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NEET • PHYSICS • WAVE OPTICS

Wave optics practice · Mixed difficulty

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91 Hard
91 of 150
Application +4 | -1
A thin transparent film with refractive index 1.4, is held on circular ring of radius 1.8 cm. The fluid in the film evaporates such that transmission through the film at wavelength 560 nm goes to a minimum every 12 seconds. Assuming that the film is flat on its two sides, the rate of evaporation is _____ π×10−13 m3/s.\pi \times 10^{-13} \mathrm{~m}^3 / \mathrm{s}.
View answer and explanation
Correct option: A
54
Option A (54) is correct. The rate of evaporation is calculated by considering the change in thickness of the film and the time interval for minimum transmission.
92 Easy
92 of 150
Application +4 | -1
If the free space between the plates is replaced by a transparent liquid of refractive index 1.51.5 and the wavelength of light remains 4200A∘4200 \mathop A\limits^ \circ the number of fringes seen will be:
View answer and explanation
Correct option: A
45
Option A (45) is correct because the number of fringes is directly proportional to the refractive index of the medium.
93 Medium
93 of 150
Analyzing +4 | -1
Statement-1 : When light is incident from air to water then Brewster's angle is θB\theta_{{B}} then if light is incident from water to air then Brewster's angle is π2−θB\frac{\pi}{2}-\theta_{{B}}.
Statement-2 : When light goes from air to any medium of refractive index is then Brewster's angle (θB)\left(\theta_{{B}}\right) is given by θB=tan⁡−1(μ)\theta_{{B}}=\tan ^{-1}(\mu).
View answer and explanation
Correct option: A
Both statement-1 and Statement-2 is true
Statement-1 and Statement-2 are both true. Brewster's angle relationship holds for both air to water and water to air incidence, and the formula for Brewster's angle is correctly stated.
94 Easy
94 of 150
Application +4 | -1
Monochromatic light of wavelength 600 nm600 \mathrm{~nm} is incident from air on a glass surface (refractive index of glass is 1.51.5). The frequency of the refracted light will be
View answer and explanation
Correct option: A
5×1014 Hz5 \times 10^{14} \mathrm{~Hz}
Option A (5×1014 Hz5 \times 10^{14} \mathrm{~Hz}) is correct because the frequency of light remains constant when it travels from one medium to another; it only depends on the source.
95 Medium
95 of 150
Remembering +4 | -1
For the constructive interference the path difference between the two interfering waves must be equal to
View answer and explanation
Correct option: C
nλn\lambda
Option C is correct. For constructive interference, the path difference must be an integer multiple of the wavelength, represented as nλn\lambda.
96 Medium
96 of 150
Analyzing +4 | -1
Assertion: Diffraction is common in sound but not common in light waves.

Reason: Wavelength of light is more than the wavelength of sound.
View answer and explanation
Correct option: C
Assertion is true and Reason is false
Option C (Assertion is true and Reason is false) is correct. Diffraction is more common in sound because sound's wavelength is *larger* than light's, not smaller.
97 Medium
97 of 150
Remembering +4 | -1
Fraunhoffer spectrum is
View answer and explanation
Correct option: A
Line absorption spectrum
Option A (Line absorption spectrum) is correct because Fraunhofer lines are dark lines in the solar spectrum, indicating absorption of specific wavelengths.
98 Hard
98 of 150
Application +4 | -1
Diameter of human eye lens is 2 mm2 {~mm}. What will be the minimum distance between two points to resolve them, which are situated at a distance of 50 m50 {~m} from eye. The wavelength of light is 5000A∘5000\mathop {\text{A}}\limits^ \circ
View answer and explanation
Correct option: C
1.25 cm1.25 {~cm}
Option C (1.25 cm1.25 {~cm}) is correct because it is calculated using the resolution limit formula.
99 Medium
99 of 150
Remembering +4 | -1
Intensity of light depends upon
View answer and explanation
Correct option: C
Amplitude
Option C is correct. The intensity of light is directly proportional to the square of the amplitude of the light wave.
100 Medium
100 of 150
Understanding +4 | -1
If the intensities of the two interfering beams in Young’s double alit experiment be I1{I_1} and I2{I_2}, then the contrast between the maximum and minimum intensity is good when
View answer and explanation
Correct option: C
I1=I2{I_1} = {I_2}
Option C is correct. The best contrast between maximum and minimum intensity occurs when the intensities of the two interfering beams are equal, I1=I2{I_1} = {I_2}.
101 Easy
101 of 150
Application +4 | -1
Two glass plates are touching at one end and separated by a thin wire at the other end. When a monochromatic parallel beam of wavelength 4200A∘4200 \mathop A\limits^ \circ incident normally on the glass plates is reflected, an interference pattern of 30 fringes are observed. If the wavelength of light used is taken 7000A∘7000 \mathop A\limits^ \circ instead of 4200A∘4200 \mathop A\limits^ \circ, the number of fringes observed will be:
View answer and explanation
Correct option: A
18
Option A (18) is correct because the number of fringes is inversely proportional to the wavelength of light used.
102 Easy
102 of 150
Understanding +4 | -1
A thin air film is formed by putting the convex surface of a plane-convex lens over a plane glass plate. With monochromatic light, this film gives an interference pattern due to light reflected from the top (convex) surface and the bottom (glass plate) surface of the film.

Statement I : When light reflects from the air-glass plate interface, the reflected wave suffers a phase change of π\pi.

Statement II : The centre of the interference pattern is dark.
View answer and explanation
Correct option: B
Statement I is true, Statement II is true; Statement II is not the correct explanation for Statement I
Option B is correct. Statement I is true because light reflecting from a denser medium (air-glass) undergoes a phase change of π\pi. Statement II is also true because the center is dark due to this phase change, but the phase change is not the explanation for statement I itself.
103 Medium
103 of 150
Application +4 | -1
Light of wavelength 550 nm falls normally on a slit of width 22.0×10−5cm22.0 \times {10^{ - 5}}cm. The angular position of the second minima from the central maximum will be (in radians)
View answer and explanation
Correct option: D
π6\frac{\pi }{6}
The correct answer is D (π6\frac{\pi}{6}) because it represents the angular position of the second minima calculated using the single-slit diffraction formula.
104 Medium
104 of 150
Remembering +4 | -1
Fraunhoffer lines are obtained in
View answer and explanation
Correct option: A
Solar spectrum
Option A (Solar spectrum) is correct because Fraunhofer lines are observed in the solar spectrum due to absorption by elements in the Sun's atmosphere.
105 Medium
105 of 150
Remembering +4 | -1
In an interference experiment, the spacing between successive maxima or minima is
View answer and explanation
Correct option: B
λD/d\lambda D/d
Option B (λD/d\lambda D/d) is correct because the fringe width in an interference experiment is given by this formula.
106 Medium
106 of 150
Remembering +4 | -1
The idea of secondary wavelets for the propagation of a wave was first given by
View answer and explanation
Correct option: B
Huygen
Option B (Huygen) is correct because Huygens' principle describes wave propagation using secondary wavelets.
107 Medium
107 of 150
Application +4 | -1
Two straight narrow slits separated by a distance 0.08m mare illuminated by a source of light of wavelength 5460 A∘5460\,\mathop \mathrm{A}\limits^ \circ. If the fringes are measured with a micrometer eyepiece it is found that 2 fringes occupy a distance of 10.92mm. The distance of the eye piece from the slits is
View answer and explanation
Correct option: A
0.8 m
Option A (0.8 m) is correct because it is calculated using the fringe width formula and the given parameters.
108 Medium
108 of 150
Remembering +4 | -1
In Huygen’s wave theory, the locus of all points in the same state of vibration is called
View answer and explanation
Correct option: C
A wave-front
Option C (A wave-front) is correct because a wavefront, according to Huygens' principle, is the locus of all points in the same phase of vibration.
109 Medium
109 of 150
Understanding +4 | -1
In Young’s double slit experiment, the separation between the slit and the screen increases. The fringe width
View answer and explanation
Correct option: A
Increases
Option A (Increases) is correct because the fringe width is directly proportional to the distance between the slit and the screen.
110 Medium
110 of 150
Analyzing +4 | -1
Given below two statements: One is labelled as Assertion (A) and other is labelled as Reason (R).
Assertion (A): Non-polar materials do not have any permanent dipole moment.
Reason (R): When a non-polar material is placed in an electric field, the centre of the positive charge distribution of it’s individual atom or molecule coincides with the centre of the negative charge distribution.
In the light of above statements, choose the most appropriate answer from the options given below.
View answer and explanation
Correct option: C
(A) is correct but (R) is not correct.
Option C ((A) is correct but (R) is not correct) is correct because while non-polar materials indeed lack permanent dipole moments, the reason provided is inaccurate; in an electric field, the charge distributions do not coincide, leading to induced polarization.
111 Medium
111 of 150
Understanding +4 | -1
If white light is used in the Newton’s rings experiment, the colour observed in the reflected light is complementary to that observed in the transmitted light is complementary to that observed in the transmitted light through the same point. This is due to
View answer and explanation
Correct option: B
180∘180^\circchange of phase in one of the reflected waves
Option B (180∘180^\circchange of phase in one of the reflected waves) is correct because the complementary colors arise due to the 180∘180^\circ phase change upon reflection from a denser medium.
112 Medium
112 of 150
Understanding +4 | -1
Resolving power of a telescope increases when
View answer and explanation
Correct option: D
Wavelength of light decreases
Option D is correct. The resolving power of a telescope is inversely proportional to the wavelength of light, so it increases when the wavelength decreases.
113 Medium
113 of 150
Understanding +4 | -1
When a compact disc is illuminated by small source of white light, coloured bands are observed. This is due to
View answer and explanation
Correct option: B
Diffraction
Option C (Interference) is correct because the colored bands observed on a compact disc are due to the interference of light waves reflected from the disc's surface.
114 Medium
114 of 150
Remembering +4 | -1
According to corpuscular theory of light, the different colours of light are due to
View answer and explanation
Correct option: C
Different size of the corpuscles
Option C (Different size of the corpuscles) is correct because, according to Newton's corpuscular theory, different colors corresponded to different sizes of light particles.
115 Medium
115 of 150
Remembering +4 | -1
The wave theory of light, in its original form, was first postulated by
View answer and explanation
Correct option: B
Christian Huygens
Option B (Christian Huygens) is correct because Christian Huygens first proposed the wave theory of light.
116 Medium
116 of 150
Analyzing +4 | -1
In a young's double slit experiment light of wavelength 600 nm600 \mathrm{~nm} is used. P\mathrm{P} is the nearest point from the central maximum where intensity is 3/4th 3 / 4^{\text {th }} of the maximum intensity. What percentage of the maximum intensity is observed at point PP if light is replaced by the wavelength 400 nm400 \mathrm{~nm} ?
View answer and explanation
Correct option: B
75%75 \%
Option B (75%) is correct because when the wavelength changes, the phase difference at point P changes, leading to a different intensity which is calculated to be 75% of the maximum intensity.
117 Hard
117 of 150
Application +4 | -1
In YDSE of equal width slits, if intensity at the centre of screen is I0I_0, then intensity at a distance of β/4\beta / 4 from the central maxima is (β\beta is the fringe width):
View answer and explanation
Correct option: B
I02\frac{I_0}{2}
Option B (I02\frac{I_0}{2}) is correct because at a distance of β/4\beta/4 from the central maxima, the phase difference is π/2\pi/2, leading to an intensity of I0/2I_0/2.
118 Hard
118 of 150
Application +4 | -1
The Brewster angle for the glass-air interface is 54.74∘{54.74^ \circ }. If a ray of light going from air to glass strikes at an angle of incidence 45∘{45^ \circ }, then the angle of refraction is [tan⁡ 54.74∘=2]\left[ {\tan \,{{54.74}^ \circ } = \sqrt 2 } \right]
View answer and explanation
Correct option: B
30∘{30^ \circ }
Option B (30∘{30^ \circ }) is correct because it is the angle of refraction calculated using Snell's law and the given Brewster's angle.
119 Medium
119 of 150
Analyzing +4 | -1
Statement I : For a given medium, the polarising angle is 60∘60^{\circ}. The critical angle for this medium is 35∘35^{\circ}.

Statement II : μ=tan⁡iρ\mu=\tan i_\rho.
View answer and explanation
Correct option: A
Statement I is true, Statement II is true; Statement II is the correct explanation for Statement 3
Option B is correct. Statement I is true because the refractive index can be found using the polarizing angle, and then the critical angle can be calculated. Statement II is also true as it defines the relationship between refractive index and polarizing angle, but it doesn't explain Statement I.
120 Hard
120 of 150
Application +4 | -1
Frequencies of light incident on a system of scattering particles are in the ratio of 1:2. Then, the intensity of scattered light in a particular direction is
View answer and explanation
Correct option: D
0.052777778
The correct answer is 0.052777778 because the intensity of scattered light is proportional to the fourth power of the frequency.
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