Question 1

In a two-slit experiment, there are 11 bright fringes within the central diffraction maximum (on each side, the first diffraction minimum coincides with where the next interference bright fringe would be). What is the ratio of the slit width to the slit separation?
A) 5
B) 1/6
C) 1/5
D) 6

Accepted Answer

The answer of In a two-slit experiment, there are 11...

Question 2

Two planes of atoms in a crystal are separated by 3.25 × 10^-10m. How many maxima will there be in the diffraction pattern for x-rays of wavelength 2.0 × 10^-10 m scattered from this crystal? A) 4 B) 2 C) 3 D) an infinite number E) 1

Accepted Answer

The condition for maxima in a diffraction pattern is given by Bragg's law: nλ = 2d sin θ. For maximum n, sin θ must be ≤ 1. Here, d = 3.25 × 10^-10 m and λ = 2.0 × 10^-10 m. The maximum n is 3, as nλ ≤ 2d.

Question 3

A canvas tent has a single, tiny hole in its side. While inside the tent, you observe a dot on the wall of the tent, which is 2.0 m away (the dot is due to a distant point source of light of wavelength 550 nm). The dot has a diameter of 2.0 mm, and there is a faint ring around it. What is the size of the hole in the tent?
A) 0.55 mm
B) 0.67 mm
C) 1.10 mm
D) 1.34 mm

Accepted Answer

We can use the formula for diffraction of light through a single slit:

sin(θ) = λ/d

where λ is the wavelength of light, d is the width of the slit, and θ is the angle between the straight-ahead direction and the first minimum of the diffraction pattern.

If we assume that the diameter of the dot on the wall is the diameter of the central maximum of the diffraction pattern, then we can use the formula:

D = 2 * (tan(θ/2)) * L

where D is the diameter of the central maximum, θ is the angle to the first minimum (which we just found), and L is the distance from the slit to the wall.

Plugging in the numbers:

θ = sin⁻¹(λ/d) = sin⁻¹(550 nm / d) 
D = 0.002 m 
L = 2.0 m

D = 2 * (tan(sin⁻¹(550 nm / d) / 2)) * 2.0 m

Solving for d:

d = 550 nm / sin(0.5 * tan⁻¹(D/4L))

Plugging in the numbers:

d = 550 nm / sin(0.5 * tan⁻¹(0.001 / 4*2.0))

d = 1.34 mm

Therefore, the size of the hole in the tent is 1.34 mm.

Question 4

Two planes of atoms in a crystal are separated by 3.25 × 10^-10 m. Light of what wavelength would show constructive interference at an angle of 17.5 degrees relative to the crystal surface? A) 3.9 × 10^-10 m B) 4.9 × 10^-9 m C) 9.75 × 10^-10 m D) 1.46 × 10^-10 m E) 1.95 × 10^-10 m

Accepted Answer

The answer of Two planes of atoms in a crystal...

Question 5

When scattering 0.328 nm x-rays off a crystal, the first interference maximum occurs at an angle of 23.0 degrees relative to the crystal surface. At what angle does the second maximum occur?
A) 51.4°
B) 46.0°
C) 34.5°
D) There is no second maximum.

Accepted Answer

The angle for the second maximum can be found using Bragg's law, $n\lambda = 2d\sin(\theta)$, where $n$ is the order of the maximum, $\lambda$ is the wavelength of the x-rays, $d$ is the spacing between the planes in the crystal, and $\theta$ is the angle of incidence. Given that the first maximum occurs at 23.0 degrees, we can solve for $d$ using $n=1$. For the second maximum ($n=2$), we solve for $\theta$ using the same equation with the found value of $d$. The angle that satisfies this condition for the second maximum is larger than the first, and among the given options, 51.4° is the only one that is plausible and follows the trend of increasing angle with increasing order of maximum.

Question 6

Light is shone through an adjustable-width slit in an otherwise opaque barrier, and a diffraction pattern is observed. If the slit is decreased to half its original size, by what factor does the width of the central diffraction maximum change?
A) Decreases by a factor of four
B) Doubles
C) Halves
D) Increases four times

Accepted Answer

The width of the central diffraction maximum is proportional to the width of the slit, according to the single-slit diffraction formula. Therefore, if the slit width is halved, the width of the central maximum will also be halved (since it is the same proportion). However, the intensity of the central maximum will quadruple, since the area of the slit has also been halved. Therefore, the correct answer is choice B, which reflects only the change in width, not intensity.

Question 7

If the distance between the slits in a double-slit apparatus is increased, while the width of each slit is kept the same, the number of interference fringes within the central diffraction maximum will
A) decrease.
B) increase.
C) remain the same.
D) may do any of the above.

Accepted Answer

Increasing the distance between the slits in a double-slit apparatus causes the interference fringes to spread out more, effectively reducing the number of fringes within the central diffraction maximum. This is because the fringe spacing is inversely proportional to the slit separation.

Question 8

A Michelson interferometer is used to measure the index of refraction of a gas by counting interference fringes as the gas is slowly leaked into a transparent vessel placed in one arm of the interferometer. The vessel initially contains a vacuum, and as the gas is let in, 397 bright fringes are counted. That is, the central region of the screen changes from bright to dark and back to bright 397 times. If the wavelength of light used is 560 nm, and the vessel containing the gas is 25.0 cm long, what is the index of refraction of the gas?
A) 1.000111
B) 1.000222
C) 1.000445
D) 1.000889
E) 1.00178

Accepted Answer

The change in optical path length is 2(n - 1)L, where n is the index of refraction of the gas, and L is the length of the vessel. The change in optical path length is also equal to mλ, where m is the number of fringes counted. Thus, 2(n - 1)L = mλ. Solving for n, we get n = 1 + mλ/2L. Plugging in the given values, we get n = 1 + 397(560 nm)/2(25.0 cm) = 1.000111.

Question 9

When scattering x-rays with a wavelength of 3.28 × 10^-10 m from a crystal, the first interference maximum occurs at an angle of 23.0 degrees relative to the crystal surface. What is the crystal spacing? A) 4.2 × 10^-10 m B) 16.8 × 10^-10 m C) 6.3 × 10^-10 m D) 8.4 × 10^-10 m

Accepted Answer

The crystal spacing can be calculated using Bragg's law, $n\lambda = 2d\sin(\theta)$, where $n$ is the order of the maximum (here, $n=1$ for the first maximum), $\lambda$ is the wavelength of the x-rays, $d$ is the crystal spacing, and $\theta$ is the angle of incidence (or diffraction). Rearranging for $d$, we get $d = \frac{\lambda}{2\sin(\theta)}$. Substituting $\lambda = 3.28 \times 10^{-10} \, \text{m}$ and $\theta = 23.0$ degrees into the equation gives a crystal spacing of approximately $4.2 \times 10^{-10} \, \text{m}$.

Question 10

Light from a HeNe laser (632.8 nm wavelength) is directed upon a straightened strand of human hair, 7.5 m from the wall. By examining the diffraction pattern produced on the wall, the diameter of the hair can be determined (the diffraction pattern thus produced has minima in the same locations as if the hair were a slit of the same width in an otherwise opaque barrier). If the central diffraction maximum is measured to be 14.6 cm wide, what is the diameter of the hair?
A) 65 µm
B) 32.5 µm
C) 43.3 µm
D) 86.6 µm

Accepted Answer

The diameter of the hair can be determined using the formula:d = (λL) / wwhere:d is the diameter of the hairλ is the wavelength of the lightL is the distance from the hair to the wallw is the width of the central diffraction maximumPlugging in the given values, we get:d = (632.8 nm * 7.5 m) / 0.146 m = 32.5 µmTherefore, the diameter of the hair is 32.5 µm, which corresponds to choice A.

Question 11

A two-slit apparatus is illuminated with red light of wavelength 660 nm, and 9 interference maxima are observed within the 4.5 cm-wide central diffraction maximum on a distant screen. If blue light of wavelength 440 nm is used instead, how many fringes will there be in the new central diffraction maximum?
A) 3
B) 13
C) 9
D) 6

Accepted Answer

The answer of A two-slit apparatus is illuminated with red...

Question 12

What kind of image does a holographic plate produce for the viewer?
A) real
B) virtual
C) may be either virtual or real

Accepted Answer

A holographic plate produces a real image for the viewer because the light from the object is recorded on the plate and then reconstructed to form a three-dimensional image that can be viewed without the need for special glasses.

Quiz 25: Interference and Diffraction

In a two-slit experiment, there are 11 bright fringes within the central diffraction maximum (on each side, the first diffraction minimum coincides with where the next interference bright fringe would be) . What is the ratio of the slit width to the slit separation?

Two planes of atoms in a crystal are separated by 3.25 × 10^-10m. How many maxima will there be in the diffraction pattern for x-rays of wavelength 2.0 × 10^-10 m scattered from this crystal?

Two planes of atoms in a crystal are separated by 3.25 × 10^-10 m. Light of what wavelength would show constructive interference at an angle of 17.5 degrees relative to the crystal surface?

When scattering 0.328 nm x-rays off a crystal, the first interference maximum occurs at an angle of 23.0 degrees relative to the crystal surface. At what angle does the second maximum occur?

Light is shone through an adjustable-width slit in an otherwise opaque barrier, and a diffraction pattern is observed. If the slit is decreased to half its original size, by what factor does the width of the central diffraction maximum change?

If the distance between the slits in a double-slit apparatus is increased, while the width of each slit is kept the same, the number of interference fringes within the central diffraction maximum will

When scattering x-rays with a wavelength of 3.28 × 10^-10 m from a crystal, the first interference maximum occurs at an angle of 23.0 degrees relative to the crystal surface. What is the crystal spacing?

What kind of image does a holographic plate produce for the viewer?