A beam of light consisting of wavelengths 650 nm and 550 nm illuminates the Young's double slits with separation of 2…

Year	2026	2025	2024	Total
Questions	11	15	8	34

Year

2026

2025

2024

Total

Questions

Solution & Explanation

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

More Wave Optics Previous-Year Questions — Page 3

Q jee_main_2025_02_april_morning Diffraction

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

30circ

in a single slit diffraction pattern recorded using

628~nm

light, then the width of the slit is

Numerical Answer. Answer: 6 to 6

Solution

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

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: Find Ratio of Orders

Step 2: Calculate Distance

Pattern Recognition

Chapter Mix

Reference Study Guides

More Wave Optics Previous-Year Questions — Page 3

Solution

Related Formula

Core Logic

Step 1: Final Conclusion

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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