18 NEET previous-year questions on Wave Optics, each with the correct answer and a step-by-step solution. Sourced directly from official NEET papers across every booklet code.
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1.25 × 10 –6 N
2.50 × 10 –6 N
1.20 × 10 –6 N
3.0 × 10 –6 N
Solution
The average force exerted on a perfectly reflecting surface is given by , where is the energy flux, is the area, and is the speed of light. Substituting the values, , so option (b) is correct.
1.2 cm
1.2 mm
2.4 cm
2.4 mm
Solution
The distance between the first dark fringes on either side of the central bright fringe is given by . Substituting the values, , so option (d) is correct.
K
K/4
K/2
Zero
Solution
The intensity in Young's double-slit experiment is given by , where . For , and . For , and . Thus, option (c) is correct.
Eptatretus
Myxine
Neomyxine
Petromyzon
Solution
Petromyzon (Lamprey) is a migratory marine water jawless fish which shows anadromous migration. It spawns in fresh water, stops feeding and dies. Its larva (Ammocoetes) after metamorphosis will return to ocean.
π 4 radian
π 2 radian
π radian
π 8 radian
Solution
For first minima AP−BP=λ AP−MP=λ 2 So phase difference = 2π λ × λ 2=π 131. On a frictionless surfaces, a block of mass M moving at speed v collides elastically with another block of same mass M which is initially at rest. After collision the first block moves at an angle θ to its initial direction and has a speed v 3. The second block’s speed after the collision is: (1) 2√2 3 v (2) 3 4v (3) 3 √2v (4) √3 2 v Solution: (1) www.vedantu.com 43 Pi⃗⃗ =Pf⃗⃗⃗ ⇒|Pi|=|Pf|⇒√(mV 3) 2 +(mV2)2 V2=2√2 3 V
sin−1( 1 4)
sin−1( 2 3 )
sin−1( 1 2 )
sin−1( 3 4 )
Solution
asin30 = λ asinθ =3λ 2 sinθ sin30 = 3 2 sinθ =3 2 × 1 2 sin𝜃 = 3 4 θ = sin(3 4)
I0
I0 4
3 4I0
I0 2
Solution
In YDSE Imax = I0 Path difference at a point in front of one of shifts is www.vedantu.com 13 ∆x = d(y D) = d( d 2 D) = d2 2D ∆x = d2 2(10d)= d 20 = 5λ 20 = λ 4 Path difference is ϕ =2π λ = (∆x)= 2π λ (λ 4) ϕ =π 2 So intensity at that pt is I = Imaxcos2 (θ 2) I = I0 cos2 (π 4) =I0 2
1.25
1.59
1.69
1.78
Solution
X 1 = X5th dark = (2 × 5 – 1) 2 D d λ X2 = X8th bright = 8 D d λ μ X1 = X2 9 82 DD dd λλ = μ 16 1.789μ= =
0 2 I
0 4 I
0 8 I
0 16 I
Solution
P1 P3 P2 I3 I2 I1I0 90° 45° 20 2 cos 452 II =° 0 1 22 I=× www.vedantu.com 10 0 4 I= 20 3 cos 454 II =° 0 3 8 II =
2·1 mm
1·9 mm
1·8 mm
1·7 mm
Solution
The angular width of the fringes is given by . Initially, . To increase to , we use , so option (b) is correct.
Electrons are the majority carries and pentavalent atoms are the dopants
Electrons are the majority carries and trivalent atoms are the dopants
Holes are the majority carri es and trivalent atoms are the dopants
Holes are the majority carries and pentavalent atoms are the dopants
Solution
Conceptual.
𝑉2> 𝑉1 and 𝑖1 > 𝑖2
𝑉2> 𝑉1 and 𝑖1=𝑖2
𝑉1= 𝑉2 and 𝑖1>𝑖2
𝑉1= 𝑉2 and 𝑖1=𝑖2
Solution
Both voltmeters are ideal. Therefore total current in circuit (2) will flow from upper 10Ω resistance.
double
half
four times
one-fourth
Solution
The fringe width is given by , where is the wavelength, is the distance to the screen, and is the separation between the slits. If is halved and is doubled, becomes . Therefore, the fringe width becomes four times, so option (c) is correct.
10 18
10 17
10 16
10 15
Solution
The energy of one photon is given by . Substituting the values, . The number of photons per second is , so option (c) is correct.
6
8
9
12
Solution
λβ= D d Let length of segment of screen = l ⇒ 1 1 88 λ= β= Dl d …(1) and 2 2 λ=β= nDln d …(2) from (1) and (2) 8λ1 = nλ2 8(600 nm) = n(400 nm) n = 12
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.
Statement I is true but Statement II is false.
Solution
Statement I is true because the angular separation is independent of the distance to the screen. Statement II is false because increasing the wavelength increases the angular separation . Therefore, option (d) is correct.
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
Solution
At central point on screen, path difference is zero for all wavelength. So, central bright fringe is white and other fringes depend on wavelength as λβ= D d . Therefore, other fringes will be coloured. - 12 - NEET (UG)-2024 (Code-Q1)
Solution
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