diffracted at a single slit,
A) the shadow is always sharp.
B) the narrower the slit, the narrower the central diffraction maximum.
C) the narrower the slit, the wider the central diffraction maximum.
D) the width of the central diffraction maximum is independent of the width of the slit.
E) None of these is correct.

Accepted Answer

The narrower the slit, the wider the central diffraction maximum. This is because diffraction effects become more pronounced as the size of the aperture approaches the wavelength of the light passing through it, leading to a wider spread of the light on the other side.

Question 19

Light of wavelength 450 nm is incident on a narrow slit. The diffraction pattern is observed on a screen 5.0 m from the slit, and the central maximum is observed to have a width of 22 cm. What is the width of the slit?
A) 4.5 µm
B) 5.0 µm
C) 10 µm
D) 20 µm
E) 0.20 µm

Accepted Answer

The answer of Light of wavelength 450 nm is incident...

Question 20

An easy way to distinguish whether the fringe pattern is the result of a single slit or from a double slit is
A) the intensity for each fringe is much stronger for a single slit than a double slit.
B) the intensity for each fringe is much stronger for a double slit than a single slit.
C) the width central maximum from a single slit is twice as width as the other wide whereas the widths of the fringes from a double slit are about the same width.
D) that there are many more fringes from a single slit than a double slit.
E) that there are many more fringes from a double slit than a single slit.

Accepted Answer

The width of the central maximum for a single slit is approximately twice as wide as the other fringes, whereas the fringes from a double slit are approximately equal in width.

Quiz 13: Interference and Diffraction

The pattern of light and dark fringes formed in Young's double-slit experiment is due to

Five coherent sources are used to produce an interference pattern. The phasor diagram shown could be used to calculate the intensity of the

Which of the phasor diagrams shows the first minimum for five equally spaced in-phase sources?

Two slits of width a = 0.020 mm are separated by a distance d = 0.05 mm and illuminated by light of wavelength $\lambda$ = 500 nm. The number of bright fringes seen in the central maximum is

In dealing with the diffraction pattern of a single slit, we are usually interested in the location of the first minimum in light intensity because

The bending of light around an obstacle such as the edge of a slit is called

Suppose you observe fifteen bright fringes in the central maximum of a double-slit interference pattern. If you know that the separation of the slits d = 0.01 mm, you can conclude that the width of each slit is

Light of wavelength 650 nm is incident on a slit of width 25.0 µm. At what angle is the second diffraction minimum observed?

$The fringes are the result of A) diffraction from a single slit. B) interference from a double slit in addition to diffraction from the two slits. C) interference from three slits. D) diffraction from three slits. E) None of these is correct.$ The fringes are the result of

Suppose you observe nineteen bright fringes in the central maximum of a double-slit interference pattern. If you know that the separation of the slits d = 0.06 mm, you can conclude that the width of each slit is

Which of the following statements concerning the interference pattern described by the phasor diagram is true?

Two slits of width a = 0.0016 mm are separated by a distance d = 0.04 mm and illuminated by light of wavelength $\lambda$ = 450 nm. The number of bright fringes seen in the central maximum is

A single-slit diffraction pattern is displayed on a screen 0.900 m away from the slit. If the wavelength of the light is 600 nm and the slit is 1.50 * 10^-4 m wide, the distance from the first minimum on the right to the first minimum on the left is

As the width of the slit producing a single-slit diffraction pattern is slowly and steadily reduced (always remaining larger than the wavelength of the light) , the diffraction pattern

When a parallel beam of light is diffracted at a single slit,

Light of wavelength 450 nm is incident on a narrow slit. The diffraction pattern is observed on a screen 5.0 m from the slit, and the central maximum is observed to have a width of 22 cm. What is the width of the slit?

An easy way to distinguish whether the fringe pattern is the result of a single slit or from a double slit is