Physics Syllabus (JEE Main)
Q. In a diffraction pattern, what does the term 'order of diffraction' refer to?
A.
The number of slits used
B.
The number of maxima observed
C.
The wavelength of light used
D.
The distance to the screen
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Solution
The order of diffraction refers to the number of maxima observed in the diffraction pattern, with the central maximum being the zeroth order.
Correct Answer: B — The number of maxima observed
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Q. In a diffraction pattern, what is the effect of increasing the slit width?
A.
Wider central maximum
B.
Narrower central maximum
C.
No effect
D.
Increased number of fringes
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Solution
Increasing the slit width results in a wider central maximum due to the decrease in diffraction effects.
Correct Answer: A — Wider central maximum
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Q. In a diffraction pattern, what is the ratio of the intensity of the central maximum to that of the first minimum?
A.
1:0
B.
1:1
C.
1:2
D.
1:4
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Solution
The intensity of the central maximum is typically much greater than that of the first minimum, often approximated as 1:4.
Correct Answer: D — 1:4
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Q. In a digital communication system, what does the term 'bit rate' refer to?
A.
The number of bits transmitted per second
B.
The number of bits stored in a device
C.
The frequency of the carrier wave
D.
The duration of each bit
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Solution
Bit rate refers to the number of bits transmitted per second in a digital communication system.
Correct Answer: A — The number of bits transmitted per second
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Q. In a diode, what is the region called where no charge carriers are present?
A.
Conduction band
B.
Valence band
C.
Depletion region
D.
N-type region
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Solution
The depletion region in a diode is the area where no charge carriers are present.
Correct Answer: C — Depletion region
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Q. In a diode, what is the region called where no current flows?
A.
Forward bias
B.
Reverse bias
C.
Depletion region
D.
Conduction region
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Solution
The depletion region is the area in a diode where no current flows under reverse bias.
Correct Answer: C — Depletion region
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Q. In a double convex lens, if the radii of curvature are 10 cm and 15 cm, what is the focal length using the lens maker's formula?
A.
6 cm
B.
10 cm
C.
12 cm
D.
8 cm
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Solution
Using the lens maker's formula, f = (R1 * R2) / (n - 1) * (1/R1 - 1/R2), we find f = 12 cm.
Correct Answer: C — 12 cm
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Q. In a double-slit experiment, if the distance between the screen and the slits is increased, what happens to the fringe separation?
A.
It increases
B.
It decreases
C.
It remains the same
D.
It becomes zero
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Solution
Fringe separation is directly proportional to the distance from the slits to the screen; increasing this distance increases fringe separation.
Correct Answer: A — It increases
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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 0.5 mm and the distance to the screen is 2 m, what is the distance between the first and second bright fringes if the wavelength of light used is 500 nm?
A.
0.5 cm
B.
1 cm
C.
1.5 cm
D.
2 cm
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Solution
Using the formula for fringe separation, y = (λD)/d, we find y = (500 x 10^-9 m * 2 m) / (0.5 x 10^-3 m) = 2 cm.
Correct Answer: B — 1 cm
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Q. In a double-slit experiment, if the distance between the slits is doubled while keeping the wavelength constant, 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 is halved.
Correct Answer: B — Halves
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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 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 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 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 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 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 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 happens to the interference pattern if the distance between the slits is increased?
A.
Pattern becomes wider
B.
Pattern becomes narrower
C.
No change
D.
Pattern disappears
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Solution
Increasing the distance between the slits decreases the fringe width, making the pattern narrower.
Correct Answer: B — Pattern becomes narrower
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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 fluid at rest, the pressure at a depth h is given by which equation?
A.
P = ρgh
B.
P = ρg/h
C.
P = ρg + h
D.
P = gh/ρ
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Solution
The pressure at a depth h in a fluid is given by P = ρgh, where ρ is the fluid density and g is the acceleration due to gravity.
Correct Answer: A — P = ρgh
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