Optics
Q. In a double-slit experiment, if the distance between the slits is doubled, what happens to the fringe width?
A.
Doubles
B.
Halves
C.
Remains the same
D.
Quadruples
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Solution
Fringe width (β) is given by β = λD/d, where d is the distance between the slits. If d is doubled, β becomes β/2, hence the fringe width halves.
Correct Answer: B — Halves
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Q. In a double-slit experiment, if the distance between the slits is doubled, what happens to the fringe separation on the screen?
A.
It doubles
B.
It halves
C.
It remains the same
D.
It quadruples
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Solution
The fringe separation is inversely proportional to the distance between the slits; thus, if the distance is doubled, the fringe separation halves.
Correct Answer: B — It halves
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Q. In a double-slit experiment, if the distance between the slits is halved, what happens to the fringe separation on the screen?
A.
It doubles
B.
It halves
C.
It remains the same
D.
It quadruples
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Solution
The fringe separation is inversely proportional to the distance between the slits; halving the distance doubles the fringe separation.
Correct Answer: A — It doubles
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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 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 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 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 distance is increased, what happens to the fringe separation?
A.
Fringe separation increases
B.
Fringe separation decreases
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), hence it increases.
Correct Answer: A — Fringe separation increases
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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 double-slit experiment, if the wavelength of light is 600 nm and the distance between the slits is 0.5 mm, what is the fringe width if the screen is 1 m away?
A.
0.12 mm
B.
0.3 mm
C.
0.6 mm
D.
0.5 mm
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Solution
Fringe width β = λD/d = (600 x 10^-9 m)(1 m)/(0.5 x 10^-3 m) = 0.12 mm.
Correct Answer: A — 0.12 mm
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Q. In a double-slit experiment, if the wavelength of light is increased, what happens to the distance between the fringes?
A.
Increases
B.
Decreases
C.
Remains the same
D.
Becomes zero
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Solution
The distance between the fringes increases with an increase in wavelength, as fringe separation is directly proportional to wavelength.
Correct Answer: A — Increases
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Q. In a double-slit experiment, what is the effect of increasing the distance between the slits on the fringe width?
A.
Fringe width increases
B.
Fringe width decreases
C.
Fringe width remains constant
D.
Fringe width becomes zero
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Solution
Increasing the distance between the slits increases the fringe width because the angle of diffraction increases.
Correct Answer: A — Fringe width increases
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Q. In a Michelson interferometer, what happens to the interference pattern if one of the mirrors is moved away from the beam splitter?
A.
Fringes move closer
B.
Fringes move apart
C.
Fringes disappear
D.
No change in pattern
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Solution
Moving one mirror changes the path length of one beam, causing the fringes to move apart or closer depending on the direction of movement.
Correct Answer: B — Fringes move apart
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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 Michelson interferometer, what happens to the interference pattern if one of the mirrors is moved?
A.
The pattern disappears
B.
The pattern shifts
C.
The pattern becomes brighter
D.
The pattern becomes dimmer
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Solution
Moving one of the mirrors changes the path length of one beam, causing a shift in the interference pattern.
Correct Answer: B — The pattern shifts
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Q. In a Michelson interferometer, what happens when one of the mirrors is moved slightly?
A.
No change in interference pattern
B.
Fringes shift
C.
Fringes disappear
D.
Fringes become brighter
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Solution
Moving one of the mirrors changes the path length, causing a shift in the interference pattern (fringes).
Correct Answer: B — Fringes shift
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Q. In a Newton's rings experiment, if the radius of the ring increases, what can be inferred about the wavelength of light used?
A.
Wavelength is increasing
B.
Wavelength is decreasing
C.
Wavelength remains constant
D.
Wavelength cannot be determined
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Solution
The radius of the rings is directly proportional to the wavelength of light used. If the radius increases, the wavelength must also be increasing.
Correct Answer: A — Wavelength is increasing
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Q. In a plane mirror, if an object is placed 10 cm in front of the mirror, where will the image be formed?
A.
5 cm
B.
10 cm
C.
15 cm
D.
20 cm
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Solution
In a plane mirror, the image is formed at the same distance behind the mirror as the object is in front, so the image is at 10 cm.
Correct Answer: B — 10 cm
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Q. In a prism, if the angle of incidence is equal to the angle of emergence, what can be said about the angle of deviation?
A.
It is zero
B.
It is equal to the angle of incidence
C.
It is equal to the angle of emergence
D.
It is equal to the angle of the prism
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Solution
When the angle of incidence equals the angle of emergence, the angle of deviation is equal to the angle of the prism.
Correct Answer: D — It is equal to the angle of the prism
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Q. In a prism, if the angle of the prism is 60 degrees, what is the minimum angle of deviation for light passing through it?
A.
30 degrees
B.
60 degrees
C.
90 degrees
D.
45 degrees
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Solution
The minimum angle of deviation (D) for a prism is given by D = A, where A is the angle of the prism. Therefore, for a 60-degree prism, the minimum angle of deviation is 30 degrees.
Correct Answer: A — 30 degrees
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Q. In a single-slit diffraction experiment, what happens to the width of the central maximum as the slit width decreases?
A.
It increases
B.
It decreases
C.
It remains the same
D.
It becomes zero
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Solution
As the slit width decreases, the central maximum becomes wider due to increased diffraction.
Correct Answer: A — It increases
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Q. In a single-slit diffraction pattern, how does the intensity of the central maximum compare to the first minimum?
A.
Equal
B.
Twice
C.
Four times
D.
Half
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Solution
The intensity of the central maximum is four times that of the first minimum in a single-slit diffraction pattern.
Correct Answer: C — Four times
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Q. In a single-slit diffraction pattern, how does the intensity of the first minimum compare to the intensity of the central maximum?
A.
It is equal
B.
It is half
C.
It is zero
D.
It is one-fourth
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Solution
The intensity at the first minimum is zero, while the central maximum has maximum intensity.
Correct Answer: C — It is zero
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Q. In a single-slit diffraction pattern, how many minima are there on either side of the central maximum?
A.
One
B.
Two
C.
Three
D.
Infinite
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Solution
In a single-slit diffraction pattern, there are theoretically infinite minima on either side of the central maximum.
Correct Answer: D — Infinite
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Q. In a single-slit diffraction pattern, the width of the central maximum is 4 mm. If the slit width is halved, what will be the new width of the central maximum?
A.
2 mm
B.
4 mm
C.
8 mm
D.
16 mm
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Solution
The width of the central maximum is inversely proportional to the slit width. Halving the slit width doubles the width of the central maximum to 8 mm.
Correct Answer: C — 8 mm
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Q. In a single-slit diffraction pattern, what is the angle for the first minimum if the slit width is 0.5 mm and the wavelength of light is 600 nm?
A.
30°
B.
60°
C.
45°
D.
15°
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Solution
For single-slit diffraction, the first minimum occurs at sin θ = λ/a. Here, sin θ = 600 x 10^-9 m / 0.5 x 10^-3 m = 0.0012, θ ≈ 0.0698 rad ≈ 4°.
Correct Answer: C — 45°
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Q. In a single-slit diffraction pattern, what is the angle for the first minimum?
A.
sin(θ) = λ/a
B.
sin(θ) = 2λ/a
C.
sin(θ) = 3λ/a
D.
sin(θ) = 0
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Solution
The angle for the first minimum in single-slit diffraction is given by sin(θ) = λ/a, where a is the slit width.
Correct Answer: A — sin(θ) = λ/a
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