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Wave Optics: JEE Main Previous Year Questions (2021-2026)

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JEE Main Previous Year Questions (2021-2026): 
Wave Optics  
 
(January 2026) 
 
Q1: Given below are two statements : 
Statement I: A plane wave after passing through prism remains as plane wave but 
passing through small pin hole may become spherical wave. 
Statement II: The curvature of a spherical wave emerging from a slit will increase for 
increasing slit width. 
In the light of the above statements, choose the correct answer from the options given 
below : 
(a) Both Statement I and Statement II are false 
(b) Both Statement I and Statement II are true 
(c) Statement I is true but Statement II is false 
(d) Statement I is false but Statement II is true 
Ans: (c) 
Sol:  
When a plane wave-front (parallel rays) enters a prism, all parts of the wave-front travel at the 
same speed through the medium. While the wave-front is tilted due to refraction, it remains flat 
and parallel, thus remaining a plane wave. 
 
According to Huygens' Principle, every point on a wave-front acts as a source of secondary 
spherical wavelets. When a plane wave encounters a very small opening (pinhole) comparable 
Page 2


JEE Main Previous Year Questions (2021-2026): 
Wave Optics  
 
(January 2026) 
 
Q1: Given below are two statements : 
Statement I: A plane wave after passing through prism remains as plane wave but 
passing through small pin hole may become spherical wave. 
Statement II: The curvature of a spherical wave emerging from a slit will increase for 
increasing slit width. 
In the light of the above statements, choose the correct answer from the options given 
below : 
(a) Both Statement I and Statement II are false 
(b) Both Statement I and Statement II are true 
(c) Statement I is true but Statement II is false 
(d) Statement I is false but Statement II is true 
Ans: (c) 
Sol:  
When a plane wave-front (parallel rays) enters a prism, all parts of the wave-front travel at the 
same speed through the medium. While the wave-front is tilted due to refraction, it remains flat 
and parallel, thus remaining a plane wave. 
 
According to Huygens' Principle, every point on a wave-front acts as a source of secondary 
spherical wavelets. When a plane wave encounters a very small opening (pinhole) comparable 
to the wavelength of light, it undergoes diffraction. The emerging light spreads out in all 
directions, forming a spherical wave-front. 
 
Hence, the statement I is true. When light passes through a slit, the degree of spreading 
(diffraction) depends on the slit width (d) relative to the wavelength (?). Smaller slit widths lead 
to greater diffraction and more pronounced curvature. As the slit width increases, the light 
behaves more like a straight beam, meaning the wave-front becomes flatter. 
 
Curvature is the reciprocal of the radius of the wave-front, i.e.,   So, a flatter wave 
has less curvature. Therefore, increasing the slit width decreases the curvature, it does not 
increase it.  
Page 3


JEE Main Previous Year Questions (2021-2026): 
Wave Optics  
 
(January 2026) 
 
Q1: Given below are two statements : 
Statement I: A plane wave after passing through prism remains as plane wave but 
passing through small pin hole may become spherical wave. 
Statement II: The curvature of a spherical wave emerging from a slit will increase for 
increasing slit width. 
In the light of the above statements, choose the correct answer from the options given 
below : 
(a) Both Statement I and Statement II are false 
(b) Both Statement I and Statement II are true 
(c) Statement I is true but Statement II is false 
(d) Statement I is false but Statement II is true 
Ans: (c) 
Sol:  
When a plane wave-front (parallel rays) enters a prism, all parts of the wave-front travel at the 
same speed through the medium. While the wave-front is tilted due to refraction, it remains flat 
and parallel, thus remaining a plane wave. 
 
According to Huygens' Principle, every point on a wave-front acts as a source of secondary 
spherical wavelets. When a plane wave encounters a very small opening (pinhole) comparable 
to the wavelength of light, it undergoes diffraction. The emerging light spreads out in all 
directions, forming a spherical wave-front. 
 
Hence, the statement I is true. When light passes through a slit, the degree of spreading 
(diffraction) depends on the slit width (d) relative to the wavelength (?). Smaller slit widths lead 
to greater diffraction and more pronounced curvature. As the slit width increases, the light 
behaves more like a straight beam, meaning the wave-front becomes flatter. 
 
Curvature is the reciprocal of the radius of the wave-front, i.e.,   So, a flatter wave 
has less curvature. Therefore, increasing the slit width decreases the curvature, it does not 
increase it.  
Hence, the statement II is false.  
Since Statement I is true and Statement II is false, so the correct option is (C). 
 
Q2: In the Young's double slit experiment the intensity produced by each one of the 
individual slits is I
o
. The distance between two slits is 2 mm. The distance of screen from 
slits is 10 m. The wavelength of light is 6000 A
°
. The intensity of light on the screen in 
front of one of the slits is _____ 
(a) I
o
/2 
(b) I
o
 
(c) 2I
o
 
(d) 4I
o
 
Ans: (b) 
Sol:  
The intensity from each slit is equal, I
1
 = I
2
 = I
0
  
The slit width is equal to distance between slits d = 2 mm = 2 × 10
-3
 m  
Distance of screen from slits D = 10 m  
Wavelength of light   
The distance from the central maximum to a point directly in front of one slit is : y = d/2  
 
The path difference for a point at height y on the screen is given by the formula 
Page 4


JEE Main Previous Year Questions (2021-2026): 
Wave Optics  
 
(January 2026) 
 
Q1: Given below are two statements : 
Statement I: A plane wave after passing through prism remains as plane wave but 
passing through small pin hole may become spherical wave. 
Statement II: The curvature of a spherical wave emerging from a slit will increase for 
increasing slit width. 
In the light of the above statements, choose the correct answer from the options given 
below : 
(a) Both Statement I and Statement II are false 
(b) Both Statement I and Statement II are true 
(c) Statement I is true but Statement II is false 
(d) Statement I is false but Statement II is true 
Ans: (c) 
Sol:  
When a plane wave-front (parallel rays) enters a prism, all parts of the wave-front travel at the 
same speed through the medium. While the wave-front is tilted due to refraction, it remains flat 
and parallel, thus remaining a plane wave. 
 
According to Huygens' Principle, every point on a wave-front acts as a source of secondary 
spherical wavelets. When a plane wave encounters a very small opening (pinhole) comparable 
to the wavelength of light, it undergoes diffraction. The emerging light spreads out in all 
directions, forming a spherical wave-front. 
 
Hence, the statement I is true. When light passes through a slit, the degree of spreading 
(diffraction) depends on the slit width (d) relative to the wavelength (?). Smaller slit widths lead 
to greater diffraction and more pronounced curvature. As the slit width increases, the light 
behaves more like a straight beam, meaning the wave-front becomes flatter. 
 
Curvature is the reciprocal of the radius of the wave-front, i.e.,   So, a flatter wave 
has less curvature. Therefore, increasing the slit width decreases the curvature, it does not 
increase it.  
Hence, the statement II is false.  
Since Statement I is true and Statement II is false, so the correct option is (C). 
 
Q2: In the Young's double slit experiment the intensity produced by each one of the 
individual slits is I
o
. The distance between two slits is 2 mm. The distance of screen from 
slits is 10 m. The wavelength of light is 6000 A
°
. The intensity of light on the screen in 
front of one of the slits is _____ 
(a) I
o
/2 
(b) I
o
 
(c) 2I
o
 
(d) 4I
o
 
Ans: (b) 
Sol:  
The intensity from each slit is equal, I
1
 = I
2
 = I
0
  
The slit width is equal to distance between slits d = 2 mm = 2 × 10
-3
 m  
Distance of screen from slits D = 10 m  
Wavelength of light   
The distance from the central maximum to a point directly in front of one slit is : y = d/2  
 
The path difference for a point at height y on the screen is given by the formula 
 
The relation between path difference and phase difference is : 
 
The formula for resultant intensity when two waves of equal intensity I
0
 interfere is: 
 
The intensity of light on the screen directly in front of one of the slits is equal to the intensity of a 
single slit, i.e., I
0
.  
Hence, the correct option is (B). 
 
Q3: When an unpolarized light falls at a particular angle on a glass plate (placed in air), it 
is observed that the reflected beam is linearly polarized. The angle of refracted beam with 
respect to the normal is _____.  
(tan
-1
   (1.52) = 57.7
°
, refractive indices of air and glass are 1.00 and 1.52, respectively)  
Page 5


JEE Main Previous Year Questions (2021-2026): 
Wave Optics  
 
(January 2026) 
 
Q1: Given below are two statements : 
Statement I: A plane wave after passing through prism remains as plane wave but 
passing through small pin hole may become spherical wave. 
Statement II: The curvature of a spherical wave emerging from a slit will increase for 
increasing slit width. 
In the light of the above statements, choose the correct answer from the options given 
below : 
(a) Both Statement I and Statement II are false 
(b) Both Statement I and Statement II are true 
(c) Statement I is true but Statement II is false 
(d) Statement I is false but Statement II is true 
Ans: (c) 
Sol:  
When a plane wave-front (parallel rays) enters a prism, all parts of the wave-front travel at the 
same speed through the medium. While the wave-front is tilted due to refraction, it remains flat 
and parallel, thus remaining a plane wave. 
 
According to Huygens' Principle, every point on a wave-front acts as a source of secondary 
spherical wavelets. When a plane wave encounters a very small opening (pinhole) comparable 
to the wavelength of light, it undergoes diffraction. The emerging light spreads out in all 
directions, forming a spherical wave-front. 
 
Hence, the statement I is true. When light passes through a slit, the degree of spreading 
(diffraction) depends on the slit width (d) relative to the wavelength (?). Smaller slit widths lead 
to greater diffraction and more pronounced curvature. As the slit width increases, the light 
behaves more like a straight beam, meaning the wave-front becomes flatter. 
 
Curvature is the reciprocal of the radius of the wave-front, i.e.,   So, a flatter wave 
has less curvature. Therefore, increasing the slit width decreases the curvature, it does not 
increase it.  
Hence, the statement II is false.  
Since Statement I is true and Statement II is false, so the correct option is (C). 
 
Q2: In the Young's double slit experiment the intensity produced by each one of the 
individual slits is I
o
. The distance between two slits is 2 mm. The distance of screen from 
slits is 10 m. The wavelength of light is 6000 A
°
. The intensity of light on the screen in 
front of one of the slits is _____ 
(a) I
o
/2 
(b) I
o
 
(c) 2I
o
 
(d) 4I
o
 
Ans: (b) 
Sol:  
The intensity from each slit is equal, I
1
 = I
2
 = I
0
  
The slit width is equal to distance between slits d = 2 mm = 2 × 10
-3
 m  
Distance of screen from slits D = 10 m  
Wavelength of light   
The distance from the central maximum to a point directly in front of one slit is : y = d/2  
 
The path difference for a point at height y on the screen is given by the formula 
 
The relation between path difference and phase difference is : 
 
The formula for resultant intensity when two waves of equal intensity I
0
 interfere is: 
 
The intensity of light on the screen directly in front of one of the slits is equal to the intensity of a 
single slit, i.e., I
0
.  
Hence, the correct option is (B). 
 
Q3: When an unpolarized light falls at a particular angle on a glass plate (placed in air), it 
is observed that the reflected beam is linearly polarized. The angle of refracted beam with 
respect to the normal is _____.  
(tan
-1
   (1.52) = 57.7
°
, refractive indices of air and glass are 1.00 and 1.52, respectively)  
(a) 36.3
°
  
(b) 39.6
°
  
(c) 42.6
° 
(d) 32.3
° 
Ans: (d) 
Sol:  
When unpolarized light strikes an interface at a specific angle called the Brewster angle (i
B
 or i
p
), 
the reflected light is completely linearly polarized. At this specific angle, the reflected ray and the 
refracted ray are perpendicular (90
°
) to each other. 
 
1. i
p
 = Angle of incidence (polarizing angle)  
2. ?
r
 = Angle of reflection  
3. ?
r
 ' = Angle of refraction  
Since the angle of reflection equals the angle of incidence:  
?
r
 = i
p
  
Since the angle between the reflected and refracted ray is 90
°
: 
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FAQs on Wave Optics: JEE Main Previous Year Questions (2021-2026)

1. What are the most important wave optics formulas for JEE Main exams?
Ans. Critical formulas include Young's double-slit interference equation (path difference = nλ), single-slit diffraction minima (a sin θ = nλ), and diffraction grating formula (d sin θ = nλ). Polarisation by reflection uses Brewster's angle (tan θ = n₂/n₁). Students should refer to mind maps and flashcards on EduRev to memorise these relationships with their physical interpretations and standard applications in previous year questions.
2. How do I identify interference vs diffraction patterns in JEE Main wave optics questions?
Ans. Interference occurs when two coherent light waves overlap, producing equally-spaced bright and dark fringes with constant intensity distribution. Diffraction happens when light bends around obstacles, creating unequally-spaced fringes with intensity variations. Interference requires two distinct sources; diffraction occurs from a single wavefront. Recognising which phenomenon applies based on experimental setup is crucial for solving JEE Main previous year problems correctly.
3. Why does path difference matter more than phase difference in solving wave optics problems?
Ans. Path difference directly translates to wavelength-dependent conditions: constructive interference occurs when path difference equals nλ, destructive when it equals (n + ½)λ. Phase difference requires converting this relationship mathematically. Since JEE Main questions involve real-world optical distances and wavelengths, path difference provides the most direct calculation route. Mastering this concept accelerates problem-solving in both main and advanced examinations significantly.
4. What's the trick to solving Young's double-slit experiment questions that appear in JEE Main?
Ans. The central maximum always lies at the geometric centre where path difference equals zero. Fringe width depends on wavelength, slit separation, and screen distance (β = λD/d). Common mistakes include confusing slit width with slit separation or forgetting wavelength changes in different media. Previous year JEE Main questions frequently test whether students account for these variables correctly when comparing bright and dark bands.
5. How should I approach single-slit diffraction problems for JEE Advanced preparation?
Ans. Single-slit diffraction minima follow a sin θ = nλ, where a is slit width. Unlike double-slit interference, diffraction produces a bright central maximum with progressively weaker secondary maxima. The intensity distribution is non-uniform-central fringe is three times wider than others. Understanding this asymmetry helps distinguish diffraction from interference patterns. JEE Advanced emphasises intensity calculations and mathematical rigour beyond recognising fringe positions alone.
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