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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 7

Q41 jee_main_2024_30_january_evening Polarisation of Light

A beam of unpolarised light of intensity

I₀

is passed through a polaroid

A

and then through another polaroid

B

which is oriented so that its principal plane makes an angle of

45^°

relative to that of

A

. The intensity of emergent light is :

A. $I₀ / 4$ $\mathrm{I_0 / 4}$
B. $I₀$ $\mathbf{I}_0$
C. $I₀ / 2$ $\mathrm{I}_0 / 2$
D. $I₀ / 8$ $\mathrm{I}_0 / 8$

Solution

Related Formula

I = Iincident ² θ (Malus's Law)

Core Logic

When unpolarised light of intensity $I₀$ $I_0$ passes through the first polaroid $A$ $\mathrm{A}$, it becomes plane-polarised, and its intensity drops by exactly half.

I₁ = (I₀)/(2)

When this polarised light passes through the second polaroid $B$ $\mathrm{B}$, the transmitted intensity is determined by Malus's Law.

Step 1: Apply Malus's Law

The angle between the principal planes of polaroids $A$ $\mathrm{A}$ and $B$ $\mathrm{B}$ is $θ = 45^°$ $\theta = 45^\circ$.

I₂ = I₁ ²(45^°)

I₂ = ((I₀)/(2)) ( 1\sqrt(2))²

I₂ = (I₀)/(2) \times (1)/(2) = (I₀)/(4)

Pattern Recognition

Unpolarised to Polarised $arrow I₀/2$ $\rightarrow I_0/2$. Polarised to Polarised $arrow I ²θ$ $\rightarrow I \cos^2\theta$. At $θ = 45^°$ $\theta = 45^\circ$, $²θ = 1/2$ $\cos^2\theta = 1/2$, resulting in a final intensity of $I₀/4$ $I_0/4$.

Chapter Mix

Class 12 Physics: Wave Optics

Q40 jee_main_2024_30_jan_morning Diffraction from a Single Slit

The diffraction pattern of a light of wavelength 400 nm diffracting from a slit of width

0.2 ~mm

is focused on the focal plane of a convex lens of focal length

100 ~cm

. The width of the

1st

secondary maxima will be :

A. $2 ~mm$ $2 \mathrm{~mm}$
B. $2 ~cm$ $2 \mathrm{~cm}$
C. $0.02 ~mm$ $0.02 \mathrm{~mm}$
D. $0.2 ~mm$ $0.2 \mathrm{~mm}$

Solution

Related Formula

Width of secondary maxima = (λ D)/(a)

Core Logic

In a single slit diffraction pattern, the linear width of any secondary maxima (fringe width of secondary bright bands) is given by $W = (λ D)/(a)$ $W = \frac{\lambda D}{a}$, whereas the central maximum is double this width ( $2(λ D)/(a)$ $2\frac{\lambda D}{a}$).

Step 1: Parameter Identification

Given values: Slit width, $a = 0.2 \times 10⁻³ ~m$ $a = 0.2 \times 10^{-3} \mathrm{~m}$ Wavelength, $λ = 400 \times 10⁻⁹ ~m$ Distance to screen (focal length of the lens), $D = 100 \times 10⁻² ~m = 1 ~m$

Step 2: Execution

Substitute these into the formula:

Width = 400 \times 10⁻⁹0.2 \times 10⁻³ \times 1

Width = (400)/(0.2) \times 10⁻⁶ ~m

Width = 2000 \times 10⁻⁶ ~m = 2 \times 10⁻³ ~m

Width = 2 ~mm

Pattern Recognition

Remember to strictly distinguish between central maximum ( $2λ D/a$ $2\lambda D/a$) and secondary maxima ( $λ D/a$ $\lambda D/a$).

Chapter Mix

Class 12 Physics: Wave Optics

Q36 jee_main_2024_31_jan_evening Polarization by Reflection (Brewster's Law)

When unpolarized light is incident at an angle of

60°

on a transparent medium from air. The reflected ray is completely polarized. The angle of refraction in the medium is

A. $30°$ $30^{\circ}$
B. $60°$ $60^{\circ}$
C. $90°$ $90^{\circ}$
D. $45°$ $45^{\circ}$

Solution

Related Formula

Brewster's Law states that at complete polarization upon reflection, the reflected and refracted rays are perpendicular to each other:

iₚ + r = 90°

Core Logic

The incident angle is given as $iₚ = 60°$ $i_p = 60^{\circ}$. At this angle, since the reflected ray is completely polarized, the geometry of Brewster's angle applies.

Polarization by Reflection (Brewster's Law) diagram for Q36 - JEE Main 2024 Evening

Step 1: Calculate Refraction Angle

60° + r = 90°

r = 90° - 60° = 30°

Pattern Recognition

The condition "reflected ray is completely polarized" is a direct trigger for Brewster's Law ( $iₚ + r = 90°$ $i_p + r = 90^{\circ}$). No refractive index ( $μ$ $\mu$) calculation is needed if only the geometric angle is asked.

Chapter Mix

Class 12 Physics: Wave Optics Class 12 Physics: Ray Optics and Optical Instruments

Q56 jee_main_2024_31_jan_morning Interference Of Waves

Two waves of intensity ratio

1:9

cross each other at a point. The resultant intensities at the point, when (a) Waves are incoherent is

I₁

(b) Waves are coherent is

I₂

and differ in phase by

60^°

. If

(I₁)/(I₂) = (10)/(x)

then

x =

Numerical Answer. Answer: 13 to 13

Solution

Related Formula

Iincoherent = IA + IB

Icoherent = IA + IB + 2\sqrt(IA IB) φ

Core Logic

Let the individual intensities be $IA = I₀$ $I_A = I_0$ and $IB = 9I₀$ $I_B = 9I_0$.

For incoherent waves, the net intensity is simply the algebraic sum:

I₁ = IA + IB = I₀ + 9I₀

$I₁ = 10I₀$ $I_1 = 10I_0$

Step 2: Coherent Waves Interference

For coherent waves with a phase difference of $φ = 60^°$ $\phi = 60^\circ$:

I₂ = IA + IB + 2\sqrt(IA IB) (60^°)

I₂ = I₀ + 9I₀ + 2\sqrt((I₀)(9I₀)) ((1)/(2))

I₂ = 10I₀ + 2(3I₀) ((1)/(2))

I₂ = 10I₀ + 3I₀ = 13I₀

Step 3: Finding x

Taking the ratio:

(I₁)/(I₂) = (10I₀)/(13I₀) = (10)/(13)

Comparing with the given expression $(10)/(x)$ $\frac{10}{x}$, we get: $x = 13$ $x = 13$

Chapter Mix

Class 12 Physics: Wave Optics

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JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%) | JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%)