Interference
Q. For destructive interference to occur in a thin film, the path difference must be equal to:
A.
nλ/2 (n is an integer)
B.
nλ (n is an integer)
C.
λ/4
D.
λ/2
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Solution
Destructive interference occurs when the path difference is an odd multiple of λ/2 (i.e., (2n+1)λ/2).
Correct Answer: A — nλ/2 (n is an integer)
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Q. If the wavelength of light used in a double-slit experiment is 600 nm and the distance between the slits is 0.3 mm, what is the distance between the first and second bright fringes on the screen placed 2 m away?
A.
0.4 m
B.
0.6 m
C.
0.8 m
D.
0.2 m
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Solution
Distance between fringes = (λD)/d = (600 x 10^-9 m * 2 m) / (0.3 x 10^-3 m) = 0.004 m = 0.4 m.
Correct Answer: A — 0.4 m
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Q. If the wavelength of light used in a double-slit experiment is increased, what happens to the position of the interference fringes?
A.
Fringes move closer together
B.
Fringes move further apart
C.
Fringes disappear
D.
Fringes become brighter
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Solution
Increasing the wavelength increases the fringe width, causing the fringes to move further apart.
Correct Answer: B — Fringes move further apart
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Q. If the wavelength of light used in an interference experiment is 500 nm, what is the fringe separation when the screen is placed 2 m away from the slits separated by 0.1 mm?
A.
0.01 m
B.
0.025 m
C.
0.05 m
D.
0.1 m
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Solution
Fringe separation β = λD/d = (500 x 10^-9 m)(2 m)/(0.1 x 10^-3 m) = 0.01 m.
Correct Answer: C — 0.05 m
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Q. If the wavelength of light used in Young's experiment is 600 nm and the distance between the slits is 0.1 mm, what is the distance between the first and second bright fringes on a screen 2 m away?
A.
0.12 m
B.
0.24 m
C.
0.36 m
D.
0.48 m
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Solution
Fringe separation (β) = λD/d. β = (600 x 10^-9 * 2) / 0.0001 = 0.012 m. Distance between first and second bright fringes = 2β = 0.024 m.
Correct Answer: B — 0.24 m
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Q. If the wavelength of light used in Young's experiment is 600 nm, what is the fringe width when the distance between the slits is 0.1 mm and the distance to the screen is 2 m?
A.
0.03 mm
B.
0.06 mm
C.
0.12 mm
D.
0.15 mm
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Solution
Fringe width (β) = (λD)/d = (600 x 10^-9 * 2)/(0.1 x 10^-3) = 0.012 mm = 0.06 mm.
Correct Answer: B — 0.06 mm
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Q. If two coherent sources of light are in phase, what type of interference pattern will be observed?
A.
No interference pattern
B.
Destructive interference
C.
Constructive interference
D.
Random interference
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Solution
When two coherent sources are in phase, they produce constructive interference, resulting in bright fringes.
Correct Answer: C — Constructive interference
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Q. If two coherent sources of light are in phase, what type of interference will occur?
A.
Destructive interference
B.
Constructive interference
C.
No interference
D.
Random interference
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Solution
When two coherent sources are in phase, they produce constructive interference, resulting in bright fringes.
Correct Answer: B — Constructive interference
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Q. If two coherent sources of light are in phase, what will be the phase difference at a point where the path difference is λ/4?
A.
0 radians
B.
π/2 radians
C.
π radians
D.
3π/2 radians
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Solution
The phase difference (Δφ) is given by (2π/λ) * path difference. For a path difference of λ/4, Δφ = (2π/λ) * (λ/4) = π/2 radians.
Correct Answer: B — π/2 radians
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Q. If two coherent sources of light are in phase, what will be the phase difference at a point where the path difference is λ/2?
A.
0 radians
B.
π/2 radians
C.
π radians
D.
2π radians
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Solution
A path difference of λ/2 corresponds to a phase difference of π radians, leading to destructive interference.
Correct Answer: C — π radians
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Q. If two coherent sources of light are in phase, what will be the result at a point where the path difference is λ/2?
A.
Constructive interference
B.
Destructive interference
C.
No interference
D.
Partial interference
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Solution
A path difference of λ/2 results in destructive interference, as the waves will be out of phase.
Correct Answer: B — Destructive interference
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Q. If two waves have a phase difference of π radians, what type of interference occurs?
A.
Constructive interference
B.
Destructive interference
C.
No interference
D.
Complete interference
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Solution
A phase difference of π radians results in destructive interference.
Correct Answer: B — Destructive interference
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Q. If two waves interfere constructively, what is the condition for the path difference?
A.
(n + 1/2)λ
B.
nλ
C.
(n - 1/2)λ
D.
n/2 λ
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Solution
Constructive interference occurs when the path difference is nλ, where n is an integer.
Correct Answer: B — nλ
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Q. If two waves of equal amplitude interfere destructively, what is the phase difference between them?
A.
0 rad
B.
π/2 rad
C.
π rad
D.
3π/2 rad
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Solution
Destructive interference occurs when the phase difference is π rad.
Correct Answer: C — π rad
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Q. If two waves of equal amplitude interfere, what is the maximum intensity observed?
A.
A^2
B.
2A^2
C.
4A^2
D.
A
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Solution
Maximum intensity (I_max) = 4A^2 for two waves of equal amplitude A.
Correct Answer: B — 2A^2
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Q. In a double-slit experiment, if the distance between the slits is 0.2 mm and the distance to the screen is 1 m, what is the fringe width for light of wavelength 600 nm?
A.
0.3 mm
B.
0.6 mm
C.
0.9 mm
D.
1.2 mm
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Solution
Fringe width β = λD/d = (600 x 10^-9 m)(1 m)/(0.2 x 10^-3 m) = 0.003 m = 0.6 mm.
Correct Answer: B — 0.6 mm
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Q. In a double-slit experiment, if the distance between the slits is 0.5 mm and the wavelength of light is 400 nm, what is the distance between the first and second bright fringes?
A.
0.4 m
B.
0.8 m
C.
1.2 m
D.
1.6 m
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Solution
Distance between fringes = β = λD/d. For first and second bright fringes, the distance is 2β.
Correct Answer: B — 0.8 m
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Q. In a double-slit experiment, if the distance between the slits is doubled, how does the fringe width change?
A.
Doubles
B.
Halves
C.
Remains the same
D.
Quadruples
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Solution
Fringe width (β) is inversely proportional to the distance between the slits (d). If d is doubled, β is halved.
Correct Answer: B — Halves
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Q. In a double-slit experiment, if the distance between the slits is increased, what happens to the number of visible fringes on the screen?
A.
Increases
B.
Decreases
C.
Remains the same
D.
Becomes zero
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Solution
Increasing the distance between the slits decreases the fringe width, which can lead to more visible fringes within a given distance on the screen.
Correct Answer: B — Decreases
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Q. In a double-slit experiment, if the distance between the slits is increased, what happens to the interference pattern?
A.
Fringes become wider
B.
Fringes become narrower
C.
Fringes disappear
D.
Fringes remain unchanged
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Solution
Increasing the distance between the slits (d) causes the fringe width (β) to decrease, making the fringes narrower.
Correct Answer: B — Fringes become narrower
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Q. In a double-slit experiment, if the distance to the screen is increased, what happens to the fringe pattern?
A.
Fringe width decreases
B.
Fringe width increases
C.
Fringe pattern disappears
D.
Fringe pattern becomes sharper
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Solution
Increasing the distance to the screen increases the fringe width, as fringe width is proportional to the distance from the slits.
Correct Answer: B — Fringe width increases
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Q. In a double-slit experiment, if the distance to the screen is increased, what happens to the fringe separation?
A.
Fringe separation decreases
B.
Fringe separation increases
C.
Fringe separation remains the same
D.
Fringe separation becomes zero
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Solution
Fringe separation is directly proportional to the distance from the slits to the screen (D), so increasing D increases the fringe separation.
Correct Answer: B — Fringe separation increases
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Q. In a double-slit experiment, if the distance to the screen is increased, what happens to the interference pattern?
A.
Fringe width decreases
B.
Fringe width increases
C.
Fringe pattern disappears
D.
Fringe spacing remains unchanged
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Solution
Increasing the distance to the screen increases the fringe width, as fringe width is proportional to the distance from the slits.
Correct Answer: B — Fringe width increases
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Q. In a double-slit experiment, if the intensity of light at the center of the fringe pattern is I0, what is the intensity at the first minimum?
A.
0
B.
I0
C.
I0/2
D.
I0/4
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Solution
At the first minimum, the intensity is 0 due to destructive interference.
Correct Answer: A — 0
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Q. In a double-slit experiment, if the screen is moved further away from the slits, what happens to the fringe pattern?
A.
Fringes become wider
B.
Fringes become narrower
C.
Fringe intensity increases
D.
Fringe intensity decreases
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Solution
Moving the screen further away increases the distance (D) in the fringe width formula, causing the fringes to become wider.
Correct Answer: A — Fringes become wider
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Q. In a double-slit experiment, if the screen is moved further away from the slits, what effect does this have on the fringe spacing?
A.
Increases
B.
Decreases
C.
Remains the same
D.
Becomes zero
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Solution
Moving the screen further away increases the fringe spacing, as fringe width is directly proportional to the distance from the slits.
Correct Answer: A — Increases
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Q. In a Michelson interferometer, what happens to the interference pattern if one of the mirrors is moved slightly?
A.
The pattern remains unchanged
B.
The pattern shifts
C.
The pattern disappears
D.
The pattern becomes brighter
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Solution
Moving one of the mirrors changes the path length for one of the beams, causing a shift in the interference pattern.
Correct Answer: B — The pattern shifts
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Q. In a thin film interference pattern, what happens to the colors observed as the angle of incidence increases?
A.
Colors become more vivid
B.
Colors disappear
C.
Colors shift
D.
Colors remain unchanged
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Solution
As the angle of incidence increases, the effective path difference changes, causing a shift in the observed colors.
Correct Answer: C — Colors shift
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Q. In a thin film interference, if the film is denser than the medium it is placed in, what happens to the phase of the reflected wave?
A.
No phase change
B.
Phase change of π
C.
Phase change of 2π
D.
Phase change of π/2
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Solution
When light reflects off a denser medium, it undergoes a phase change of π (180 degrees).
Correct Answer: B — Phase change of π
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Q. In a thin film interference, if the film is of thickness t and the refractive index is n, what is the condition for the first dark fringe?
A.
2nt = (m + 1/2)λ
B.
2nt = mλ
C.
2nt = (m - 1/2)λ
D.
2nt = 0
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Solution
For the first dark fringe in thin film interference, the condition is 2nt = (m + 1/2)λ, where m = 0.
Correct Answer: A — 2nt = (m + 1/2)λ
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