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JEE • PHYSICS • WAVE OPTICS

Wave optics practice · Mixed difficulty

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These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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Showing 151-180 of 418 questions

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151 Medium
151 of 418
Understanding +4 | -1
When unpolarised light beam is incident from air onto glass (n=1.5)\left( {n = 1.5} \right) at the polarizing angle
View answer and explanation
Correct option: A
Reflected beam is polarized 100 percent
Option A (Reflected beam is polarized 100 percent) is correct because at the polarizing angle (Brewster's angle), the reflected light is completely polarized.
152 Medium
152 of 418
Application +4 | -1
In a Young's double slit experiment in front of one of the slits a thin transparent sheet of thickness tt and refractive index μ\mu is inserted. The entire fringe pattern will be shifted by a distance (dd - separation between slits; DD - distance of screen from the slits and λ\lambda - wavelength of light used)
View answer and explanation
Correct option: C
(μ−1)tDd\frac{(\mu-1) t D}{d}
Option C ((μ−1)tDd\frac{(\mu-1) t D}{d}) is correct because the fringe shift is proportional to the path difference introduced by the sheet, the distance to the screen, and inversely proportional to the slit separation.
153 Medium
153 of 418
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
Both statement-1 and Statement-2 are true. Statement 2 is the definition of Brewster's angle, and statement 1 correctly relates Brewster's angles for air-water and water-air interfaces.
154 Medium
154 of 418
Application +4 | -1
The limit of resolution of microscope, if the numerical aperture of microscope is 0.12, and the wavelength of light used is 600 nm, is
View answer and explanation
Correct option: D
3μm3\mu m
Option D (3μm3\mu m) is correct because it's calculated using the formula for the limit of resolution of a microscope, considering the numerical aperture and wavelength of light.
155 Medium
155 of 418
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 A (Statement I is true, Statement II is true; Statement II is the correct explanation for Statement 3) is correct because the polarizing angle and critical angle are related through the refractive index, and the given values satisfy this relationship.
156 Medium
156 of 418
Application +4 | -1
The wave front of a light beam is given by the equation x + 2y + 3z = constant. The angle made by the direction of light with the y-axis is
View answer and explanation
Correct option: C
cos⁡−1(214){\cos ^{ - 1}}\left( {\frac{2}{{\sqrt {14} }}} \right)
Option C (cos⁡−1(214)){\cos ^{ - 1}}(\frac{2}{{\sqrt {14} }})) is correct because the angle is found using the direction vector of the plane and the dot product formula.
157 Medium
157 of 418
Application +4 | -1
Ratio of intensities of two waves from slits S1S_1 and S2S_2 is 4:14: 1 respectively. Then the ratio of maximum to that of minimum intensities in YDSEYDSE is
View answer and explanation
Correct option: B
9:19: 1
Option B (9:19: 1) is correct because it is the ratio of the maximum to minimum intensities calculated from the given ratio of intensities of the two waves.
158 Medium
158 of 418
Understanding +4 | -1
The rectilinear propagation of light in a medium is due to
View answer and explanation
Correct option: C
Its high velocity
Option C (Its high velocity) is correct because light travels at a very high speed, causing it to experience negligible gravitational force and thus propagate in a straight line.
159 Medium
159 of 418
Understanding +4 | -1
Two beams of light will not give rise to an interference pattern, if
View answer and explanation
Correct option: D
They are not monochromatic
The correct answer is D (They are not monochromatic) because interference requires coherent and monochromatic light sources.
160 Medium
160 of 418
Application +4 | -1
Two points, separated by a distance of 0.1  mm\mathrm{0}\mathrm{.1}\,\,\mathrm{mm}, can just be inspected on a microscope when light of wavelength 6000 A∘6000\,\mathop \mathrm{A}\limits^ \circ is used. If the light of wavelength 4800 A∘4800\,\mathop \mathrm{A}\limits^ \circ is used, the limit of resolution is
View answer and explanation
Correct option: B
0.08  mm\mathrm{0}\mathrm{.08}\,\,\mathrm{mm}
Option B (0.08 mm\mathrm{0}\mathrm{.08}\,\mathrm{mm}) is correct because the limit of resolution is directly proportional to the wavelength of light used.
161 Medium
161 of 418
Application +4 | -1
Two waves of amplitude 3 mm3 \mathrm{~mm} and 2 mm2 \mathrm{~mm} reach a point in the same phase. The resultant amplitude is
View answer and explanation
Correct option: B
5 mm5 \mathrm{~mm}
Option B (5 mm) is correct because when waves are in phase, their amplitudes add directly.
162 Medium
162 of 418
Remembering +4 | -1
According to Huygens, medium through which light waves travel is
View answer and explanation
Correct option: B
luminiferous ether
Option B (luminiferous ether) is correct because Huygens proposed that light waves travel through a hypothetical medium called luminiferous ether.
163 Medium
163 of 418
Remembering +4 | -1
A wave can transmit ………. from one place to another
View answer and explanation
Correct option: B
Amplitude
Option A (Energy) is correct because waves transfer energy from one location to another without transferring matter.
164 Hard
164 of 418
Remembering +4 | -1
In Young’s double slit experiment, a minimum is obtained when the phase difference of superimposing waves is
View answer and explanation
Correct option: B
(2n−1)π\left( {2n - 1} \right)\pi
Option B ((2n−1)π\left( {2n - 1} \right)\pi) is correct because a minimum (destructive interference) in Young's double-slit experiment occurs when the phase difference is an odd multiple of π\pi.
165 Medium
165 of 418
Application +4 | -1
Unpolarised light incident on transparent glass at incident angle 60∘60^{\circ}. If reflected ray is completely polarised, then angle of refraction is
View answer and explanation
Correct option: C
30∘30^{\circ}
Option C (30∘30^{\circ}) is correct because, according to Brewster's law, the angle of incidence plus the angle of refraction equals 90 degrees when the reflected ray is completely polarized.
166 Medium
166 of 418
Application +4 | -1
In a diffraction pattern due to a single slit of width ‘a’, the first minimum is observed at an angle 30o{30^\mathrm{o}} when light of wavelength 5000 Ao\mathop \mathrm{A}\limits^\mathrm{o} is incident on the slit. The first secondary maximum is observed at an angle of:
View answer and explanation
Correct option: D
sin⁡−1 (34){\sin ^{ - 1}}\,\left( {\frac{3}{4}} \right)
Option D (sin⁡−1 (34){\sin ^{ - 1}}\,\left( {\frac{3}{4}} \right)) is correct because it satisfies the condition for the first secondary maximum in a single-slit diffraction pattern.
167 Medium
167 of 418
Application +4 | -1
A Young's double slit apparatus has slits separated by 0.28 mm0.28 \mathrm{~mm} and a screen 48 cm48 \mathrm{~cm} away from the slits. The whole apparatus is immersed in water and the slits are illuminated by the red light ( λ=700 nm\lambda=700 \mathrm{~nm} in vacuum). Find the fringe-width of the pattern formed on the screen.
View answer and explanation
Correct option: A
0.90 mm
Option A (0.90 mm) is correct because it is calculated using the formula for fringe width in Young's double-slit experiment, considering the refractive index of water.
168 Medium
168 of 418
Understanding +4 | -1
If the sodium light in Young’s double slit experiment is replaced by red light, the fringe width will
View answer and explanation
Correct option: B
Increase
Option B is correct because the fringe width in Young's double-slit experiment is directly proportional to the wavelength of light, and red light has a longer wavelength than sodium light.
169 Medium
169 of 418
Understanding +4 | -1
To observe diffraction, the size of an aperture
View answer and explanation
Correct option: A
Should be of the same orders wavelength should be much larger than the wavelength
The correct answer is A (Should be of the same orders wavelength should be much larger than the wavelength) because diffraction is most prominent when the aperture size is comparable to the wavelength of the wave.
170 Medium
170 of 418
Application +4 | -1
Two incoherent waves of intensities I0I_0 and 4I04 I_0 superimpose at a place. What is resultant intensity at that point?
View answer and explanation
Correct option: B
5I05I_0
Option B (5I05I_0) is correct because the resultant intensity of incoherent waves is the sum of their individual intensities.
171 Medium
171 of 418
Understanding +4 | -1
Resolving power of reflecting type telescope increases with
View answer and explanation
Correct option: D
Both 1 &\& 3
Option D is correct because the resolving power of a telescope is inversely proportional to the wavelength of light and directly proportional to the diameter of the objective mirror.
172 Medium
172 of 418
Understanding +4 | -1
The phenomenon of interference is shown by
View answer and explanation
Correct option: D
All the above types of waves
Option D (All the above types of waves) is correct because interference is a phenomenon exhibited by all types of waves, including longitudinal and transverse mechanical waves, as well as electromagnetic waves.
173 Hard
173 of 418
Application +4 | -1
In young’s single slit experiment with sodium vapour lamp of wavelength 589 nm and the slits 0.589 mm apart, the half angular width of the central maximum is
View answer and explanation
Correct option: C
sin⁡−1[0.001]{\sin ^{ - 1}}\left[ {0.001} \right]
Option C (sin⁡−1[0.001]{\sin ^{ - 1}}\left[ {0.001} \right]) is correct because the half angular width of the central maximum in single-slit diffraction is given by sin⁡θ=λd\sin \theta = \frac{\lambda}{d}.
174 Medium
174 of 418
Understanding +4 | -1
Statement-I: The sound waves do not exhibit the property of polarization.
Statement-II: The sound waves do not have the transverse nature.
View answer and explanation
Correct option: C
Both statement-I &\& II are correct
Option C (Both statement-I &IIarecorrect)iscorrectbecausesoundwavesarelongitudinalandcannotbepolarized.\& II are correct) is correct because sound waves are longitudinal and cannot be polarized.
175 Medium
175 of 418
Understanding +4 | -1
Angular width (β) of central maximum of a diffraction pattern on a single slit does not depend upon
View answer and explanation
Correct option: A
Distance between slit and source
Option A (Distance between slit and source) is correct because the angular width of the central maximum in single-slit diffraction depends on the wavelength of light and the width of the slit, but not on the distance between the slit and the source.
176 Medium
176 of 418
Understanding +4 | -1
What is the effect on Fresnel’s biprism experiment when the use of white light is made
View answer and explanation
Correct option: C
Central fringe is white and all are coloured
Option C (Central fringe is white and all are coloured) is correct because with white light, the central fringe has zero path difference, resulting in white, while other fringes are colored due to different wavelengths.
177 Medium
177 of 418
Understanding +4 | -1
In order to see diffraction the thickness of the film is
View answer and explanation
Correct option: B
10000  A∘10000\;\mathop \mathrm{A}\limits^ \circ
Option B is correct because for diffraction to be observable, the size of the obstacle (film thickness) should be comparable to the wavelength of visible light, which is around 10000A∘10000 \mathop \mathrm{A}\limits^ \circ.
178 Medium
178 of 418
Understanding +4 | -1
A parallel monochromatic beam of light is incident normally on a narrow slit. A diffraction pattern is formed on a screen placed perpendicular to the direction of the incident beam. At the first minimum of the diffraction pattern, the phase difference between the rays coming from the two edges of the slit is
View answer and explanation
Correct option: D
2π2\pi
Option D (2π2\pi) is correct because for the first minimum in a single-slit diffraction pattern, the path difference between the rays from the edges of the slit is one wavelength, which corresponds to a phase difference of 2π2\pi.
179 Medium
179 of 418
Application +4 | -1
In a single slit diffraction experiment first minimum for red light (660  nm)\left( {660\;nm} \right) coincides with first maximum of some other wavelength λ{\lambda}. The value of λ{\lambda} is
View answer and explanation
Correct option: A
4400  A∘4400\;\mathop \mathrm{A}\limits^ \circ
Option A (4400  A∘4400\;\mathop \mathrm{A}\limits^ \circ) is correct because it is the wavelength calculated by equating the position of the first minimum of red light to the first maximum of the other wavelength.
180 Medium
180 of 418
Understanding +4 | -1
If n represents the order of a half period zone, the area of this zone is approximately proportional to nm{n^m} where m is equal to
View answer and explanation
Correct option: A
Zero
Option A (Zero) is correct because the area of a half-period zone is approximately constant and independent of the order of the zone.
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