If the measured angular separation between the second minimum to the left of the central maximum and the third minimum…

Year	2026	2025	2024	Total
Questions	11	15	8	34

Year

2026

2025

2024

Total

Questions

Solution & Explanation

Related Formula

a θₙ = n λ

θₐₚₚᵣₒₓ = n (λ)/(a) (for small angles)

Core Logic

Let the width of the slit be $a$ $a$, and the light wavelength be $λ = 628~nm = 628 \times 10⁻⁹~m$ .

The second minimum ( $n = 2$ $n = 2$) is located at angular position:

θ₁ = (2λ)/(a)

The third minimum ( $n = 3$ $n = 3$) is located at angular position:

θ₂ = (3λ)/(a)

The total angular separation is $30°$ $30^{\circ}$:

θ₁ + θ₂ = 30° = (π)/(6)~rad

Using the small-angle approximation (where $θ \approx θ$ $\theta \approx \sin\theta$):

θ₁ + θ₂ \approx (2λ)/(a) + (3λ)/(a) = (5λ)/(a)

Equating to the given separation:

(5λ)/(a) = (π)/(6) a = (30λ)/(π)

Substitute the given values (using $π \approx 3.14$ $\pi \approx 3.14$):

a = 30 \times 628 \times 10⁻⁹~m3.14 = 30 \times 200 \times 10⁻⁹~m = 6 \times 10⁻⁶~m = 6~μ m

Step 1: Final Conclusion

The width of the slit is $6~μ m$ $6\mathrm{~\mu m}$.

Pattern Recognition

In single slit diffraction, the position of minima is $a θ = n λ$ $a \sin\theta = n \lambda$. The angular spread from the $n₁$ $n_1$-th minimum on one side to the $n₂$ $n_2$-th minimum on the other is $(n₁ + n₂) (λ)/(a)$ $(n_1 + n_2) \frac{\lambda}{a}$. Since $π \approx 3.14$ $\pi \approx 3.14$, note how $628 / 3.14 = 200$ $628 / 3.14 = 200$, resolving to a neat integer.

Chapter Mix

Class 12 Physics: Wave Optics

More Wave Optics Previous-Year Questions — Page 3

Q49 jee_main_2026_28_january_evening YDSE Missing Wavelength

A beam of light consisting of wavelengths

650 nm

and

550 nm

illuminates the Young's double slits with separation of

2 mm

such that the interference fringes are formed on a screen, placed at a distance of

1.2 m

from the slits. The least distance of a point from the central maximum, where the bright fringes due to both the wavelengths coincide, is ____

\times 10⁻⁵

Numerical Answer. Answer: 429 to 429

Solution

Related Formula

y = n (λ D)/(d)

Core Logic

For bright fringes to coincide, the position $y$ $y$ of the $n₁$ $n_1$-th maxima of $λ₁$ $\lambda_1$ must equal the position of the $n₂$ $n_2$-th maxima of $λ₂$ $\lambda_2$: $y₁ = y₂$ $y_1 = y_2$

n₁ λ₁ (D)/(d) = n₂ λ₂ (D)/(d)

Step 1: Find Ratio of Orders

Given $λ₁ = 650 nm$ $\lambda_1 = 650 \text{ nm}$ and $λ₂ = 550 nm$ $\lambda_2 = 550 \text{ nm}$.

(n₁)/(n₂) = (λ₂)/(λ₁) = (550)/(650) = (11)/(13)

For the least distance, we take the smallest integer values for $n₁$ $n_1$ and $n₂$ $n_2$. Thus, $n₁ = 11$ $n_1 = 11$ and $n₂ = 13$ $n_2 = 13$.

Step 2: Calculate Distance

Using $n₁ = 11$ $n_1 = 11$ for $λ₁ = 650 nm$ $\lambda_1 = 650 \text{ nm}$:

y = 11 \times (λ₁ D)/(d)

y = 11 \times 650 \times 10⁻⁹ \times 1.22 \times 10⁻³

y = 11 \times 650 \times 1.2 \times 10⁻⁶2

y = 11 \times 650 \times 0.6 \times 10⁻⁶

y = 11 \times 390 \times 10⁻⁶

y = 4290 \times 10⁻⁶ m

y = 429 \times 10⁻⁵ m

Pattern Recognition

Whenever two wavelengths are sent through YDSE, the condition for overlapping maxima is simply $n₁ λ₁ = n₂ λ₂$ $n_1 \lambda_1 = n_2 \lambda_2$. Pick the smallest integers for $n₁, n₂$ $n_1, n_2$ to find the first coincidental fringe distance.

Chapter Mix

Class 12 Physics: Wave Optics

Q1 jee_main_2025_02_april_morning Wavefronts and Huygen's Principle

A light wave is propagating with plane wave fronts of the type

x+y+z = constant

. The angle made by the direction of wave propagation with the

x

-axis is:

A. $⁻¹( 1\sqrt(3))$
B. $⁻¹((2)/(3))$ $\cos^{-1}\left(\frac{2}{3}\right)$
C. $⁻¹((1)/(3))$ $\cos^{-1}\left(\frac{1}{3}\right)$
D. $⁻¹(\sqrt((2)/(3)))$

Solution

Related Formula

²α + ²β + ²γ = 1

Core Logic

The direction of propagation of a light wave is always perpendicular to its plane wavefronts.

For wavefronts of the form $x + y + z = constant$ $x + y + z = \text{constant}$, the normal vector representing the direction of wave propagation is given by:

n = i + j + k

Since the coefficients of $x$ $x$, $y$ $y$, and $z$ $z$ are all equal, the propagation direction is symmetric with respect to all three axes:

α = β = γ

Substituting this into the direction cosine identity:

²α + ²α + ²α = 1 3 ²α = 1

α = 1\sqrt(3)

α = ⁻¹( 1\sqrt(3))

Step 1: Final Conclusion

The angle made by the direction of wave propagation with the $x$ $x$-axis is $⁻¹( 1\sqrt(3))$ .

Pattern Recognition

Sees: plane wavefront $Ax + By + Cz = d$ $Ax + By + Cz = \text{d}$ → propagation vector is parallel to the normal $k = A i + B j + C k$ . Use standard direction cosines to find the angle with any axis.

Chapter Mix

Class 12 Physics: Wave Optics Class 12 Mathematics: Three Dimensional Geometry

Q4 jee_main_2025_03_april_evening Young's Double Slit Experiment

Width of one of the two slits in a Young's double slit interference experiment is half of the other slit. The ratio of the maximum to the minimum intensity in the interference pattern is :

A. $(2\sqrt(2)+1):(2\sqrt(2)-1)$ $(2\sqrt{2}+1):(2\sqrt{2}-1)$
B. $(3+2\sqrt(2)):(3-2\sqrt(2))$ $(3+2\sqrt{2}):(3-2\sqrt{2})$
C. 9:1
D. 3:1

Solution

Related Formula

The intensity of a slit is proportional to its width:

$I \propto w$ $I \propto w$

The ratio of maximum to minimum intensity in an interference pattern is given by:

II = (\sqrt(I₁) + \sqrt(I₂))²(\sqrt(I₁) - \sqrt(I₂))²

Core Logic

Let the width of the larger slit be $w₂ = w$ $w_2 = w$ and the smaller slit be $w₁ = w/2$ $w_1 = w/2$. Thus:

I₂ = I₀, I₁ = (I₀)/(2)

\sqrt(I₂) = \sqrt(I₀), \sqrt(I₁) = \sqrt(I₀)\sqrt(2)

Step 1: Substitute values into intensity ratio

II = (\sqrt(I₀) + \sqrt(I₀)\sqrt(2))²(\sqrt(I₀) - \sqrt(I₀)\sqrt(2))²

II = (1 + 1\sqrt(2))²(1 - 1\sqrt(2))² = ( \sqrt(2) + 1\sqrt(2))²( \sqrt(2) - 1\sqrt(2))² = (\sqrt(2) + 1)²(\sqrt(2) - 1)²

Step 2: Expand the terms

II = 2 + 1 + 2\sqrt(2)2 + 1 - 2\sqrt(2) = 3 + 2\sqrt(2)3 - 2\sqrt(2)

Pattern Recognition

When dealing with fractional width ratio $β = w₂/w₁$ $\beta = w_2/w_1$, the ratio $\sqrt(I₂/I₁) = \sqrt(β)$ $\sqrt{I_2/I_1} = \sqrt{\beta}$. The intensity ratio formula can be rewritten as $(\sqrt(β)+1)² / (\sqrt(β)-1)²$ . For $β = 2$ $\beta = 2$, this directly resolves to $(\sqrt(2)+1)²/(\sqrt(2)-1)² = (3+2\sqrt(2))/(3-2\sqrt(2))$ .

Chapter Mix

Class 12 Physics: Wave Optics

Q6 jee_main_2025_03_april_evening Interference of Light and Intensity Ratio

Two monochromatic light beams have intensities in the ratio

1:9

. An interference pattern is obtained by these beams. The ratio of the intensities of maximum to minimum is :

A. 8:1
B. 9:1
C. 3:1
D. 4:1

Solution

Related Formula

The intensity limits for two overlapping coherent light sources of intensities $I₁$ $I_1$ and $I₂$ $I_2$ are:

II = ( \sqrt(I₂) + \sqrt(I₁)\sqrt(I₂) - \sqrt(I₁) )²

Core Logic

Given ratio:

(I₁)/(I₂) = (1)/(9)

If we let $I₁ = I₀$ $I_1 = I_0$, then $I₂ = 9I₀$ $I_2 = 9I_0$. Taking square roots:

\sqrt(I₁) = \sqrt(I₀), \sqrt(I₂) = 3\sqrt(I₀)

Step 1: Calculate the ratio

II = ( 3\sqrt(I₀) + \sqrt(I₀)3\sqrt(I₀) - \sqrt(I₀) )² = ( (4)/(2) )² = (2)² = 4

Thus, the ratio is $4:1$ $4:1$.

Pattern Recognition

A classic shortcut is using amplitude ratios. If intensities are in ratio $1:9$ $1:9$, amplitudes $a₁:a₂$ $a_1:a_2$ are in ratio $1:3$ $1:3$.

aa = (3+1)/(3-1) = 2 \Rightarrow I:I = 4:1

Chapter Mix

Class 12 Physics: Wave Optics

Solution & Explanation

Related Formula

Core Logic

Step 1: Final Conclusion

Pattern Recognition

Chapter Mix

More Wave Optics Previous-Year Questions — Page 3

Solution

Related Formula

Core Logic

Step 1: Find Ratio of Orders

Step 2: Calculate Distance

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Final Conclusion

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Substitute values into intensity ratio

Step 2: Expand the terms

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate the ratio

Pattern Recognition

Chapter Mix

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