Question 1

At most, how many bright fringes can be formed on each side of the central bright fringe (not counting the central bright fringe) when light of 625 nm falls on a double slit whose spacing is

Accepted Answer

C

Question 2

Two radio antennas are 120 m apart on a north-south line, and they radiate in phase at a frequency of 3.4 MHz. All radio measurements are made far from the antennas. If the east-west reference line passes midway between the two antennas, what is the smallest angle from the antennas, measured north of east, at which constructive interference of two radio waves occurs? (c = 3.00 × 10⁸ m/s)

Accepted Answer

First, we can find the wavelength of the radio waves:
λ = c/f = 3.00 × 10^8 m/s / 3.4 MHz = 88.2 m

Now, we can consider the two antennas as sources of waves that interfere with each other. We want to find the angle at which the waves from the two antennas interfere constructively. This means that the path difference between the waves from the two antennas must be an integer multiple of the wavelength.

Let's consider a point P north of the antennas, at a distance y from the line connecting the antennas (see diagram below). The path difference between the waves from the two antennas to point P is:

Δx = d sinθ
Δy = y

where d = 120 m is the distance between the antennas, θ is the angle we want to find (measured from east, counterclockwise), and y is the distance from the reference line.

The total path difference is then:

Δ = Δx + Δy = d sinθ + y

For constructive interference to occur, Δ must be an integer multiple of the wavelength λ. In symbols:

Δ = nλ

where n is an integer.

We want to find the smallest angle θ for which this equation is satisfied. To do that, we can try different values of n and solve for θ. However, we can also notice that the smallest angle corresponds to the smallest value of n that satisfies the equation.

Let's start with n = 1. Then we have:

d sinθ + y = λ

Plugging in the values, we get:

120 sinθ + y = 88.2

Now, we can try to eliminate y from the equation. Notice that if we subtract the same equation with y = -d/2 (which corresponds to a point midway between the two antennas), we get:

120 sinθ = 88.2 + d/2

Substituting d = 120 m, we get:

120 sinθ = 148.2

Solving for sinθ, we get:

sinθ = 1.235

However, this value is too large, because the maximum value of sinθ is 1. So, there is no solution for n = 1.

Let's try n = 2. Then we have:

d sinθ + y = 2λ

Plugging in the values, we get:

120 sinθ + y = 176.4

Subtracting the same equation with y = -d/2, we get:

120 sinθ = 236.4

Solving for sinθ, we get:

sinθ = 1.97

Again, this value is too large.

Let's try n = 3. Then we have:

d sinθ + y = 3λ

Plugging in the values, we get:

120 sinθ + y = 264.6

Subtracting the same equation with y = -d/2, we get:

120 sinθ = 324.6

Solving for sinθ, we get:

sinθ = 2.71

This value is also too large.

Finally, let's try n = 4. Then we have:

d sinθ + y = 4λ

Plugging in the values, we get:

120 sinθ + y = 352.8

Subtracting the same equation with y = -d/2, we get:

120 sinθ = 412.8

Solving for sinθ, we get:

sinθ = 3.44

This value is too large, but notice that it is very close to 1. So, we can assume that sinθ ≈ 1 and solve for θ:

120 sinθ = 120

sinθ = 1

θ = 90°

This is the maximum value of θ, but we want the minimum value, which corresponds to the other side of the reference line. Therefore, the answer is:

θ = 90° - 47° = 43°

Therefore, the answer is D) 47°, and the explanation is as shown above.

Question 3

In a double-slit experiment, if the slit separation is increased, which of the following happens to the interference pattern shown on the screen?

Accepted Answer

When the slit separation in a double-slit experiment is increased, the interference pattern on the screen changes such that the minima get closer together. This is because the fringe width is inversely proportional to the slit separation; as the separation increases, the fringe width decreases, causing the minima to move closer together.

Question 4

A single-slit diffraction pattern is formed on a distant screen. Assuming the angles involved are small, by what factor will the width of the central bright spot on the screen change if the slit width is doubled?

Accepted Answer

According to the single slit diffraction formula, the central bright spot width is directly proportional to the slit width (w) and inversely proportional to the distance to the screen (D). Therefore, if slit width is doubled while keeping the distance to the screen constant, the width of the central bright spot will be halved.

Question 5

Light from a 600 nm source goes through two slits 0.080 mm apart. What is the angular separation of the two first order maxima occurring on a screen 2.0 m from the slits?

Accepted Answer

The angular separation between two adjacent maxima is given by:
$	heta = \frac{\lambda}{d}$,
where $\lambda$ is the wavelength of light and $d$ is the distance between the two slits.
For first order maxima, $m=1$, so the distance from the central maximum to the first order maximum is:
$y_1 = \frac{m\lambda L}{d}$,
where $L$ is the distance between the slits and the screen.
Substituting the given values, we have:
$	heta = \frac{\lambda}{d} = \frac{600\ 	ext{nm}}{0.080\ 	ext{mm}} = 7.5	imes10^{-3}\ 	ext{rad}$
$y_1 = \frac{(1)(600\ 	ext{nm})(2.0\ 	ext{m})}{0.080\ 	ext{mm}} = 0.015\ 	ext{m}$
The distance from the central maximum to the first order maximum on either side is half of this value, so the angular separation between the two first order maxima is given by:
$	heta_{	ext{separation}} = \frac{y_1}{L} = \frac{0.015\ 	ext{m}}{2.0\ 	ext{m}} = 0.0075\ 	ext{rad}$
Converting to degrees and rounding to two decimal places, we get:
$	heta_{	ext{separation}} = 0.43^\circ \approx 0.86^\circ.$
Therefore, the answer is B.

Question 6

In a double slit experiment, the slit separation is constructed to be exactly 4 times the wavelength of the light passing through the slits. At what angles from the center of the pattern will the third bright fringes on both sides of the central fringe occur?

Accepted Answer

The answer of In a double slit experiment, the slit...

Question 7

Two sources of light illuminate a double slit simultaneously. One has wavelength 570 nm and the second has an unknown wavelength. The  bright fringe of the unknown wavelength overlaps the  bright fringe of the light of 570 nm wavelength. What is the unknown wavelength?

Accepted Answer

The answer of Two sources of light illuminate a double...

Question 8

In the two-slit experiment, monochromatic light of frequency 5.00 × 10¹⁴ Hz passes through a pair of slits separated by 2.20 × 10^-5 m. (c = 3.00 × 10⁸ m/s) (a) At what angle away from the central bright spot does the third bright fringe past the central bright spot occur? (b) At what angle does the second dark fringe occur?

Accepted Answer

The answer of In the two-slit experiment, monochromatic light of...

Question 9

A light beam shines through a thin slit and illuminates a distant screen. The central bright fringe on the screen is 1.00 cm wide, as measured between the dark fringes that border it on either side. Which of the following actions would decrease the width of the central bright fringe? (There may be more than one correct choice.)

Accepted Answer

The answer of A light beam shines through a thin...

Question 10

A double slit illuminated with light of wavelength 588 nm forms a diffraction pattern on a screen 11.0 cm away. The slit separation is 2464 nm. What is the distance between the third and fourth bright fringes away from the central fringe?

Accepted Answer

The answer of A double slit illuminated with light of...

Question 11

In a single-slit diffraction experiment, the width of the slit through which light passes is reduced. What happens to the width of the central bright fringe?

Accepted Answer

When the width of the slit in a single-slit diffraction experiment is reduced, the diffraction effect becomes more pronounced, leading to a wider central bright fringe. This is because the angle of diffraction increases as the slit width decreases, spreading the light over a larger area.

Question 12

Light from a monochromatic source shines through a double slit onto a screen 5.00 m away. The slits are 0.180 mm apart. The dark bands on the screen are measured to be 1.70 cm apart. What is the wavelength of the incident light?

Accepted Answer

The answer of Light from a monochromatic source shines through...

Question 13

In a two-slit experiment, the slit separation is 3.00 ×  m. The interference pattern is recorded on a flat screen-like detector that is 2.00 m away from the slits. If the seventh bright fringe on the detector is 10.0 cm away from the central fringe, what is the wavelength of the light passing through the slits?

Accepted Answer

The answer of In a two-slit experiment, the slit separation...

Question 14

In a double-slit experiment, the slit separation is 2.0 mm, and two wavelengths, 750 nm and 900 nm, illuminate the slits simultaneously. A screen is placed 2.0 m from the slits. At what distance from the central maximum on the screen will a bright fringe from one pattern first coincide with a bright fringe from the other?

Accepted Answer

The first bright fringe of one pattern will coincide with the first bright fringe of the other pattern when the path difference between the two waves is equal to one wavelength. The path difference can be calculated using the formula:

$d \sin 	heta = m \lambda$

where $d$ is the slit separation, $	heta$ is the angle between the central maximum and the fringe, $m$ is the order of the fringe (1 for the first bright fringe), and $\lambda$ is the wavelength.

For the 750 nm wavelength, the angle for the first bright fringe is:

$\sin 	heta = \frac{m \lambda}{d} = \frac{1 \cdot 750 	imes 10^{-9} \  	ext{m}}{2.0 	imes 10^{-3} \ 	ext{m}} = 0.375 	imes 10^{-3}$

$	heta = \sin^{-1} (0.375 	imes 10^{-3}) = 0.0214 \ 	ext{rad}$

The distance from the central maximum to the first bright fringe for the 750 nm wavelength is:

$y_{750} = D 	an 	heta = 2.0 \ 	ext{m} 	an(0.0214 \ 	ext{rad}) = 0.045 \ 	ext{m} = 4.5 \ 	ext{mm}$

Similarly, for the 900 nm wavelength:

$	heta = \sin^{-1} (0.450 	imes 10^{-3}) = 0.0259 \ 	ext{rad}$

$y_{900} = D 	an 	heta = 2.0 \ 	ext{m} 	an(0.0259 \ 	ext{rad}) = 0.052 \ 	ext{m} = 5.2 \ 	ext{mm}$

Therefore, the first coincidence of bright fringes will occur when $y_{750} = y_{900}$, which gives:

$0.045 \ 	ext{m} = 0.052 \ 	ext{m} - y$

$y = 0.007 \ 	ext{m} = 7.0 \ 	ext{mm}$

However, this is the distance from the central maximum to the first coincidence of dark fringes. To find the bright fringes, we need to add half a wavelength (450 nm) to $y$:

$y_{bright} = y + \frac{\lambda}{2} = 0.007 \ 	ext{m} + 0.450 	imes 10^{-9} \ 	ext{m} = 7.45 \ 	ext{mm}$

Therefore, the first coincidence of bright fringes from the two patterns will occur at a distance of approximately 7.5 mm from the central maximum on the screen.

The best choice is C, which is the closest to the calculated value of 7.45 mm.

Question 15

Coherent monochromatic light of wavelength 632.8 nm passes through a pair of thin parallel slits. The figure shows the central portion of the pattern of bright fringes viewed on a screen 1.40 m beyond the slits. What is the distance between the two slits?

Accepted Answer

The answer of Coherent monochromatic light of wavelength 632.8 nm...

Question 16

In a double slit experiment, if the separation between the two slits is 0.050 mm and the distance from the slits to a screen is 2.5 m, find the spacing between the first-order and second-order bright fringes when coherent light of wavelength 600 nm illuminates the slits.

Accepted Answer

The spacing between consecutive bright fringes is given by the formula:

d = λD/d

where d is the slit separation, λ is the wavelength of light, and D is the distance from the slits to the screen.

For the first-order bright fringe, we have:

d1 = λD/d = (600 nm)(2.5 m)/(0.050 mm) = 30 mm

For the second-order bright fringe, we have:

d2 = 2d1 = 2(30 mm) = 60 mm

The spacing between the first-order and second-order bright fringes is:

d2 - d1 = 60 mm - 30 mm = 30 mm = 3.0 cm

Therefore, the best choice is B.

Question 17

Light of wavelength 575 nm passes through a double-slit and the third order bright fringe is seen at an angle of 6.5° away from the central fringe. What is the separation between the double slits?

Accepted Answer

We can use the equation:

d*sin(θ) = m*λ

Where d is the separation between the double slits, θ is the angle of the fringe, m is the order of the fringe (in this case, m = 3), and λ is the wavelength of the light.

Rearranging this equation, we can solve for d:

d = m*λ/sin(θ)

Plugging in the values given in the problem:

d = 3*(575 nm)/(sin(6.5°))

d = 15 µm

Therefore, the separation between the double slits is 15 µm (option C).

Question 18

Light of wavelength 519 nm passes through two slits. In the interference pattern on a screen 4.6 m away, adjacent bright fringes are separated by 5.2 mm in the general vicinity of the center of the pattern. What is the separation of the two slits?

Accepted Answer

The answer of Light of wavelength 519 nm passes through...

Question 19

A pair of narrow slits, separated by 1.8 mm, is illuminated by a monochromatic light source. Light waves arrive at the two slits in phase, and a fringe pattern is observed on a screen 4.8 m from the slits. If there are 5.0 complete bright fringes per centimeter on the screen near the center of the pattern, what is the wavelength of the monochromatic light?

Accepted Answer

The answer of A pair of narrow slits, separated by...

Question 20

Two small forward-facing speakers are 2.50 m apart. They are both emitting, in phase with each other, a sound of frequency 1100 Hz in a room where the speed of sound is 344 m/s. A woman is standing opposite the midpoint between the speakers and is initially 35.0 m from the midpoint. As she slowly walks parallel to the line connecting the speakers, at what angle θ (relative to the centerline coming outward from the midpoint between the speakers) will she first hear no sound?

Accepted Answer

The first position where the woman hears no sound is when there is a path difference of half a wavelength ($\lambda/2$) between the sound waves reaching her from the two speakers. The wavelength ($\lambda$) of the sound can be calculated using the formula $\lambda = v/f$, where $v$ is the speed of sound (344 m/s) and $f$ is the frequency (1100 Hz). This gives $\lambda = 344 / 1100 = 0.3127$ m. Therefore, $\lambda/2 = 0.1564$ m. The path difference corresponds to the difference in distances from each speaker to the woman's position, forming a right triangle with the line connecting the speakers and the line from the midpoint to the woman's position. Using the sine rule, $\sin(\theta) = \frac{\lambda/2}{d}$, where $d = 2.50$ m is the distance between the speakers. Substituting the values, $\sin(\theta) = \frac{0.1564}{2.50}$, giving $\theta = \arcsin(0.06256)$, which is approximately $3.6^\circ$.

Quiz 33: Wave Optics

At most, how many bright fringes can be formed on each side of the central bright fringe (not counting the central bright fringe) when light of 625 nm falls on a double slit whose spacing is

In a double-slit experiment, if the slit separation is increased, which of the following happens to the interference pattern shown on the screen?

A single-slit diffraction pattern is formed on a distant screen. Assuming the angles involved are small, by what factor will the width of the central bright spot on the screen change if the slit width is doubled?

Light from a 600 nm source goes through two slits 0.080 mm apart. What is the angular separation of the two first order maxima occurring on a screen 2.0 m from the slits?

In a double slit experiment, the slit separation is constructed to be exactly 4 times the wavelength of the light passing through the slits. At what angles from the center of the pattern will the third bright fringes on both sides of the central fringe occur?

Two sources of light illuminate a double slit simultaneously. One has wavelength 570 nm and the second has an unknown wavelength. The bright fringe of the unknown wavelength overlaps the bright fringe of the light of 570 nm wavelength. What is the unknown wavelength?

A double slit illuminated with light of wavelength 588 nm forms a diffraction pattern on a screen 11.0 cm away. The slit separation is 2464 nm. What is the distance between the third and fourth bright fringes away from the central fringe?

In a single-slit diffraction experiment, the width of the slit through which light passes is reduced. What happens to the width of the central bright fringe?

Light from a monochromatic source shines through a double slit onto a screen 5.00 m away. The slits are 0.180 mm apart. The dark bands on the screen are measured to be 1.70 cm apart. What is the wavelength of the incident light?

Coherent monochromatic light of wavelength 632.8 nm passes through a pair of thin parallel slits. The figure shows the central portion of the pattern of bright fringes viewed on a screen 1.40 m beyond the slits. What is the distance between the two slits?

In a double slit experiment, if the separation between the two slits is 0.050 mm and the distance from the slits to a screen is 2.5 m, find the spacing between the first-order and second-order bright fringes when coherent light of wavelength 600 nm illuminates the slits.

Light of wavelength 575 nm passes through a double-slit and the third order bright fringe is seen at an angle of 6.5° away from the central fringe. What is the separation between the double slits?

Light of wavelength 519 nm passes through two slits. In the interference pattern on a screen 4.6 m away, adjacent bright fringes are separated by 5.2 mm in the general vicinity of the center of the pattern. What is the separation of the two slits?