JEE Main Previous Year Questions (2025): Wave Optics

 Page 1


JEE Main Previous Year Questions 
(2025): Wave Optics 
Q1: A double slit interference experiment performed with a light of wavelength 600 
nm forms an interference fringe pattern on a screen with 10 th bright fringe having its 
centre at a distance of 10 mm from the central maximum. Distance of the centre of 
the same ???? th bright fringe from the central maximum when the source of light is 
replaced by another source of wavelength 660 nm would be____ mm . 
JEE Main 2025 (Online) 28th January Morning Shift 
Ans: 11 
Solution: 
In Young's double slit experiment, the distance of ?? th 
 maxima from central maxima is given by, 
?? =
?????? ?? 
Here, ?? ,?? &?? are same for both the cases. 
So, ?? ? ?? 
?
?? 1
?? 2
=
?? 1
?? 2
?
10
?? 2
=
600
660
=
1
11
 
? ?? 2
= 11 mm . 
Q2: 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 
?????? ???? goes to a minimum every 12 seconds. Assuming that the film is flat on its two sides, 
the rate of evaporation is ____ ?? × ????
-????
 ?? ?? /?? . 
JEE Main 2025 (Online) 28th January Evening Shift 
Ans: 54 
Solution: 
Maxima condition 
2?? t = n?? ? t =
n?? 2?? ? t =
?? 2?? ,
2?? 2?? ,…… 
Minima condition 2?? t = (2n- 1)?? /2 
 ? t =
(2n- 1)?? 4?? ? t =
?? 4?? ,
3?? 4?? ,……
?t =
2?? 4?? 
Rate of evaporation =
A(?t)
 time 
= 54× 10
-13
 m
3
/s 
Page 2


JEE Main Previous Year Questions 
(2025): Wave Optics 
Q1: A double slit interference experiment performed with a light of wavelength 600 
nm forms an interference fringe pattern on a screen with 10 th bright fringe having its 
centre at a distance of 10 mm from the central maximum. Distance of the centre of 
the same ???? th bright fringe from the central maximum when the source of light is 
replaced by another source of wavelength 660 nm would be____ mm . 
JEE Main 2025 (Online) 28th January Morning Shift 
Ans: 11 
Solution: 
In Young's double slit experiment, the distance of ?? th 
 maxima from central maxima is given by, 
?? =
?????? ?? 
Here, ?? ,?? &?? are same for both the cases. 
So, ?? ? ?? 
?
?? 1
?? 2
=
?? 1
?? 2
?
10
?? 2
=
600
660
=
1
11
 
? ?? 2
= 11 mm . 
Q2: 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 
?????? ???? goes to a minimum every 12 seconds. Assuming that the film is flat on its two sides, 
the rate of evaporation is ____ ?? × ????
-????
 ?? ?? /?? . 
JEE Main 2025 (Online) 28th January Evening Shift 
Ans: 54 
Solution: 
Maxima condition 
2?? t = n?? ? t =
n?? 2?? ? t =
?? 2?? ,
2?? 2?? ,…… 
Minima condition 2?? t = (2n- 1)?? /2 
 ? t =
(2n- 1)?? 4?? ? t =
?? 4?? ,
3?? 4?? ,……
?t =
2?? 4?? 
Rate of evaporation =
A(?t)
 time 
= 54× 10
-13
 m
3
/s 
 
Q3: If the measured angular separation between the second minimum to the left of 
the central maximum and the third minimum to the right of the central maximum is 
????
°
 in a single slit diffraction pattern recorded using 628 nm light, then the width of 
the slit is ____ ?? m. 
JEE Main 2025 (Online) 2nd April Morning Shift 
Ans: 6 
Solution: 
 
?? 1
= sin
-1
 (
2?? ?? )
?? 2
= sin
-1
 (
3?? ?? )
 ? ?? 1
+ ?? 2
= 30
°
 ? sin
-1
 (
2?? a
)+ sin
-1
 (
3?? a
)=
?? 6
 ?
2?? a
v
1- (
3?? a
)
2
+
3?? a
v
1+ (
2?? a
)
2
= sin 
?? 6
 
Here ?? = 628 nm 
After solving 
A = 6.07?? m 
Approximate Method : 
?? = ?? 1
+ ?? 2
 ?
?? 6
=
2?? ?? +
3?? ?? ?
?? 6
=
5
?? (628 nm )
 ? ?? = 6?? m
 
Page 3


JEE Main Previous Year Questions 
(2025): Wave Optics 
Q1: A double slit interference experiment performed with a light of wavelength 600 
nm forms an interference fringe pattern on a screen with 10 th bright fringe having its 
centre at a distance of 10 mm from the central maximum. Distance of the centre of 
the same ???? th bright fringe from the central maximum when the source of light is 
replaced by another source of wavelength 660 nm would be____ mm . 
JEE Main 2025 (Online) 28th January Morning Shift 
Ans: 11 
Solution: 
In Young's double slit experiment, the distance of ?? th 
 maxima from central maxima is given by, 
?? =
?????? ?? 
Here, ?? ,?? &?? are same for both the cases. 
So, ?? ? ?? 
?
?? 1
?? 2
=
?? 1
?? 2
?
10
?? 2
=
600
660
=
1
11
 
? ?? 2
= 11 mm . 
Q2: 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 
?????? ???? goes to a minimum every 12 seconds. Assuming that the film is flat on its two sides, 
the rate of evaporation is ____ ?? × ????
-????
 ?? ?? /?? . 
JEE Main 2025 (Online) 28th January Evening Shift 
Ans: 54 
Solution: 
Maxima condition 
2?? t = n?? ? t =
n?? 2?? ? t =
?? 2?? ,
2?? 2?? ,…… 
Minima condition 2?? t = (2n- 1)?? /2 
 ? t =
(2n- 1)?? 4?? ? t =
?? 4?? ,
3?? 4?? ,……
?t =
2?? 4?? 
Rate of evaporation =
A(?t)
 time 
= 54× 10
-13
 m
3
/s 
 
Q3: If the measured angular separation between the second minimum to the left of 
the central maximum and the third minimum to the right of the central maximum is 
????
°
 in a single slit diffraction pattern recorded using 628 nm light, then the width of 
the slit is ____ ?? m. 
JEE Main 2025 (Online) 2nd April Morning Shift 
Ans: 6 
Solution: 
 
?? 1
= sin
-1
 (
2?? ?? )
?? 2
= sin
-1
 (
3?? ?? )
 ? ?? 1
+ ?? 2
= 30
°
 ? sin
-1
 (
2?? a
)+ sin
-1
 (
3?? a
)=
?? 6
 ?
2?? a
v
1- (
3?? a
)
2
+
3?? a
v
1+ (
2?? a
)
2
= sin 
?? 6
 
Here ?? = 628 nm 
After solving 
A = 6.07?? m 
Approximate Method : 
?? = ?? 1
+ ?? 2
 ?
?? 6
=
2?? ?? +
3?? ?? ?
?? 6
=
5
?? (628 nm )
 ? ?? = 6?? m
 
Q4: Two coherent monochromatic light beams of intensities 4I and 9I are 
superimposed. The difference between the maximum and minimum intensities in the 
resulting interference pattern is ?? ?? . The value of ?? is ____ . 
JEE Main 2025 (Online) 3rd April Morning Shift 
Ans: 24 
Solution: 
Calculate Maximum Intensity ( ?? max 
 ): 
?? max 
= (v?? 1
+ v?? 2
)
2
 
Substituting the values, we have: 
?? max
= (v4?? + v9?? )
2
 
?? max
= (2v?? + 3v?? )
2
= (5v?? )
2
= 25?? 
Calculate Minimum Intensity ( ?? min 
 ): 
?? min 
= (v?? 1
- v?? 2
)
2
 
Substituting the values, we have: 
?? min
= (v4?? - v9?? )
2
 
?? min 
= (2v?? - 3v?? )
2
= (-1v?? )
2
= ?? 
Calculate the Difference Between Maximum and Minimum Intensities: 
?? max
- ?? min
= 25?? - ?? = 24?? 
Thus, the value of ?? , which represents the difference between the maximum and minimum 
intensities, is 24 . 
Q5: In a Young's double slit experiment, two slits are located 1.5 mm apart. The 
distance of screen from slits is 2 m and the wavelength of the source is ?????? ???? . If the 
???? maxima of the double slit pattern are contained within the central maximum of 
the single slit diffraction pattern, then the width of each slit is ?? × ????
-?? ???? , where ?? -
value is ____ . 
JEE Main 2025 (Online) 4th April Evening Shift 
Ans: 15 
Solution: 
Width of 20 maxima of double slit = width of central maxima of single slit 
20????
?? =
2????
?? 10
?? =
1
?? ?? =
?? 10
=
1.5× 10
-1
10
 cm= 15× 10
-3
 cm
 
Value of ?? is 15 
Answer is 15 
Page 4


JEE Main Previous Year Questions 
(2025): Wave Optics 
Q1: A double slit interference experiment performed with a light of wavelength 600 
nm forms an interference fringe pattern on a screen with 10 th bright fringe having its 
centre at a distance of 10 mm from the central maximum. Distance of the centre of 
the same ???? th bright fringe from the central maximum when the source of light is 
replaced by another source of wavelength 660 nm would be____ mm . 
JEE Main 2025 (Online) 28th January Morning Shift 
Ans: 11 
Solution: 
In Young's double slit experiment, the distance of ?? th 
 maxima from central maxima is given by, 
?? =
?????? ?? 
Here, ?? ,?? &?? are same for both the cases. 
So, ?? ? ?? 
?
?? 1
?? 2
=
?? 1
?? 2
?
10
?? 2
=
600
660
=
1
11
 
? ?? 2
= 11 mm . 
Q2: 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 
?????? ???? goes to a minimum every 12 seconds. Assuming that the film is flat on its two sides, 
the rate of evaporation is ____ ?? × ????
-????
 ?? ?? /?? . 
JEE Main 2025 (Online) 28th January Evening Shift 
Ans: 54 
Solution: 
Maxima condition 
2?? t = n?? ? t =
n?? 2?? ? t =
?? 2?? ,
2?? 2?? ,…… 
Minima condition 2?? t = (2n- 1)?? /2 
 ? t =
(2n- 1)?? 4?? ? t =
?? 4?? ,
3?? 4?? ,……
?t =
2?? 4?? 
Rate of evaporation =
A(?t)
 time 
= 54× 10
-13
 m
3
/s 
 
Q3: If the measured angular separation between the second minimum to the left of 
the central maximum and the third minimum to the right of the central maximum is 
????
°
 in a single slit diffraction pattern recorded using 628 nm light, then the width of 
the slit is ____ ?? m. 
JEE Main 2025 (Online) 2nd April Morning Shift 
Ans: 6 
Solution: 
 
?? 1
= sin
-1
 (
2?? ?? )
?? 2
= sin
-1
 (
3?? ?? )
 ? ?? 1
+ ?? 2
= 30
°
 ? sin
-1
 (
2?? a
)+ sin
-1
 (
3?? a
)=
?? 6
 ?
2?? a
v
1- (
3?? a
)
2
+
3?? a
v
1+ (
2?? a
)
2
= sin 
?? 6
 
Here ?? = 628 nm 
After solving 
A = 6.07?? m 
Approximate Method : 
?? = ?? 1
+ ?? 2
 ?
?? 6
=
2?? ?? +
3?? ?? ?
?? 6
=
5
?? (628 nm )
 ? ?? = 6?? m
 
Q4: Two coherent monochromatic light beams of intensities 4I and 9I are 
superimposed. The difference between the maximum and minimum intensities in the 
resulting interference pattern is ?? ?? . The value of ?? is ____ . 
JEE Main 2025 (Online) 3rd April Morning Shift 
Ans: 24 
Solution: 
Calculate Maximum Intensity ( ?? max 
 ): 
?? max 
= (v?? 1
+ v?? 2
)
2
 
Substituting the values, we have: 
?? max
= (v4?? + v9?? )
2
 
?? max
= (2v?? + 3v?? )
2
= (5v?? )
2
= 25?? 
Calculate Minimum Intensity ( ?? min 
 ): 
?? min 
= (v?? 1
- v?? 2
)
2
 
Substituting the values, we have: 
?? min
= (v4?? - v9?? )
2
 
?? min 
= (2v?? - 3v?? )
2
= (-1v?? )
2
= ?? 
Calculate the Difference Between Maximum and Minimum Intensities: 
?? max
- ?? min
= 25?? - ?? = 24?? 
Thus, the value of ?? , which represents the difference between the maximum and minimum 
intensities, is 24 . 
Q5: In a Young's double slit experiment, two slits are located 1.5 mm apart. The 
distance of screen from slits is 2 m and the wavelength of the source is ?????? ???? . If the 
???? maxima of the double slit pattern are contained within the central maximum of 
the single slit diffraction pattern, then the width of each slit is ?? × ????
-?? ???? , where ?? -
value is ____ . 
JEE Main 2025 (Online) 4th April Evening Shift 
Ans: 15 
Solution: 
Width of 20 maxima of double slit = width of central maxima of single slit 
20????
?? =
2????
?? 10
?? =
1
?? ?? =
?? 10
=
1.5× 10
-1
10
 cm= 15× 10
-3
 cm
 
Value of ?? is 15 
Answer is 15 
Q6: Given below are two statements : one is labelled as Assertion (A) and the other is 
labelled as Reason (R). 
Assertion-(A) : If Young's double slit experiment is performed in an optically denser 
medium than air, then the consecutive fringes come closer. 
Reason-(R) : The speed of light reduces in an optically denser medium than air while 
its frequency does not change. 
In the light of the above statements, choose the most appropriate answer from the 
options given below : 
JEE Main 2025 (Online) 22nd January Morning Shift 
Options: 
A. (A) is false but (R) is true 
B. Both (A) and (R) are true but (R) is not the correct explanation of (A) 
C. Both (A) and (R) are true and (R) is the correct explanation of (A) 
D. (A) is true but (R) is false 
Ans: C 
Solution: 
We know, in YDSE, 
?? ( fringe width )=
????
?? 
In denser medium, ?? ?? ?? ? 
? fringes come closer. 
Also, ?? =
?? ?? ? ?? =
?? ?? 
? ?? ?? ?? ?? speed of light reduces. 
frequency remains same, 
? ?? =
?? vac 
.?? ?? med 
.?? ? ?? med 
=
?? vac 
.
?? ( as ?? = ???? ) 
Hence, option C is correct. 
Q7: Given below are two statements. One is labelled as Assertion (A) and the other is 
labelled as Reason (R). 
Assertion (A) : In Young's double slit experiment, the fringes produced by red light are 
closer as compared to those produced by blue light. 
Reason ( ?? ) : The fringe width is directly proportional to the wavelength of light. 
In the light of the above statements, choose the correct answer from the options 
given below : 
JEE Main 2025 (Online) 22nd January Evening Shift 
Options: 
A. (A) is true but (R) is false 
B. Both (A) and (R) are true and (R) is the correct explanation of (A) 
Page 5


JEE Main Previous Year Questions 
(2025): Wave Optics 
Q1: A double slit interference experiment performed with a light of wavelength 600 
nm forms an interference fringe pattern on a screen with 10 th bright fringe having its 
centre at a distance of 10 mm from the central maximum. Distance of the centre of 
the same ???? th bright fringe from the central maximum when the source of light is 
replaced by another source of wavelength 660 nm would be____ mm . 
JEE Main 2025 (Online) 28th January Morning Shift 
Ans: 11 
Solution: 
In Young's double slit experiment, the distance of ?? th 
 maxima from central maxima is given by, 
?? =
?????? ?? 
Here, ?? ,?? &?? are same for both the cases. 
So, ?? ? ?? 
?
?? 1
?? 2
=
?? 1
?? 2
?
10
?? 2
=
600
660
=
1
11
 
? ?? 2
= 11 mm . 
Q2: 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 
?????? ???? goes to a minimum every 12 seconds. Assuming that the film is flat on its two sides, 
the rate of evaporation is ____ ?? × ????
-????
 ?? ?? /?? . 
JEE Main 2025 (Online) 28th January Evening Shift 
Ans: 54 
Solution: 
Maxima condition 
2?? t = n?? ? t =
n?? 2?? ? t =
?? 2?? ,
2?? 2?? ,…… 
Minima condition 2?? t = (2n- 1)?? /2 
 ? t =
(2n- 1)?? 4?? ? t =
?? 4?? ,
3?? 4?? ,……
?t =
2?? 4?? 
Rate of evaporation =
A(?t)
 time 
= 54× 10
-13
 m
3
/s 
 
Q3: If the measured angular separation between the second minimum to the left of 
the central maximum and the third minimum to the right of the central maximum is 
????
°
 in a single slit diffraction pattern recorded using 628 nm light, then the width of 
the slit is ____ ?? m. 
JEE Main 2025 (Online) 2nd April Morning Shift 
Ans: 6 
Solution: 
 
?? 1
= sin
-1
 (
2?? ?? )
?? 2
= sin
-1
 (
3?? ?? )
 ? ?? 1
+ ?? 2
= 30
°
 ? sin
-1
 (
2?? a
)+ sin
-1
 (
3?? a
)=
?? 6
 ?
2?? a
v
1- (
3?? a
)
2
+
3?? a
v
1+ (
2?? a
)
2
= sin 
?? 6
 
Here ?? = 628 nm 
After solving 
A = 6.07?? m 
Approximate Method : 
?? = ?? 1
+ ?? 2
 ?
?? 6
=
2?? ?? +
3?? ?? ?
?? 6
=
5
?? (628 nm )
 ? ?? = 6?? m
 
Q4: Two coherent monochromatic light beams of intensities 4I and 9I are 
superimposed. The difference between the maximum and minimum intensities in the 
resulting interference pattern is ?? ?? . The value of ?? is ____ . 
JEE Main 2025 (Online) 3rd April Morning Shift 
Ans: 24 
Solution: 
Calculate Maximum Intensity ( ?? max 
 ): 
?? max 
= (v?? 1
+ v?? 2
)
2
 
Substituting the values, we have: 
?? max
= (v4?? + v9?? )
2
 
?? max
= (2v?? + 3v?? )
2
= (5v?? )
2
= 25?? 
Calculate Minimum Intensity ( ?? min 
 ): 
?? min 
= (v?? 1
- v?? 2
)
2
 
Substituting the values, we have: 
?? min
= (v4?? - v9?? )
2
 
?? min 
= (2v?? - 3v?? )
2
= (-1v?? )
2
= ?? 
Calculate the Difference Between Maximum and Minimum Intensities: 
?? max
- ?? min
= 25?? - ?? = 24?? 
Thus, the value of ?? , which represents the difference between the maximum and minimum 
intensities, is 24 . 
Q5: In a Young's double slit experiment, two slits are located 1.5 mm apart. The 
distance of screen from slits is 2 m and the wavelength of the source is ?????? ???? . If the 
???? maxima of the double slit pattern are contained within the central maximum of 
the single slit diffraction pattern, then the width of each slit is ?? × ????
-?? ???? , where ?? -
value is ____ . 
JEE Main 2025 (Online) 4th April Evening Shift 
Ans: 15 
Solution: 
Width of 20 maxima of double slit = width of central maxima of single slit 
20????
?? =
2????
?? 10
?? =
1
?? ?? =
?? 10
=
1.5× 10
-1
10
 cm= 15× 10
-3
 cm
 
Value of ?? is 15 
Answer is 15 
Q6: Given below are two statements : one is labelled as Assertion (A) and the other is 
labelled as Reason (R). 
Assertion-(A) : If Young's double slit experiment is performed in an optically denser 
medium than air, then the consecutive fringes come closer. 
Reason-(R) : The speed of light reduces in an optically denser medium than air while 
its frequency does not change. 
In the light of the above statements, choose the most appropriate answer from the 
options given below : 
JEE Main 2025 (Online) 22nd January Morning Shift 
Options: 
A. (A) is false but (R) is true 
B. Both (A) and (R) are true but (R) is not the correct explanation of (A) 
C. Both (A) and (R) are true and (R) is the correct explanation of (A) 
D. (A) is true but (R) is false 
Ans: C 
Solution: 
We know, in YDSE, 
?? ( fringe width )=
????
?? 
In denser medium, ?? ?? ?? ? 
? fringes come closer. 
Also, ?? =
?? ?? ? ?? =
?? ?? 
? ?? ?? ?? ?? speed of light reduces. 
frequency remains same, 
? ?? =
?? vac 
.?? ?? med 
.?? ? ?? med 
=
?? vac 
.
?? ( as ?? = ???? ) 
Hence, option C is correct. 
Q7: Given below are two statements. One is labelled as Assertion (A) and the other is 
labelled as Reason (R). 
Assertion (A) : In Young's double slit experiment, the fringes produced by red light are 
closer as compared to those produced by blue light. 
Reason ( ?? ) : The fringe width is directly proportional to the wavelength of light. 
In the light of the above statements, choose the correct answer from the options 
given below : 
JEE Main 2025 (Online) 22nd January Evening Shift 
Options: 
A. (A) is true but (R) is false 
B. Both (A) and (R) are true and (R) is the correct explanation of (A) 
C. (A) is false but (R) is true 
D. Both (A) and (R) are true but (R) is NOT the correct explanation of (A) 
Ans: C 
Solution: 
In Young's double-slit experiment, the width of the fringes, denoted as ?? , can be calculated 
using the formula: 
?? =
????
?? 
In this formula: 
?? represents the wavelength of the light used. 
?? is the distance from the slits to the screen. 
?? is the distance between the slits. 
According to this relationship, the fringe width ?? is directly proportional to the wavelength ?? . 
This means that as the wavelength increases, the fringe width also increases. 
When comparing red and blue light: 
Red light has a longer wavelength ( ?? ?? ) compared to blue light ( ?? ?? ). 
Therefore, the fringe width ?? ?? for red light is greater than the fringe width ?? ?? for blue light, as 
?? ?? > ?? ?? . 
This shows that with red light, the fringes are actually farther apart compared to the fringes 
produced by blue light. Thus, the assertion that red light fringes are closer than those produced 
by blue light is false, while the reason about fringe width being proportional to the wavelength 
is true. 
Q8: A transparent film of refractive index, 2.0 is coated on a glass slab of refractive 
index, 1.45 . What is the minimum thickness of transparent film to be coated for the 
maximum transmission of Green light of wavelength ?????? ???? . [Assume that the light 
is incident nearly perpendicular to the glass surface.] 
JEE Main 2025 (Online) 22nd January Evening Shift 
Options: 
A. 68.7 nm 
B. 137.5 nm 
C. 94.8 nm 
D. 275 nm 
Ans: B 
Solution: