NEET 2025: Wave Optics PYQs with Answers

Q 1 :
Two slits in Young’s double slit experiment are 1.5 mm apart and the screen is placed at a distance of 1 m from the slits. If the wavelength of light used is $600 \times 10^{- 9}$ m, then the fringe separation is [2024]

$4 \times 10^{- 5}$ m
$9 \times 10^{- 8}$ m
$4 \times 10^{- 7}$ m
$4 \times 10^{- 4}$ m

1

2

3

4

(4)

Slit separation distance, $d = 1.5 mm = 1.5 \times 10^{- 3} m$

Slit and screen separation distance, $D = 1$ $m$

Wavelength, $λ = 600 \times 10^{- 9}$ $m$

Fringe separation $= \frac{λ D}{d} = \frac{600 \times 10^{- 9} \times 1}{1.5 \times 10^{- 3}} = 4 \times 10^{- 4}$ $m$

Q 2 :
If the monochromatic source in Young’s double slit experiment is replaced by white light, then [2024]

interference pattern will disappear.
there will be a central dark fringe surrounded by a few coloured fringes.
there will be a central bright white fringe surrounded by a few coloured fringes.
all bright fringes will be of equal width.

1

2

3

4

(3)

If the monochromatic source in YDSE is replaced by white light, the central fringe will be a bright white fringe. This fringe will be surrounded by a few coloured fringes.

Q 3 :
For Young’s double slit experiment, two statements are given below:

Statement I: If screen moved away from the plane of slits, angular separation of the fringes remains constant.
Statement II: If the monochromatic source is replaced by another monochromatic source of larger wavelength, the angular separation of fringes decreases.

In the light of the above statements, choose the correct answer from the options given below: [2023]

Statement I is true but Statement II is false.
Statement I is false but Statement II is true.
Both Statement I and Statement II are true.
Both Statement I and Statement II are false.

1

2

3

4

(1)

Angular separation, $θ = \frac{λ}{d}$

So, it is independent of D which is the distance between screen and slit.

As $λ$ increases, $θ$ also increases.

Thus, statement I is true but statement II is false.

Q 4 :
In a Young’s double slit experiment, a student observes 8 fringes in a certain segment of screen when a monochromatic light of 600 nm wavelength is used. If the wavelength of light is changed to 400 nm, then the number of fringes he would observe in the same region of the screen is [2022]

6
8
9
12

1

2

3

4

(4)

As we know, wavelength of light is inversely proportional to number of fringes observed.

$∴$ $λ_{1} n_{1} = λ_{2} n_{2}$

$n_{2} = \frac{λ_{1} n_{1}}{λ_{2}} = \frac{600 \times 8}{400} = 12$

Q 5 :
In Young’s double slit experiment, if the separation between coherent sources is halved and the distance of the screen from the coherent sources is doubled, then the fringe width becomes [2020]

double
half
four times
one-fourth

1

2

3

4

(3)

Fringe width, $β = \frac{D λ}{d}$

$d becomes half \Rightarrow d' = \frac{d}{2}$

$D doubles, so \Rightarrow D' = 2 D$

New fringe width, $β' = \frac{2 D λ}{(\frac{d}{2})} = 4 β$

Q 6 :
In a double slit experiment, when light of wavelength 400 nm was used, the angular width of the first minima formed on a screen placed 1 m away, was found to be $0 . 2^{o}$ . What will be the angular width of the first minima, if the entire experimental apparatus is immersed in water? $(μ_{water} = \frac{4}{3})$ [2019]

$0 . 1^{o}$
$0 . 266^{o}$
$0 . 15^{o}$
$0 . 05^{o}$

1

2

3

4

(3)

Angular width for first minima in Young's double slit experiment, $θ = \frac{λ}{a}$

For given value of $a, θ \propto λ$

$\frac{θ}{θ_{w}} = \frac{λ}{λ_{w}} = \frac{λ}{\frac{λ}{μ}} = μ \Rightarrow θ_{w} = \frac{θ}{μ} = \frac{0.2 °}{\frac{4}{3}} = 0.15 °$

Q 7 :
In a Young's double slit experiment, if there is no initial phase difference between the light from the two slits, a point on the screen corresponding to the fifth minimum has path difference. [2019]

$5 \frac{λ}{2}$
$10 \frac{λ}{2}$
$9 \frac{λ}{2}$
$11 \frac{λ}{2}$

1

2

3

4

(3)

Given, there is no initial phase difference.

$∴$ Initial phase = $δ$ = 0

Again, Phase difference $= \frac{2 π}{λ} \times$ Path difference

$\Rightarrow δ' = \frac{2 π}{λ} \times Δ x \Rightarrow Δ x = \frac{λ}{2 π} \times δ'$

Now, for the fifth minima we will consider $n = 4$ as initial phase difference is zero.

$∴$ For fifth minimum, $δ' = (8 + 1) π = 9 π$

$∴$ Path difference, $Δ x = \frac{λ}{2 π} \times 9 π = \frac{9 λ}{2}$

Q 8 :
In Young's double slit experiment, the separation $d$ between the slits is 2 mm, the wavelength $λ$ of the light used is 5896 $Å$ and the distance D between the screen and the slits is 100 cm. It is found that the angular width of the fringes is $0 . 20^{o}$ . To increase the fringe angular width to $0 . 21^{o}$ (with same $λ$ and D), the separation between the slits needs to be changed to [2018]

1.8 mm
1.9 mm
2.1 mm
1.7 mm

1

2

3

4

(2)

Angular width $= \frac{λ}{d}$

$0.20 ° = \frac{λ}{2 mm} and 0.21 ° = \frac{λ}{d}$

On dividing we get, $\frac{0.20}{0.21} = \frac{d}{2 mm}$ $∴$ $d = 1.9$ $mm$

Q 9 :
Young’s double slit experiment is first performed in air and then in a medium other than air. It is found that $8^{th}$ bright fringe in the medium lies where $5^{th}$ dark fringe lies in air. The refractive index of the medium is nearly [2017]

1.59
1.69
1.78
1.25

1

2

3

4

(3)

Position of $8^{t h}$ bright fringe in medium,

$x = \frac{8 λ_{m} D}{d}$

Position of $5^{t h}$ dark fringe in air,

$x' = \frac{(5 - \frac{1}{2}) λ_{air} D}{d} = \frac{4.5 λ_{air} D}{d}$

Given $x = x' \Rightarrow \frac{8 λ_{m} D}{d} = \frac{4.5 λ_{air} D}{d}$

$μ_{m} = \frac{λ_{air}}{λ_{m}} = \frac{8}{4.5} = 1.78$

Q 10 :
The intensity at the maximum in a Young’s double slit experiment is $I_{0}$ . Distance between two slits is $d = 5 λ$ , where $λ$ is the wavelength of light used in the experiment. What will be the intensity in front of one of the slits on the screen placed at a distance $D = 10 d$ ? [2016]

$\frac{3}{4} I_{0}$
$\frac{I_{0}}{2}$
$I_{0}$
$\frac{I_{0}}{4}$

1

2

3

4

(2)

Here, $d = 5 λ, D = 10 d, y = \frac{d}{2}$

Resultant Intensity at $y = \frac{d}{2},$ $I_{y} = ?$

The path difference between two waves at $y = \frac{d}{2}$

$Δ x = d \tan θ = d \times \frac{y}{D} = \frac{d \times \frac{d}{2}}{10 d} = \frac{d}{20} = \frac{5 λ}{20} = \frac{λ}{4}$

Corresponding phase difference, $ϕ = \frac{2 π}{λ} Δ x = \frac{π}{2}$

Now, maximum intensity in Young’s double slit experiment,

$I_{max} = I_{1} + I_{2} + 2 I_{1} I_{2} or I_{0} = 4 I (∵ I_{1} = I_{2} = I)$

$∴$ $I = \frac{I_{0}}{4}$

Required intensity,

$I_{y} = I_{1} + I_{2} + 2 I_{1} I_{2} \cos \frac{π}{2} = 2 I = \frac{I_{0}}{2}$

Topic Question Set

Q 1 :
Two slits in Young’s double slit experiment are 1.5 mm apart and the screen is placed at a distance of 1 m from the slits. If the wavelength of light used is $600 \times 10^{- 9}$ m, then the fringe separation is [2024]

Q 2 :
If the monochromatic source in Young’s double slit experiment is replaced by white light, then [2024]

Q 4 :
In a Young’s double slit experiment, a student observes 8 fringes in a certain segment of screen when a monochromatic light of 600 nm wavelength is used. If the wavelength of light is changed to 400 nm, then the number of fringes he would observe in the same region of the screen is [2022]

Q 5 :
In Young’s double slit experiment, if the separation between coherent sources is halved and the distance of the screen from the coherent sources is doubled, then the fringe width becomes [2020]

Q 7 :
In a Young's double slit experiment, if there is no initial phase difference between the light from the two slits, a point on the screen corresponding to the fifth minimum has path difference. [2019]

Q 9 :
Young’s double slit experiment is first performed in air and then in a medium other than air. It is found that $8^{th}$ bright fringe in the medium lies where $5^{th}$ dark fringe lies in air. The refractive index of the medium is nearly [2017]

Want Us to Email you About Special Offers & Updates?

Topic Question Set

Q 1 : Two slits in Young’s double slit experiment are 1.5 mm apart and the screen is placed at a distance of 1 m from the slits. If the wavelength of light used is 600×10-9 m, then the fringe separation is [2024]

Q 2 : If the monochromatic source in Young’s double slit experiment is replaced by white light, then [2024]

Q 4 : In a Young’s double slit experiment, a student observes 8 fringes in a certain segment of screen when a monochromatic light of 600 nm wavelength is used. If the wavelength of light is changed to 400 nm, then the number of fringes he would observe in the same region of the screen is [2022]

Q 5 : In Young’s double slit experiment, if the separation between coherent sources is halved and the distance of the screen from the coherent sources is doubled, then the fringe width becomes [2020]

Q 7 : In a Young's double slit experiment, if there is no initial phase difference between the light from the two slits, a point on the screen corresponding to the fifth minimum has path difference. [2019]

Q 9 : Young’s double slit experiment is first performed in air and then in a medium other than air. It is found that 8th bright fringe in the medium lies where 5th dark fringe lies in air. The refractive index of the medium is nearly [2017]

Q 10 : The intensity at the maximum in a Young’s double slit experiment is I0. Distance between two slits is d=5λ, where λ is the wavelength of light used in the experiment. What will be the intensity in front of one of the slits on the screen placed at a distance D=10d ? [2016]

Want Us to Email you About Special Offers & Updates?

Q 1 :
Two slits in Young’s double slit experiment are 1.5 mm apart and the screen is placed at a distance of 1 m from the slits. If the wavelength of light used is $600 \times 10^{- 9}$ m, then the fringe separation is [2024]

Q 2 :
If the monochromatic source in Young’s double slit experiment is replaced by white light, then [2024]

Q 4 :
In a Young’s double slit experiment, a student observes 8 fringes in a certain segment of screen when a monochromatic light of 600 nm wavelength is used. If the wavelength of light is changed to 400 nm, then the number of fringes he would observe in the same region of the screen is [2022]

Q 5 :
In Young’s double slit experiment, if the separation between coherent sources is halved and the distance of the screen from the coherent sources is doubled, then the fringe width becomes [2020]

Q 7 :
In a Young's double slit experiment, if there is no initial phase difference between the light from the two slits, a point on the screen corresponding to the fifth minimum has path difference. [2019]

Q 9 :
Young’s double slit experiment is first performed in air and then in a medium other than air. It is found that $8^{th}$ bright fringe in the medium lies where $5^{th}$ dark fringe lies in air. The refractive index of the medium is nearly [2017]