Question 1

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 2

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 3

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

,C
,E

Question 4

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 5

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 distance between the central maximum and the first bright fringe in the double-slit pattern is given by:
$x=\frac{\lambda L}{d}$ where $\lambda$ is the wavelength, $L$ is the distance between the slits and the screen, and $d$ is the slit separation.

For the 750 nm light, we have:
$x_1=\frac{750	imes10^{-9}	imes2.0}{2.0	imes10^{-3}}=0.75$ mm

For the 900 nm light, we have:
$x_2=\frac{900	imes10^{-9}	imes2.0}{2.0	imes10^{-3}}=0.9$ mm

The first coincidence of bright fringes from both patterns will occur when the difference between the positions is equal to an integer number of fringes, i.e.
$x_2-x_1=n\lambda L/d$

where n is an integer.

Solving for n, we get:
$n=\frac{(x_2-x_1)d}{\lambda L}=\frac{(0.9-0.75)	imes10^{-3}	imes2.0	imes10^{-3}}{(750+900)/2	imes10^{-9}	imes2.0}=3$

So the first coincidence of bright fringes from both patterns will occur at a distance of:
$x=(x_2-x_1)n=4.5$ mm.

Therefore, the correct choice is (C) 4.5 mm.

Question 6

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 in a double-slit experiment is given by:

d sinθ = mλ

where d is the slit separation, λ is the wavelength of light, θ is the angle between the line connecting the slit and the fringe and the axis perpendicular to the screen, and m is the order of the bright fringe.

For the first-order fringe m=1, so we have:

d sinθ = λ

For the second-order fringe m=2, so we have:

d sinθ = 2λ

Subtracting the two equations, we get:

d (sinθ2 - sinθ1) = λ

Using the small angle approximation sinθ ≈ θ (valid for small values of θ), we have:

θ2 - θ1 ≈ (λ/d)

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

Δy = L (θ2 - θ1) ≈ L(λ/d)

where L is the distance from the slits to the screen.

Substituting the given values, we get:

Δy ≈ (2.5 m)(600 nm)/(0.050 mm) ≈ 30 cm

Therefore, the spacing between the first-order and second-order bright fringes is 30 cm, which corresponds to choice (B).

Question 7

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

Question 8

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 use the formula for the position of the bright fringes on the screen:

y = (mλL)/d

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

We are given that m = 3, λ = 575 nm, y = 6.5°, and we can assume that L is much greater than d (i.e. the screen is far away).

We first need to convert the angle to meters using the small angle approximation:

y = Ltanθ ≈ L(θ in radians)

y = Ltan(6.5°) ≈ L(0.113 radians)

Now we can solve for d:

d = (mλL)/y

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

d ≈ 15 µm

Question 9

Monochromatic light strikes a metal surface and electrons are ejected from the metal.If the intensity of the light is increased,what will happen to the ejection rate and maximum energy of the electrons?

Accepted Answer

Increasing the intensity of monochromatic light increases the number of photons hitting the metal surface per unit time, which increases the rate of electron ejection. However, the maximum energy of the ejected electrons is determined by the frequency of the light, not its intensity, and remains unchanged.

Question 10

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 11

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 formula for finding the angular separation of the two first order maxima is:

θ = λ/d    (where λ is the wavelength of light and d is the distance between the slits)

θ = (600 nm)/(0.080 mm)
θ = 0.0075 radians

We can convert this to degrees by multiplying by 180/π:

θ = 0.0075 x (180/π)
θ = 0.43 degrees

Since there are two first order maxima, the angular separation between them would be twice this value:

θ = 2 x 0.43
θ = 0.86 degrees

Therefore, the best choice is B, with an angular separation of 0.86 degrees.

Question 12

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 13

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 14

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 the bright fringes on the detector screen:
y = mλL/d
where y is the distance from the central fringe, m is the order of the fringe, λ is the wavelength of the light, L is the distance from the slits to the detector screen, and d is the slit separation.

We are given d = 3.00 × 10^-5 m, L = 2.00 m, and m = 7. We can solve for λ:
λ = md/(Ly) = 7(3.00 × 10^-5 m)/(2.00 m × 0.100 m) = 2.10 × 10^-7 m = 210 nm

Therefore, the best choice is C, 214 nm.

Question 15

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. Doubling the width of the slit will halve the width of the central bright spot. This is because the central bright spot is formed by constructive interference, and doubling the width of the slit will increase the number of wavelengths that can be in phase at the center, resulting in a stronger interference pattern with a narrower central peak.

Question 16

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 separation...

Question 17

A nonrelativistic electron and a nonrelativistic proton have the same de Broglie wavelength.Which of the following statements about these particles are accurate? (There may be more than one correct choice.)

Accepted Answer

,D

Question 18

A beam of red light and a beam of violet light each deliver the same power on a surface.For which beam is the number of photons hitting the surface per second the greatest?

Accepted Answer

The energy of a photon is given by $E = hf$, where $h$ is Planck's constant and $f$ is the frequency of the light. Since violet light has a higher frequency than red light, each photon of violet light carries more energy. Therefore, for two beams delivering the same power, the beam with the lower energy photons (red light) must have a greater number of photons hitting the surface per second.

Question 19

A lens is designed to work in the visible,near-infrared,and near-ultraviolet.The best resolution of this lens from a diffraction standpoint is

Accepted Answer

The resolution of a lens from a diffraction standpoint improves with shorter wavelengths. Near-ultraviolet light has shorter wavelengths compared to visible and near-infrared light, thus offering the best resolution.

Question 20

If the diameter of a radar dish is doubled,what happens to its resolving power assuming that all other factors remain unchanged?

Accepted Answer

Resolving power is directly proportional to the diameter of the radar dish. When the diameter is doubled, the resolving power also doubles.

Quiz 1: A: Foundations