Question 1

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

A

Question 2

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 3

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 the bright fringes can be calculated using the equation:
d sinθ = mλ
where d is the slit separation, θ is the angle between the line normal to the screen and the line from the slit to the bright fringe, m is the order of the bright fringe (m = 1 for first order and m = 2 for second order), and λ is the wavelength of light.

For first order bright fringe:
d sinθ = λ
sinθ = λ/d
θ = sin^(-1)(λ/d)

For second order bright fringe:
d sinθ = 2λ
sinθ = 2λ/d
θ = sin^(-1)(2λ/d)

The spacing between the first and second order bright fringes can be found by taking the difference between the angles θ for each fringe and using the distance formula:
spacing = distance from screen × tan(θ2 - θ1)

Plugging in the given values:
d = 0.050 mm = 5.0×10^(-5) m
λ = 600 nm = 6.0×10^(-7) m
distance from screen = 2.5 m

For first order bright fringe:
θ1 = sin^(-1)(λ/d) = sin^(-1)(6.0×10^(-7)/5.0×10^(-5)) ≈ 0.012°

For second order bright fringe:
θ2 = sin^(-1)(2λ/d) = sin^(-1)(2×6.0×10^(-7)/5.0×10^(-5)) ≈ 0.024°

spacing = 2.5 m × tan(θ2 - θ1) ≈ 3.0 cm

Therefore, the answer is B: 3.0 cm.

Question 4

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 5

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

The width of the central bright spot is proportional to the width of the slit, with a proportionality constant of lambda*f/D, where lambda is the wavelength of the light, f is the distance from the slit to the screen, and D is the width of the slit. Since the angles involved are small, we can assume that the distance from the slit to the screen and the wavelength of the light are constant. Therefore, doubling the width of the slit will result in halving the width of the central bright spot.

Question 6

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 1.97 × 10^-6 m?

Accepted Answer

The number of bright fringes on each side of the central bright fringe is given by the equation:$$N = \frac{aW}{\lambda}-1$$where:* N is the number of bright fringes* a is the slit spacing* W is the width of the screen* λ is the wavelength of lightIn this problem, we are given the following information:* a = 1.97 × 10^-6 m* W = 1.00 m* λ = 625 nm = 6.25 × 10^-7 mPlugging these values into the equation, we get:$$N = \frac{(1.97 	imes 10^{-6} 	ext{ m})(1.00 	ext{ m})}{6.25 	imes 10^{-7} 	ext{ m}}-1 = 3$$Therefore, there will be 3 bright fringes on each side of the central bright fringe.

Question 7

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

Let the distance between the two slits be d.

From the given diagram, we can see that the distance between two consecutive bright fringes is given by:

y = (λD) / d

where λ is the wavelength of the light, D is the distance between the slits and the screen, and y is the distance between two adjacent bright fringes.

We are given λ = 632.8 nm, D = 1.40 m, and y = 1.2 mm (the distance between the first and seventh bright fringes in the central portion of the pattern).

Substituting these values in the above equation, we get:

1.2 x 10^-3 m = (632.8 x 10^-9 m)(1.40 m) / d

Simplifying and solving for d, we get:

d = (632.8 x 10^-9 m)(1.40 m) / (1.2 x 10^-3 m) = 0.281 mm

Therefore, the distance between the two slits is 0.281 mm, which is option C.

Question 8

Two sources of light illuminate a double slit simultaneously. One has wavelength 570 nm and the second has an unknown wavelength. The m = 5 bright fringe of the unknown wavelength overlaps the
M = 4 bright fringe of the light of 570 nm wavelength. What is the unknown wavelength?

Accepted Answer

The condition for bright fringes is given by the equation $ m \lambda = d \sin \theta $. For the two wavelengths to overlap, $ m_1 \lambda_1 = m_2 \lambda_2 $. Given $ m_1 = 5 $, $ \lambda_1 = \lambda $, $ m_2 = 4 $, and $ \lambda_2 = 570 $ nm, solving $ 5 \lambda = 4 \times 570 $ gives $ \lambda = 456 $ nm.

Question 9

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 10

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 $3.6^\circ$.

Question 11

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 12

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 answer of In a double-slit experiment, the slit separation...

Question 13

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 in a double-slit experiment is given by the equation, 
$$	heta = \frac{\lambda}{d}$$
where $\lambda$ is the wavelength of the light and $d$ is the distance between the slits.

In this case, $\lambda=600$ nm $= 6	imes10^{-7}$ m and $d = 0.080$ mm $ = 8	imes10^{-5}$ m.

The distance between the screen and the slits, $L$ is 2.0 m.

For the first order maxima, the angle for the central maximum is zero and the angles for the first order maxima can be found using the equation, 
$$	heta_{1}=\frac{\lambda}{d}\sin^{-1}\left(\frac{\lambda}{d}ight)= \frac{\lambda}{d}\sin^{-1}\left(\frac{1}{2}ight)$$

Plugging in the values, we get 
$$	heta_{1}=\frac{6	imes10^{-7}}{8	imes10^{-5}}\sin^{-1}\left(\frac{1}{2}ight)\approx 0.86^\circ$$

Therefore, the angular separation of the two first order maxima on the screen is $2	heta_1\approx1.72^\circ$.

The best choice is (B) 0.86°, which is twice the angle for the first order maximum.

Question 14

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 15

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 16

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

For the third order bright fringe, we have:
$$d\sin	heta = \frac{(m+1/2)\lambda}{2}$$
where $d$ is the separation between the double slits, $	heta$ is the angle of the fringe from the central fringe, $m$ is the order of the bright fringe, and $\lambda$ is the wavelength of the light.

Plugging in the values given, we get:
$$d\sin(6.5^\circ) = \frac{(3+1/2)(575~\mathrm{nm})}{2}$$
Solving for $d$, we get:
$$d = \frac{(3+1/2)(575~\mathrm{nm})}{2\sin(6.5^\circ)} \approx 14.9~\mu\mathrm{m}$$

Therefore, the closest answer choice is C: 15 µm.

Question 17

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

The condition for constructive interference is that the path difference between the two waves from the antennas should be an integer multiple of the wavelength. Let's assume that the angle we are looking for is θ.

The path difference between the two waves arriving at a point P north of the antennas is given by:
Δx = d sinθ
where d is the distance between the antennas, and θ is the angle between the line connecting the antennas and the line connecting one of the antennas to point P.

The wavelength of the radio waves is given by:
λ = c/f = 88 m
where c is the speed of light and f is the frequency of the waves.

Constructive interference occurs when:
Δx = nλ
where n is an integer.

Substituting the values we have:
120 sinθ = n × 88
or
sinθ = n×88/120

Since we want the smallest angle, we take n = 1, which gives:
sinθ = 88/120
or
θ = sin−1(88/120) = 47° (rounded to the nearest degree) north of east.

Therefore, the correct answer is D.

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

In a two-slit experiment, the slit separation is 3.00 × 10^-5 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

We can use the formula for the position of bright fringes in the interference pattern:

y = (mλL) / d

where y is the distance from the central fringe to the m-th bright fringe, λ is the wavelength of the light, L is the distance from the slits to the screen, d is the slit separation, and m is the order of the fringe.

In this case, we have y = 0.1 m, m = 7, L = 2.00 m, and d = 3.00 × 10^-5 m. Solving for λ, we get:

λ = (y d) / (m L) = (0.1 m) (3.00 × 10^-5 m) / (7) (2.00 m) = 2.14 × 10^-7 m = 214 nm

Therefore, the wavelength of the light passing through the slits is 214 nm. The answer is (C).

Question 20

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...

Quiz 22: Wave Optics