Question 1

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.)
A)increase the wavelength of the light
B)decrease the wavelength of the light
C)increase the width of the slit
D)decrease the width of the slit
E)put the apparatus all under water

Accepted Answer

B,C,E

Question 2

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

A

Question 3

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

We can use the formula $d\sin	heta = m\lambda$ to solve this problem, where $d$ is the distance between the two slits, $	heta$ is the angular separation, $m$ is the order of the maximum, and $\lambda$ is the wavelength of light.

For the first order maximum, $m=1$. Plugging in the values we have, we get:
$$0.080	ext{ mm} \cdot \sin	heta = 1 \cdot 600	ext{ nm}$$
Solving for $\sin	heta$, we get:
$$\sin	heta = \frac{600	ext{ nm}}{0.080	ext{ mm}} = 0.0075$$
Taking the inverse sine of both sides gives us:
$$	heta = \sin^{-1}(0.0075) \approx 0.86°$$
Therefore, the answer is B.

Question 4

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 5

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

The equation for the position of the bright fringe on the screen is given by:

y = (mλL) / d

Where:
y = distance from the central maximum to the mth bright fringe
m = order of the bright fringe
λ = wavelength of light
L = distance from double slit to the screen
d = separation between the double slits

We are given:
λ = 575 nm
m = 3
y = 6.5° = 0.113 radians

We need to solve for d. Rearranging the equation above, we get:

d = (mλL) / y

Substituting the values we know:

d = (3 * 575 nm * L) / 0.113

We don't have the value of L, but we can make use of the fact that for small angles, the distance between the bright fringes on the screen is approximately:

Δy ≈ Lθ

Where θ is the angle from the central maximum. For the third order bright fringe, we know that Δy = 3y. Therefore:

Δy = 3y ≈ Lθ

Solving for L:

L ≈ (3y) / θ
L ≈ (3 * 0.113) / (6.5°)
L ≈ 0.052 m

Now we can substitute this into the equation we derived earlier to solve for d:

d = (3 * 575 nm * 0.052 m) / 0.113
d ≈ 15 µm

Therefore, the correct answer is (C) 15 µm.

Question 6

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 frequency...

Question 7

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 8

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 1.8...

Question 9

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 equation, the central bright spot has a width of approximately (2λD)/w, where λ is the wavelength of the light, D is the distance from the slit to the screen, and w is the width of the slit. Doubling the width of the slit (i.e. changing w to 2w) will result in a decrease in the width of the central bright spot by a factor of 1/2, according to the equation. Therefore, choice B is correct.

Question 10

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

The number of bright fringes on each side of the central bright fringe can be found using the formula:

y = mλL/d

where y is the distance from the central bright fringe to the mth bright fringe, λ is the wavelength of light, L is the distance from the double slit to the screen, d is the spacing between the slits, and m is the order of the bright fringe.

Plugging in the given values, we get:

y = (1)(625 nm)(1 m)/(11eab929_2916_f7ad_99e6_c374e964ed20_TB4470_11 m) ≈ 56.8 nm

The distance between adjacent bright fringes is roughly equal to y. Therefore, the number of bright fringes that can be formed on each side of the central bright fringe is:

(56.8 nm)/(625 nm) ≈ 0.091

Since we can't have a fractional number of fringes, we round down to get:

Number of bright fringes = 0

Therefore, we can only count the central bright fringe.

However, if we look carefully, we can see that there is a dim fringe on either side of the central bright fringe. These are known as the first-order or secondary maxima. Using the same formula, we find that the distance from the central fringe to the first-order maxima is:

y = (1)(625 nm)(1 m)/(2*11eab929_2916_f7ad_99e6_c374e964ed20_TB4470_11 m) ≈ 28.4 nm

Therefore, we can count three bright fringes on each side of the central bright fringe, giving a total of seven bright fringes:

Central bright fringe, two first-order maxima, and four more bright fringes.

In conclusion, the best choice is C) 3 because we can count three bright fringes on each side of the central bright fringe, giving a total of seven bright fringes.

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 is reduced, the diffraction effect becomes more pronounced, leading to a wider central bright fringe. This is because diffraction is inversely related to the slit width; narrower slits spread the light more.

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 is...

Question 14

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 15

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 angle for the nth bright fringe in a double slit experiment is given by $\theta = \sin^{-1}(\frac{n\lambda}{d})$, where $n$ is the fringe order, $\lambda$ is the wavelength, and $d$ is the slit separation. For the third bright fringe ($n = 3$) and $d = 4\lambda$, $\theta = \sin^{-1}(\frac{3\lambda}{4\lambda}) = \sin^{-1}(0.75)$, which is approximately $48.6^\circ$.

Question 16

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 is...

Question 17

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 woman will first hear no sound when there is a path difference of half a wavelength ($\lambda/2$) between the sound waves reaching her from the two speakers, leading to destructive interference. The wavelength $\lambda$ can be calculated using the speed of sound $v$ and the frequency $f$ of the sound: $\lambda = v/f = 344 \, \text{m/s} / 1100 \, \text{Hz} = 0.3127 \, \text{m}$. Thus, $\lambda/2 = 0.1564 \, \text{m}$. The path difference corresponds to the difference in distances from each speaker to the point where the woman first hears no sound, which forms a right-angled triangle with the line connecting the speakers and the line from the midpoint to the woman. Using the geometry of the situation and the small angle approximation, $\theta \approx \sin(\theta) = \frac{\lambda/2}{d} = \frac{0.1564}{2.5} = 0.06256$. Converting this to degrees gives $\theta \approx 3.6^\circ$.

Question 18

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 19

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 two consecutive bright fringes can be calculated using the formula:
d*sinθ = mλ
where d is the separation between the slits, θ is the angle between the central maximum and the mth bright fringe, m is the order of the bright fringe, and λ is the wavelength of the light.

For the first-order bright fringe, m=1, and θ  can be approximated as θ = tan(θ) = L/x, where L is the distance from the slits to the screen, and x is the distance from the central maximum to the first-order bright fringe.

Using these values, we get: 
d*sinθ = mλ
0.050*10^-3 *sin(tan^-1(2.5/ x)) = 1*600*10^-9 
x = 0.0015 m or 1.5 cm for the first-order bright fringe.

For the second-order bright fringe, m=2, and we can calculate the spacing in the same way: 
d*sinθ = mλ
0.050*10^-3 *sin(tan^-1(2.5/ y)) = 2*600*10^-9 
y = 0.0030 m or 3.0 cm for the second-order bright fringe.

The spacing between the first-order and second-order bright fringes is the difference between these two values: 
y-x = 0.0030-0.0015 = 0.0015 m or 3.0 cm.
Hence, the correct option is B.

Question 20

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...

Quiz 33: Wave Optics

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?