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JEE • PHYSICS • WAVE OPTICS

Wave optics practice · Medium

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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 121-150 of 396 questions

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121 Medium
121 of 396
Understanding +4 | -1
Which of the following phenomenon exhibits particle’s nature of light?
View answer and explanation
Correct option: D
Photoelectric effect
Option D (Photoelectric effect) is correct because the photoelectric effect demonstrates the particle nature of light, where photons transfer energy to electrons.
122 Medium
122 of 396
Application +4 | -1
Red light of wavelength 625  nm625\;nm is incident normally on an optical diffraction grating with 2×105  lines/m2 \times {10^5}\;lines/m. Including central principal maxima, how many maxima may be observed on a screen which is for from the grating?
View answer and explanation
Correct option: B
17
Option B (17) is correct because the number of maxima is determined by the grating equation and the condition that sin⁡θ\sin \theta must be less than or equal to 1.
123 Medium
123 of 396
Analyzing +4 | -1
Assertion: In Young's double slit experiment, the fringe width for dark fringes is same as that for white fringes.

Reason: In Young's double slit experiment, when the fringes are performed with a source of white light, then only dark and bright fringes are observed.

In the light of the 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 the fringe width is the same for dark and bright fringes, but white light produces colored fringes, not just dark and bright.
124 Medium
124 of 396
Application +4 | -1
A rocket is going towards moon with a speed vv. The astronaut in the rocket sends signals of frequency vv towards the moon and receives them back on reflection from the moon. What will be the frequency of the signal received by the astronaut (Take v≪cv \ll c )
View answer and explanation
Correct option: B
cc−2vv\frac{c}{{c - 2v}}v
Option B, cc−2vv\frac{c}{{c - 2v}}v, is correct because it accounts for the Doppler effect twice: once as the signal travels to the moon and again as it returns to the astronaut.
125 Medium
125 of 396
Application +4 | -1
Unpolarised light is incident on a dielectric of refractive index 3\sqrt{3}. What is the angle of incidence if the reflected beam is completely polarised?
View answer and explanation
Correct option: C
60∘60^{\circ}
Option C (60∘60^{\circ}) is correct because the angle of incidence for complete polarization is the Brewster's angle, where the tangent equals the refractive index.
126 Medium
126 of 396
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 absorption lines observed in the solar spectrum, caused by elements in the Sun's atmosphere absorbing specific wavelengths of light.
127 Medium
127 of 396
Understanding +4 | -1
Diffraction pattern of a single slit consists of a central bright band which is
View answer and explanation
Correct option: A
Wide, and is flanked by alternate dark and bright bands of decreasing intensity
Option A (Wide, and is flanked by alternate dark and bright bands of decreasing intensity) is correct because the central bright fringe in single-slit diffraction is wider and more intense than the subsequent fringes, which decrease in intensity as you move away from the center.
128 Medium
128 of 396
Application +4 | -1
S1  and  S2{S_1}\;and\;{S_2} are two coherent sources. The intensity of both sources are same. If the intensity at the point of maxima is 4  Wm−2\mathrm{4}\;\mathrm{W}{\mathrm{m}^{ - 2}}, the intensity of each source is
View answer and explanation
Correct option: A
1  Wm−2\mathrm{1}\;\mathrm{W}{\mathrm{m}^{ - 2}}
Option A (1  Wm−2\mathrm{1}\;\mathrm{W}{\mathrm{m}^{ - 2}}) is correct because the maximum intensity is four times the intensity of each individual source when they are coherent and have the same intensity.
129 Medium
129 of 396
Application +4 | -1
Critical angle for certain medium is sin⁡−1(0.6).{\sin ^{ - 1}}\left( {0.6} \right). The polarizing angle of that medium is
View answer and explanation
Correct option: C
tan⁡−1[1.6667]{\tan ^{ - 1}}\left[ {1.6667} \right]
Option C (tan⁡−1[1.6667]{\tan ^{ - 1}}\left[ {1.6667} \right]) is correct because it's derived from calculating the refractive index using the critical angle and then applying Brewster's law.
130 Medium
130 of 396
Understanding +4 | -1
A single slit is used to observe diffraction pattern with red light. On replacing the red light with violet light the diffraction pattern would
View answer and explanation
Correct option: B
Become narrower
Option B is correct because the fringe width in a diffraction pattern is directly proportional to the wavelength of light, and violet light has a shorter wavelength than red light.
131 Medium
131 of 396
Application +4 | -1
The central fringe of the interference pattern produced by light of wavelength 6000Ao6000 \mathop {\text{A}}\limits^{\text{o}} is found to shift to the position of 4th bright fringe after a glass plate of refractive index 1.5 is introduced in front of one of slits in Young's experiment. The thickness of the glass plate will be
View answer and explanation
Correct option: A
4.8μm4.8 \mu \mathrm{m}
Option A (4.8μm4.8 \mu \mathrm{m}) is correct because the shift in the central fringe is related to the thickness and refractive index of the glass plate.
132 Medium
132 of 396
Application +4 | -1
Light of wavelength 6000 A∘6000 ~\mathop {\text{A}}\limits^ \circ is incident normally on a single slit of width 24×10−5 cm24 \times 10^{-5} \mathrm{~cm}. Find out the angular position of second minimum from central maxima?
View answer and explanation
Correct option: B
30∘30^{\circ}
Option B (30∘30^{\circ}) is correct because it is calculated using the formula for the angular position of the second minimum in a single-slit diffraction pattern.
133 Medium
133 of 396
Understanding +4 | -1
Statement I : In Young's double slit experiment, the number of fringes observed in the field of view is small with longer wavelength of light and is large with shorter wavelength of light.

Statement II : In the double slit experiment the fringe width depends directly on the wavelength of light.
View answer and explanation
Correct option: A
Statement I is true, Statement II is true; Statement II is the correct explanation for Statement 4
Option A (Statement I is true, Statement II is true; Statement II is the correct explanation for Statement 4) is correct because fringe width is directly proportional to wavelength, affecting the number of fringes observed.
134 Medium
134 of 396
Application +4 | -1
At two points P  and  QP\;and\;Q on screen in Young's double slit experiment. Waves from slits
S1  and  S2{S_1}\;and\;{S_2} have a path difference of 0  and  λ4.0\;\mathrm{and}\;\frac{\lambda }{4}\mathrm{.} respectively. The ratio of intensities at
P  and  QP\;and\;Q will be
View answer and explanation
Correct option: B
2:12:1
Option B (2:1) is correct because the ratio of intensities is calculated using the path difference and the formula for intensity in interference.
135 Medium
135 of 396
Understanding +4 | -1
To increase the resolving power of a telescope
View answer and explanation
Correct option: C
Both (1) & (2)
Option C (Both (1) & (2)) is correct because the resolving power of a telescope can be increased by decreasing the wavelength of light or increasing the aperture of the objective lens.
136 Medium
136 of 396
Understanding +4 | -1
The main difference between the phenomena of interference and diffraction is that
View answer and explanation
Correct option: C
Diffraction is due to interaction of light from wavefront, whereas the interference is the interaction of two waves derived from the same source
Option C (Diffraction is due to interaction of light from wavefront, whereas the interference is the interaction of two waves derived from the same source) is correct because diffraction involves the interaction of waves from the same wavefront, while interference involves the interaction of waves from two different sources.
137 Medium
137 of 396
Application +4 | -1
A two-slit Young's interference experiment is done with monochromatic light of wavelength 6000   A∘6000{\mkern 1mu} \;\mathop {\text{A}}\limits^ \circ. Those slits are 2  mm2 \,{~mm} apart. The fingers are observed on a screen placed 10  cm10\, {~cm} away from the slits. A transparent plate of thickness of 0.5  mm0.5 \,{~mm} is placed in front of one of the slits. It is found that the interference pattern shifts by 5  mm5 \,{~mm}. What is the refractive index of the transparent plate?
View answer and explanation
Correct option: A
1.2
Option A (1.2) is correct because it is the refractive index of the transparent plate calculated using the shift in the interference pattern formula.
138 Medium
138 of 396
Analyzing +4 | -1
Statement-1: Two thirds of possible values for phase difference result in constructive interference and one third results in destructive interference.
Statement-2: Interference is called constructive if the amplitude of the superposed waves exceeds that of one individual wave.
View answer and explanation
Correct option: A
Statement-1 is true, Statement-2 is true, Statement-2 is the correct explanation for Statement-1
Option A (Statement-1 is true, Statement-2 is true, Statement-2 is the correct explanation for Statement-1) is correct. Constructive interference occurs for two-thirds of the possible phase differences, and this is because the resulting amplitude exceeds that of the individual waves.
139 Medium
139 of 396
Application +4 | -1
Plane wave front is incident on a single slit of width 0.2 mm0.2 \mathrm{~mm} by which Fraunhoffer diffraction pattern is formed by lens of focal length 1 m1 \mathrm{~m}. If the second minima is formed at a distance of 3 mm3 \mathrm{~mm} from the centre then find the wavelength of light.
View answer and explanation
Correct option: B
3000 A∘3000 ~\mathop {\text{A}}\limits^ \circ
Option B (3000 A∘3000 ~\mathop {\text{A}}\limits^ \circ) is correct because it is the wavelength calculated using the formula for the position of minima in Fraunhofer diffraction.
140 Medium
140 of 396
Application +4 | -1
In YDSE setup, light of wavelength 640 nm640 \mathrm{~nm} is used with d=0.8 mmd=0.8 \mathrm{~mm} and D=1 mD=1 \mathrm{~m}. If intensity at central maxima is I0I_0 and it's position is y=0y=0
View answer and explanation
Correct option: A
Intensity at y=0.4 mmy=0.4 \mathrm{~mm} is zero
Intensity at y=0.4 mmy=0.4 \mathrm{~mm} is zero because at this point, the phase difference corresponds to destructive interference.
141 Medium
141 of 396
Analyzing +4 | -1
Assertion : Light ray passing through water droplet is partially polarized.
Reason : Water is a polar molecule.
View answer and explanation
Correct option: A
If both Assertion and Reason are true and the Reason is a correct explanation of the Assertion
Option A is correct because both Assertion and Reason are true and the Reason is a correct explanation of the Assertion. Light ray passing through water droplet is partially polarized due to water being a polar molecule.
142 Medium
142 of 396
Understanding +4 | -1
Which of the following statements is true, when spherical waves fall on a plane refracting surface, separating two media
View answer and explanation
Correct option: A
The reflected waves form spherical wave fronts
Option A (The reflected waves form spherical wave fronts) is correct because reflection from a plane surface preserves the spherical nature of the wavefront.
143 Medium
143 of 396
Understanding +4 | -1
Which of the following can be used to control the intensity, in sunglasses, window-panes, etc.?
View answer and explanation
Correct option: B
Polaroids
Option B (Polaroids) is correct because polaroids selectively absorb light waves based on their polarization, thus controlling intensity.
144 Medium
144 of 396
Application +4 | -1
The equation of two light waves are

y1=A1sin⁡2ωty2=A2cos⁡(2ωt+π6)\begin{aligned} & y_1=A_1 \sin 2 \omega t \\ & y_2=A_2 \cos \left(2 \omega t+\frac{\pi}{6}\right) \end{aligned}

What will be the value of phase difference at time tt
View answer and explanation
Correct option: B
2π3\frac{2 \pi}{3}
Option B (2π3\frac{{2 \pi}}{{3}}) is correct because the phase difference is calculated by converting cosine to sine and then finding the difference in phase angles.
145 Medium
145 of 396
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.
146 Medium
146 of 396
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.
147 Medium
147 of 396
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.
148 Medium
148 of 396
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.
149 Medium
149 of 396
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.
150 Medium
150 of 396
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.
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