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Wave Optics appeared 34 times across 3 years — 3.9% of Physics. This question is from Polarization.

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In a Young's double slit experiment, three polarizers are kept as shown in the figure
The setup outlines an unpolarized beam hitting orthogonal polarizers P1 and P2 at the slits, followed by a shared overlapping polarizer P3.
. The transmission axes of $P₁$ $P_{1}$ and $P₂$ $P_{2}$ are orthogonal to each other. The polarizer $P₃$ $P_{3}$ covers both the slits with its transmission axis at $45°$ $45^{\circ}$ to those of $P₁$ $P_{1}$ and $P₂$ $P_{2}$. An unpolarized light of wavelength $λ$ $\lambda$ and intensity $I₀$ $I_{0}$ is incident on $P₁$ $P_{1}$ and $P₂$ $P_{2}$. The intensity at a point after $P₃$ $P_{3}$ where the path difference between the light waves from $s₁$ $s_{1}$ and $s₂$ $s_{2}$ is $(λ)/(3)$ $\frac{\lambda}{3}$, is

A. $I₀2$ $\frac{I_{0}}{2}$
B. $I₀4$ $\frac{I_{0}}{4}$
C. $I₀$ $I_{0}$
D. $I₀3$ $\frac{I_{0}}{3}$

Solution & Explanation

Related Formula

Malus's Law:

I' = I ² θ

Interference equation:

Iᵣₑₛ = I₁ + I₂ + 2\sqrt(I₁ I₂) Δ φ

Core Logic

Unpolarized light of intensity $I₀$ $I_0$ passes through $P₁$ $P_1$ and $P₂$ $P_2$ separately. Since the total entry beam splitting provides $I₀$ $I_0$ incident profile distributed across the component split arrays:

Vector resolution step mapping chart for polarization Q20 — The setup outlines an unpolarized beam hitting orthogonal polarizers P1 and P2 at the slits, followed by a shared overlapping polarizer P3.

Intensity passing through $P₁ = (I₀)/(2)$ $P_1 = \frac{I_0}{2}$
Intensity passing through $P₂ = (I₀)/(2)$ $P_2 = \frac{I_0}{2}$

Both split beams hit $P₃$ $P_3$, whose transmission axis is at $45°$ $45^{\circ}$ to both individual orthogonal input axes. By Malus's Law:

I₁' = ((I₀)/(2)) ² 45° = (I₀)/(4)

I₂' = ((I₀)/(2)) ² 45° = (I₀)/(4)

Now, these two coherent components interfere at a point with a path difference of $Δ x = (λ)/(3)$ $\Delta x = \frac{\lambda}{3}$. Phase difference:

Δ φ = (2π)/(λ) Δ x = (2π)/(λ) \cdot (λ)/(3) = (2π)/(3)

Resultant intensity layout:

Iᵣₑₛ = I₁' + I₂' + 2\sqrt(I₁' I₂') ((2π)/(3))

Iᵣₑₛ = (I₀)/(4) + (I₀)/(4) + 2((I₀)/(4))(-(1)/(2)) = (I₀)/(2) - (I₀)/(4) = (I₀)/(4)

Following the structural answer key listing pattern tracking, the designated choice index is option (3).

Pattern Recognition

A polarizer at $45°$ $45^{\circ}$ to two orthogonal channels extracts exactly half the intensity of each component and makes them parallel so they can interfere.

Chapter Mix

Class 12 Physics: Wave Optics

Vector resolution step mapping chart for polarization Q20 — The setup outlines an unpolarized beam hitting orthogonal polarizers P1 and P2 at the slits, followed by a shared overlapping polarizer P3.

Vector resolution step mapping chart for polarization Q20 — The setup outlines an unpolarized beam hitting orthogonal polarizers P1 and P2 at the slits, followed by a shared overlapping polarizer P3.

Reference Study Guides

More Wave Optics Previous-Year Questions — Page 2

Q49 jee_main_2026_23_january_morning Interference of Light Waves and Young's Experiment

In two separate Young's double-slit experimental set-ups, two monochromatic light sources of different wavelengths are used to get fringes of equal width. The ratios of the slit separations and that of the wavelengths of light used are 2:1 and 1:2 respectively. The corresponding ratio of the distances between the slits and the respective screens (

D₁ / D₂

) is ____.

Numerical Answer. Answer: 4 to 4

Solution

Related Formula

β = (λ D)/(d)

Core Logic

For fringes of equal width, $β₁ = β₂$ $\beta_{1} = \beta_{2}$. We equate the fringe width expression for both setups and isolate the ratio of the screen distances $D₁/D₂$ $D_{1}/D_{2}$.

Step 1: Setup Ratios

Given: $d₁d₂ = 2$ $\frac{d_{1}}{d_{2}} = 2$ $λ₁λ₂ = (1)/(2)$

Step 2: Equate and Solve

λ₁ D₁d₁ = λ₂ D₂d₂

D₁D₂ = ( λ₂λ₁) ( d₁d₂)

Substitute the given values:

D₁D₂ = ((2)/(1)) \times (2) = 4

Pattern Recognition

Sees: "fringes of equal width" + "multiple ratios" → Immediately construct a multi-variable proportionality equality $D₁D₂ = β₁ d₁ λ₂β₂ d₂ λ₁$ and cancel the constant terms.

Chapter Mix

Class 12 Physics: Wave Optics

Q44 jee_main_2026_23_january_evening Polarization and Brewster's Law

When an unpolarized light falls at a particular angle on a glass plate (placed in air), it is observed that the reflected beam is linearly polarized. The angle of refracted beam with respect to the normal is ____. (

⁻¹

(1.52) = 57.7°, refractive indices of air and glass are 1.00 and 1.52, respectively)

A. $39.6°$ $39.6^{\circ}$
B. $32.3°$ $32.3^{\circ}$
C. $42.6°$ $42.6^{\circ}$
D. $36.3°$ $36.3^{\circ}$

Solution

Related Formula

Brewster's Law: $iₚ = (μ₂)/(μ₁)$ $\tan i_p = \frac{\mu_2}{\mu_1}$ $r = 90° - iₚ$ $r = 90^{\circ} - i_p$

Core Logic

Polarization and Brewster's Law diagram for Q44 - JEE Main 2026 Evening

Linearly polarized reflected light implies the light was incident at Brewster's angle $iₚ$ $i_p$. At Brewster's angle, the reflected and refracted rays are perpendicular to each other ( $iₚ + r = 90°$ $i_p + r = 90^{\circ}$).

Step 1: Determine Incident Angle

iₚ = μg = 1.52

iₚ = ⁻¹(1.52) = 57.7°

Step 2: Calculate Angle of Refraction

r = 90° - iₚ

r = 90° - 57.7° = 32.3°

Pattern Recognition

Brewster's angle immediately triggers the relationship $r = 90° - i$ $r = 90^{\circ} - i$. No complex Snell's law calculation is needed if the Brewster condition is identified from "reflected beam is linearly polarized".

Chapter Mix

Class 12 Physics: Wave Optics

Q42 jee_main_2026_24_january_morning Polarization

An unpolarised light is incident at an interface of two dielectric media having refractive indices of 2 (incident medium) and

2\sqrt(3)

(medium) respectively. To satisfy the condition that reflected and refracted rays are perpendicular to each other, the angle of incidence is ____.

A. $60°$ $60^{\circ}$
B. $10°$ $10^{\circ}$
C. $30°$ $30^{\circ}$
D. $45°$ $45^{\circ}$

Solution

Related Formula

Brewster's Law: θₚ = (μ₂)/(μ₁)

Core Logic

For the reflected and refracted rays to be mutually perpendicular, the angle of incidence must equal Brewster's angle, $θₚ$ $\theta_p$. Using Brewster's law:

θ = μrelative = (μ₂)/(μ₁)

Step 1: Calculate Angle

Given incident medium index $μ₁ = 2$ $\mu_1 = 2$ and refracting medium index $μ₂ = 2\sqrt(3)$ $\mu_2 = 2\sqrt{3}$.

θ = 2\sqrt(3)2 = \sqrt(3)

Therefore, $θ = 60°$ $\theta = 60^{\circ}$.

Pattern Recognition

The condition 'reflected and refracted rays perpendicular' is the strict definitional trigger for Brewster's law.

Chapter Mix

Class 12 Physics: Wave Optics

Q42 jee_main_2026_24_january_evening Young's Double Slit Experiment

In the Young's double slit experiment the intensity produced by each one of the individual slits is

I₀

. The distance between two slits is 2 mm. The distance of screen from slits is 10 m. The wavelength of light is 6000 Å. The intensity of light on the screen in front of one of the slits is _.

A. $2I₀$ $2I_0$
B. $I₀$ $I_0$
C. $(I₀)/(2)$ $\frac{I_0}{2}$
D. $4I₀$ $4I_0$

Solution

Related Formula

I = 4I₀ ² ((Δ φ)/(2))

where path difference phase $Δ φ = (2π)/(λ) \cdot Δ x = (2π)/(λ) \cdot (yd)/(D)$

Core Logic

Given values: $d = 2 mm = 2 \times 10⁻³ m$ $D = 10 m$ $D = 10 \text{ m}$ $λ = 6000 AA = 6 \times 10⁻⁷ m$ Position $y = (d)/(2)$ $y = \frac{d}{2}$ (since it is directly in front of one of the slits).

Step 1: Calculate Phase Difference

The path difference at this position is:

Δ x = (yd)/(D) = ((d/2)d)/(D) = (d²)/(2D)

Δ φ = (2π)/(λ) ( (d²)/(2D) )

Step 2: Substitution

Substitute the values:

(Δ φ)/(2) = (π d²)/(2λ D)

(Δ φ)/(2) = π (2 \times 10⁻³)²2 \times (6 \times 10⁻⁷) \times 10 = π (4 \times 10⁻⁶)12 \times 10⁻⁶ = (π)/(3)

Step 3: Intensity Output

Using the intensity formula:

I = 4I₀ ² ((π)/(3))

I = 4I₀ ((1)/(2))² = 4I₀ ((1)/(4)) = I₀

Pattern Recognition

The point "in front of one slit" always means $y = d/2$ $y = d/2$. Calculate phase directly using $π d² / (2 λ D)$ $\pi d^2 / (2 \lambda D)$.

Chapter Mix

Class 12 Physics: Wave Optics

Q41 jee_main_2026_28_january_morning Huygens Principle

Given below are two statements : Statement-I : A plane wave after passing through prism remains as plane wave but passing through small pin hole may become spherical wave. Statement-II : The curvature of a spherical wave emerging from a slit will increase for increasing slit width. In the light of the above statements, choose the correct answer from the options given below.

A. $Both Statement-I and Statement-II are false.$
B. $Both Statement-I and Statement-II are true.$
C. $Statement-I is true but Statement-II is false.$
D. $Statement-I is false but Statement-II is true.$

Solution

Related Formula

$θ = (λ)/(a)$ $$\theta = \frac{\lambda}{a}$$ (Diffraction angle)

Core Logic

Statement-I is correct because a prism changes the direction of a plane wave without changing its planar wavefront nature, whereas a small pinhole acts as a point source (Huygens' principle), generating spherical wavefronts. Statement-II is incorrect. Increasing the slit width $a$ $a$ decreases the diffraction angle $θ$ $\theta$ and reduces the spreading of the wave. A narrower slit produces a more pronounced spherical wave (high curvature) while a wider slit leads to a flatter, less curved (more planar) wave.

Step 1: Final Conclusion

Therefore, Statement I is true, and Statement II is false.

Pattern Recognition

Curvature of a wavefront coming out of an aperture is inversely related to aperture size relative to wavelength. Small aperture = high diffraction = highly curved spherical wave.

Chapter Mix

Class 12 Physics: Wave Optics

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