A double slit interference experiment performed with a light of wavelength 600nm forms an interference fringe pattern…

Year	2026	2025	2024	Total
Questions	11	15	8	34

Q24 jee_main_2025_04_april_evening Diffraction and Interference

In a Young's double slit experiment, two slits are located 1.5 mm apart. The distance of screen from slits is 2 m and the wavelength of the source is 400 nm. If the 20 maxima of the double slit pattern are contained within the centre maximum of the single slit diffraction pattern, then the width of each slit is

x \times 10⁻³ cm

, where x-value is ________.

Numerical Answer. Answer: 15 to 15

Solution

Related Formula

Width of central maximum in single-slit diffraction:

Δ ydiff = (2λ D)/(a)

Fringe width in double-slit interference:

β = (λ D)/(d)

Core Logic

Given condition: $20$ $20$ interference fringes fit inside the central diffraction envelope:

20 \times β = Δ ydiff

20 \times (λ D)/(d) = (2λ D)/(a)

Cancel common parameters:

(10)/(d) = (1)/(a) a = (d)/(10)

Step 1: Substitute Given Parameters

Slit separation $d = 1.5 mm = 0.15 cm$ $d = 1.5\text{ mm} = 0.15\text{ cm}$.

a = 0.15 cm10 = 0.015 cm = 15 \times 10⁻³ cm

Comparing with $x \times 10⁻³ cm$ $x \times 10^{-3}\text{ cm}$, the value of $x$ $x$ is 15.

Pattern Recognition

Envelope matching conditions rely strictly on the geometric ratio of slit separation ( $d$ $d$) to individual slit width ( $a$ $a$). Wavelength ( $λ$ $\lambda$) and screen distance ( $D$ $D$) cancel out completely.

Chapter Mix

Class 12 Physics: Wave Optics

Q jee_main_2025_04_april_morning Young's Double Slit Experiment

In a Young's double slit experiment, the slits are separated by

0.2~mm

. If the slits separation is increased to

0.4~mm

, the percentage change of the fringe width is:

A. $0%$ $0\%$
B. $100%$ $100\%$
C. $50%$ $50\%$
D. $25%$ $25\%$

Solution

Related Formula

β = (Dλ)/(d)

Therefore:

β \propto (1)/(d)

where:

$β$ $\beta$ = fringe width
$d$ $d$ = slit separation width
$D$ $D$ = distance to screen
$λ$ $\lambda$ = wavelength

Core Logic

Given data:

Initial slit separation, $d₁ = 0.2~mm$ $d_1 = 0.2\mathrm{~mm}$
Final slit separation, $d₂ = 0.4~mm$ $d_2 = 0.4\mathrm{~mm}$ ( $d$ $d$ is doubled).

Step 1: Calculate Percentage Change

Since $d₂ = 2d₁$ $d_2 = 2d_1$, the new fringe width becomes:

β₂ = (β₁)/(2)

Percentage change formulation:

Percentage Change = | (β₂ - β₁)/(β₁) | \times 100 = | (0.5β₁ - β₁)/(β₁) | \times 100 = 50%

Pattern Recognition

Doubling the denominator of an inversely proportional relationship halves the primary value, yielding an absolute $50%$ $50\%$ decrease.

Chapter Mix

Class 12 Physics: Wave Optics

Q1 jee_main_2025_24_jan_evening Young's Double Slit Experiment

Young's double slit interference apparatus is immersed in a liquid of refractive index

1.44

. It has slit separation of

1.5 mm

. The slits are illuminated by a parallel beam of light whose wavelength in air is

690 nm

. The fringe-width on a screen placed behind the plane of slits at a distance of

0.72 m

, will be:

A. $0.23 mm$ $0.23\ \mathrm{mm}$
B. $0.33 mm$ $0.33\ \mathrm{mm}$
C. $0.63 mm$ $0.63\ \mathrm{mm}$
D. $0.46 mm$ $0.46\ \mathrm{mm}$

Solution

Related Formula

β = ((λ₀)/(μ)) \times (D)/(d)

Core Logic

Given data:

Refractive index of liquid, $μ = 1.44$ $\mu = 1.44$
Slit separation, $d = 1.5 mm = 1.5 \times 10⁻³ m$
Wavelength in air, $λ₀ = 690 nm = 690 \times 10⁻⁹ m$
Distance of screen, $D = 0.72 m$ $D = 0.72\ \mathrm{m}$

Step 1: Calculation

Substituting the given values into the formula:

β = 690 \times 10⁻⁹ \times 0.721.44 \times 1.5 \times 10⁻³

β = 0.23 mm

Pattern Recognition

When a YDSE apparatus is immersed in a medium of refractive index $μ$ $\mu$, the fringe width decreases by a factor of $μ$ $\mu$, i.e., $β' = (β)/(μ)$ $\beta' = \frac{\beta}{\mu}$.

Chapter Mix

Class 12 Physics: Wave Optics

Q20 jee_main_2025_24_jan_evening Polarization

In a Young's double slit experiment, three polarizers are kept as shown in the figure

Polarizer placement in YDSE slits geometry schematic Q20 — 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

and

P₂

are orthogonal to each other. The polarizer

P₃

covers both the slits with its transmission axis at

45°

to those of

P₁

and

P₂

. An unpolarized light of wavelength

λ

and intensity

I₀

is incident on

P₁

and

P₂

. The intensity at a point after

P₃

where the path difference between the light waves from

s₁

and

s₂

is

(λ)/(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

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

Q9 jee_main_2025_24_jan_morning Young's Double Slit Experiment

The Young's double slit interference experiment is performed using light consisting of 480 nm and 600 nm wavelengths to form interference patterns. The least number of the bright fringes of 480 nm light that are required for the first coincidence with the bright fringes formed by 600 nm light is :-

A. 4
B. 8
C. 6
D. 5

Solution

Related Formula

The position $y$ $y$ of the $n$ $n$-th bright fringe from the central maximum in a YDSE setup is:

y = (nλ D)/(d)

For two wavelengths to overlap, their linear coordinates must match identically:

n₁λ₁ = n₂λ₂

Core Logic

Equating the respective path positions:

n₁ \times 480 nm = n₂ \times 600 nm

n₁n₂ = (600)/(480) = (5)/(4)

Step 1: Evaluating the Least Count

To satisfy the smallest integer ratio requirement for first spatial coincidence, the numerator must scale up to its base irreducible integer divisor:

$n1, = 5$ $n_{1,\min} = 5$

Pattern Recognition

Overlap occurs whenever indices inversely mimic their wavelength factors. The shorter wavelength always maps to a higher fringe index count.

Chapter Mix

Class 12 Physics: Wave Optics

Solution & Explanation

Related Formula

Core Logic

Step 1: Final Numerical Value

Pattern Recognition

Chapter Mix

Reference Study Guides

More Wave Optics Previous-Year Questions — Page 5

Solution

Related Formula

Core Logic

Step 1: Substitute Given Parameters

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate Percentage Change

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculation

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Evaluating the Least Count

Pattern Recognition

Chapter Mix

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