Optics
Q. What happens to the interference pattern if the two slits in a double-slit experiment are no longer coherent?
A.
The pattern becomes sharper
B.
The pattern disappears
C.
The pattern becomes brighter
D.
The pattern remains unchanged
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Solution
If the two slits are not coherent, the interference pattern will disappear as the waves will not maintain a constant phase relationship.
Correct Answer: B — The pattern disappears
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Q. What happens to the interference pattern if the wavelength of light is increased?
A.
Fringe width decreases
B.
Fringe width increases
C.
Fringe pattern disappears
D.
Fringe width remains the same
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Solution
Increasing the wavelength increases the fringe width, as fringe width is directly proportional to the wavelength.
Correct Answer: B — Fringe width increases
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Q. What happens to the light intensity when it passes through a polarizer at an angle of 60 degrees?
A.
It is halved
B.
It is reduced to one quarter
C.
It remains the same
D.
It is doubled
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Solution
The transmitted intensity is given by I = I_0 * cos²(θ), which results in one quarter of the original intensity at 60 degrees.
Correct Answer: B — It is reduced to one quarter
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Q. What happens to the light intensity when it passes through two crossed polarizers?
A.
It doubles
B.
It is halved
C.
It becomes zero
D.
It remains the same
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Solution
When light passes through two crossed polarizers (90 degrees apart), no light is transmitted, resulting in zero intensity.
Correct Answer: C — It becomes zero
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Q. What happens to the light intensity when it passes through two polarizers aligned at an angle of 30 degrees?
A.
It remains the same
B.
It is halved
C.
It is reduced to one-fourth
D.
It is reduced to three-fourths
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Solution
Using Malus's law, the transmitted intensity I = I_0 * cos²(θ). For θ = 30 degrees, I = I_0 * (√3/2)² = (3/4)I_0.
Correct Answer: D — It is reduced to three-fourths
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Q. What happens to the light ray when it hits the boundary at an angle greater than the critical angle?
A.
It is refracted
B.
It is absorbed
C.
It is reflected back into the denser medium
D.
It passes into the rarer medium
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Solution
When the angle of incidence exceeds the critical angle, the light ray undergoes total internal reflection and is reflected back into the denser medium.
Correct Answer: C — It is reflected back into the denser medium
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Q. What happens to the light ray when it hits the boundary at the critical angle?
A.
It is refracted into the second medium.
B.
It is reflected back into the first medium.
C.
It travels along the boundary.
D.
It is absorbed.
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Solution
At the critical angle, the light ray is refracted at 90°, traveling along the boundary between the two media.
Correct Answer: C — It travels along the boundary.
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Q. What happens to the polarization of light when it passes through a polarizer at an angle of 45 degrees?
A.
It becomes unpolarized
B.
It is completely absorbed
C.
It is partially polarized
D.
It is fully polarized
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Solution
When light passes through a polarizer at an angle of 45 degrees, it becomes partially polarized.
Correct Answer: C — It is partially polarized
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Q. What happens to the speed of light as it passes from air into a denser medium like glass?
A.
It increases
B.
It decreases
C.
It remains the same
D.
It becomes zero
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Solution
The speed of light decreases when it passes from a less dense medium (air) to a denser medium (glass) due to the change in refractive index.
Correct Answer: B — It decreases
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Q. What is Brewster's angle for a medium with a refractive index of 1.5?
A.
30 degrees
B.
45 degrees
C.
60 degrees
D.
53 degrees
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Solution
Brewster's angle can be calculated using the formula tan(θ) = n, which gives approximately 53 degrees for n = 1.5.
Correct Answer: D — 53 degrees
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Q. What is Brewster's angle?
A.
The angle at which light is completely absorbed
B.
The angle at which light is reflected with maximum polarization
C.
The angle at which light refracts without any reflection
D.
The angle at which light travels fastest
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Solution
Brewster's angle is the angle of incidence at which light is reflected with maximum polarization.
Correct Answer: B — The angle at which light is reflected with maximum polarization
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Q. What is the angle of incidence at which light is completely polarized upon reflection?
A.
0 degrees
B.
45 degrees
C.
60 degrees
D.
Brewster's angle
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Solution
Light is completely polarized upon reflection at Brewster's angle, which is given by the formula tan(θ_B) = n2/n1.
Correct Answer: D — Brewster's angle
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Q. What is the Brewster's angle for a medium with a refractive index of 1.5?
A.
30 degrees
B.
45 degrees
C.
60 degrees
D.
53 degrees
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Solution
Brewster's angle can be calculated using the formula tan(θ_B) = n, where n is the refractive index. For n = 1.5, θ_B = arctan(1.5) ≈ 53 degrees.
Correct Answer: D — 53 degrees
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Q. What is the Brewster's angle for light entering a medium with a refractive index of 1.5?
A.
30 degrees
B.
45 degrees
C.
60 degrees
D.
53 degrees
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Solution
Brewster's angle can be calculated using the formula tan(θ_B) = n, where n is the refractive index. For n = 1.5, θ_B = arctan(1.5) ≈ 53 degrees.
Correct Answer: D — 53 degrees
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Q. What is the Brewster's angle for light in air (n=1) reflecting off glass (n=1.5)?
A.
30 degrees
B.
45 degrees
C.
60 degrees
D.
53 degrees
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Solution
Brewster's angle can be calculated using the formula tan(θ_B) = n2/n1, which gives approximately 53 degrees.
Correct Answer: D — 53 degrees
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Q. What is the Brewster's angle for light traveling from air (n1 = 1) to glass (n2 = 1.5)?
A.
30 degrees
B.
45 degrees
C.
60 degrees
D.
53 degrees
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Solution
Brewster's angle θ_B can be calculated using θ_B = arctan(n2/n1) = arctan(1.5) ≈ 56.31 degrees.
Correct Answer: D — 53 degrees
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Q. What is the Brewster's angle?
A.
The angle at which light is completely absorbed
B.
The angle at which light is reflected with maximum polarization
C.
The angle at which light refracts without any reflection
D.
The angle at which light intensity is halved
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Solution
Brewster's angle is the angle of incidence at which light is reflected with maximum polarization.
Correct Answer: B — The angle at which light is reflected with maximum polarization
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Q. What is the condition for constructive interference in a double-slit experiment?
A.
Path difference is an odd multiple of lambda/2
B.
Path difference is an even multiple of lambda
C.
Path difference is an odd multiple of lambda
D.
Path difference is zero
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Solution
Constructive interference occurs when the path difference between the two waves is an even multiple of the wavelength (nλ, where n is an integer).
Correct Answer: B — Path difference is an even multiple of lambda
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Q. What is the condition for constructive interference in a thin film?
A.
2t = (m + 1/2)λ
B.
2t = mλ
C.
t = mλ/2
D.
t = (m + 1/2)λ/2
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Solution
For constructive interference, the condition is 2t = mλ, where t is the thickness of the film and m is an integer.
Correct Answer: B — 2t = mλ
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Q. What is the condition for constructive interference in two waves?
A.
Path difference = (n + 1/2)λ
B.
Path difference = nλ
C.
Path difference = (n - 1/2)λ
D.
Path difference = 0
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Solution
Constructive interference occurs when the path difference is nλ, where n is an integer.
Correct Answer: B — Path difference = nλ
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Q. What is the condition for light to be completely polarized by reflection?
A.
Angle of incidence equals angle of refraction
B.
Angle of incidence equals Brewster's angle
C.
Angle of incidence is 45 degrees
D.
Light must be monochromatic
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Solution
Light is completely polarized when it is reflected at Brewster's angle, which is given by θ_B = arctan(n2/n1).
Correct Answer: B — Angle of incidence equals Brewster's angle
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Q. What is the condition for light to be polarized by reflection?
A.
Angle of incidence equals angle of refraction
B.
Angle of incidence equals Brewster's angle
C.
Angle of incidence is 45 degrees
D.
Light must be monochromatic
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Solution
Light is polarized by reflection when the angle of incidence equals Brewster's angle.
Correct Answer: B — Angle of incidence equals Brewster's angle
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Q. What is the condition for the first minimum in single-slit diffraction?
A.
a = λ
B.
a = 2λ
C.
a = λ/2
D.
a = 3λ
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Solution
The first minimum occurs at a = λ, where 'a' is the slit width and 'λ' is the wavelength.
Correct Answer: A — a = λ
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Q. What is the critical angle for a glass-air interface if the refractive index of glass is 1.5?
A.
30 degrees
B.
45 degrees
C.
60 degrees
D.
90 degrees
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Solution
Critical angle θc = sin^(-1)(1/n) = sin^(-1)(1/1.5) = 41.81 degrees, approximately 60 degrees.
Correct Answer: C — 60 degrees
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Q. What is the critical angle for total internal reflection from glass to air if the refractive index of glass is 1.5?
A.
30 degrees
B.
41.8 degrees
C.
48.6 degrees
D.
60 degrees
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Solution
Critical angle (C) is given by sin(C) = n2/n1. Thus, sin(C) = 1/1.5, C = sin^(-1)(0.6667) ≈ 41.8 degrees.
Correct Answer: B — 41.8 degrees
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Q. What is the critical angle for total internal reflection from water (n = 1.33) to air (n = 1)?
A.
48.6 degrees
B.
53.1 degrees
C.
60 degrees
D.
90 degrees
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Solution
The critical angle (θc) can be calculated using sin(θc) = n2/n1. Thus, sin(θc) = 1/1.33, giving θc ≈ 53.1 degrees.
Correct Answer: B — 53.1 degrees
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Q. What is the critical angle for total internal reflection if the refractive index of the medium is 1.5?
A.
30 degrees
B.
45 degrees
C.
60 degrees
D.
90 degrees
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Solution
The critical angle (θc) can be calculated using sin(θc) = 1/n. Here, n = 1.5, so θc = sin^(-1)(1/1.5) ≈ 41.81 degrees, which is approximately 42 degrees.
Correct Answer: C — 60 degrees
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Q. What is the critical angle for total internal reflection in a medium with a refractive index of 1.5?
A.
30 degrees
B.
45 degrees
C.
60 degrees
D.
90 degrees
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Solution
The critical angle (θc) can be calculated using sin(θc) = 1/n. For n = 1.5, sin(θc) = 1/1.5 = 2/3. Therefore, θc = sin^(-1)(2/3) which is approximately 41.81 degrees, closest to 45 degrees.
Correct Answer: C — 60 degrees
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Q. What is the critical angle for total internal reflection when light travels from glass (n = 1.5) to air (n = 1)?
A.
30 degrees
B.
41.8 degrees
C.
48.6 degrees
D.
60 degrees
Show solution
Solution
The critical angle θc is given by sin(θc) = n2/n1. Thus, θc = sin^(-1)(1/1.5) ≈ 41.8 degrees.
Correct Answer: B — 41.8 degrees
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Q. What is the critical angle for total internal reflection when light travels from water (n = 1.33) to air (n = 1)?
A.
48.6 degrees
B.
53.1 degrees
C.
60 degrees
D.
90 degrees
Show solution
Solution
The critical angle (θc) can be calculated using sin(θc) = n2/n1. Here, n1 = 1.33 (water) and n2 = 1 (air). Thus, sin(θc) = 1/1.33, giving θc ≈ 53.1 degrees.
Correct Answer: B — 53.1 degrees
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