Wave Optics JEE Notes 2026 | Huygens' Principle, YDSE, Diffraction, PYQs

1

Introduction & Wavefronts

Wave optics (also known as physical optics) treats light as a wave, explaining phenomena like interference, diffraction, and polarization that cannot be explained by ray optics (geometrical optics). The cornerstone of wave optics is the concept of the wavefront.

1.1 What is a Wavefront?

A wavefront is defined as the locus of all points in a medium that are vibrating in the same phase. Alternatively, it is the surface of constant phase at any instant.

The direction of propagation of light is always perpendicular to the wavefront. The speed with which the wavefront moves forward is the speed of light in that medium.

1.2 Types of Wavefronts

Most Repeated PYQ — Types of Wavefronts

Spherical Wavefront: Produced by a point source of light. The intensity falls off as \(I \propto 1/r^2\). Area \(\propto r^2\).
Cylindrical Wavefront: Produced by a line source of light (e.g., a slit illuminated from behind). The intensity falls off as \(I \propto 1/r\). Area \(\propto r\).
Plane Wavefront: Produced by a source at infinity. The rays are parallel, and intensity is constant. A small portion of a spherical or cylindrical wavefront at a large distance appears plane.

1.3 Relation between Phase Difference and Path Difference

If two points on a wave are separated by a path difference \(\Delta x\), the corresponding phase difference \(\Delta \phi\) is:

\Delta \phi = \frac{2\pi}{\lambda} \Delta x

where \(\lambda\) is the wavelength of light. A path difference of \(\lambda\) corresponds to a phase difference of \(2\pi\) (or 360°).

PYQ — JEE Main 2024 (Jan 31 Shift 2)

Question: The shape of the wavefront emitted by a point source of light at a finite distance is:

(a) Plane

(b) Cylindrical

(c) Spherical

(d) Hyperbolic

Solution:

A point source at a finite distance emits light radially in all directions. The locus of points vibrating in the same phase forms a sphere centered at the source.

\text{Shape is Spherical}

Answer: (c) Spherical

2

Huygens' Principle

Proposed by Dutch physicist Christiaan Huygens in 1678, this principle is a geometrical method to determine the new position of a wavefront from its earlier position. It forms the foundational theoretical framework for wave optics.

2.1 Postulates of Huygens' Principle

Most Repeated PYQ — Huygens' Postulates

Huygens' principle is based on the following postulates:

Secondary Wavelets: Every point on a given primary wavefront acts as a fresh source of new disturbance called secondary wavelets. These wavelets travel in all directions with the speed of light in that medium.
Envelope (Tangent Plane): The forward envelope (common tangent) drawn to all these secondary wavelets at any later instant \(t\) gives the new position of the wavefront.
Backward Wave: Huygens assumed that the backward wavefront does not exist. Though his construction couldn't explain this, it was later justified by Kirchhoff who proved mathematically that the amplitude of the secondary wavelet in the backward direction is zero.

2.2 Construction of Wavefronts

Key Concept

For a spherical wavefront, the secondary wavelets are spheres. The new wavefront is another sphere of larger radius.
For a plane wavefront, the secondary wavelets are spheres, but the common tangent forms a new plane parallel to the original.
The line joining the source to the point on the wavefront represents a ray of light, always perpendicular to the wavefront.

2.3 Validity and Limitations

Huygens' principle successfully explained reflection, refraction, and interference. However, it failed to explain:

The exact nature of light (transverse vs longitudinal). Huygens assumed light is longitudinal, which was later proven wrong by polarization experiments.
Why light travels in straight lines (rectilinear propagation).
The phenomenon of diffraction explicitly (though later augmented by Fresnel).

PYQ — JEE Main 2023 (Apr 13 Shift 1)

Question: According to Huygens' principle, the amplitude of secondary wavelets is zero in the backward direction. Who mathematically proved this?

(a) Christiaan Huygens

(b) Augustin-Jean Fresnel

(c) Gustav Kirchhoff

(d) Thomas Young

Solution:

While Huygens proposed the principle and assumed no backward wave exists, it was Gustav Kirchhoff who later provided the rigorous mathematical proof using his diffraction theory (Kirchhoff's diffraction formula), showing the obliquity factor causes zero amplitude in the backward direction.

\text{Gustav Kirchhoff}

Answer: (c) Gustav Kirchhoff

3

Laws of Reflection & Refraction by Huygens

Huygens' principle can be used to derive the basic laws of geometrical optics: the law of reflection and the law of refraction (Snell's Law). This derivation proves that wave optics reduces to ray optics when the wavelength is much smaller than the dimensions of the obstacles/apertures.

3.1 Law of Reflection

Most Repeated PYQ — Reflection Proof

Statement: The angle of incidence (\(i\)) equals the angle of reflection (\(r\)), and the incident ray, reflected ray, and normal lie in the same plane.

Proof: Let a plane wavefront \(AB\) incident at an angle \(i\) on a reflecting surface. If \(v\) is the speed of light, in time \(\Delta t\), the wavefront travels \(v\Delta t\). By the time point \(B\) reaches \(C\), secondary wavelets from \(A\) have spread out to a radius \(AD = v\Delta t\). The new wavefront is \(CD\).

\sin i = \frac{BC}{AC} = \frac{v\Delta t}{AC} \quad \text{and} \quad \sin r = \frac{AD}{AC} = \frac{v\Delta t}{AC}

Since \(\sin i = \sin r\), we get \(i = r\).

3.2 Law of Refraction (Snell's Law)

Most Repeated PYQ — Snell's Law Proof

Statement: \(\sin i / \sin r = v_1/v_2 = \mu_2/\mu_1 = n_{21}\), where \(\mu_1, \mu_2\) are refractive indices of medium 1 and 2.

Proof: A plane wavefront \(AB\) passes from medium 1 (speed \(v_1\)) to medium 2 (speed \(v_2\)). In time \(\Delta t\), point \(B\) travels \(v_1\Delta t\) to \(C\). Meanwhile, secondary wavelets from \(A\) travel \(v_2\Delta t\) into medium 2, reaching \(D\). The new wavefront is \(CD\).

\sin i = \frac{BC}{AC} = \frac{v_1 \Delta t}{AC} \quad \text{and} \quad \sin r = \frac{AD}{AC} = \frac{v_2 \Delta t}{AC}

\frac{\sin i}{\sin r} = \frac{v_1}{v_2} = \frac{c/v_2}{c/v_1} = \frac{\mu_2}{\mu_1} = ^1\mu_2

This proves Snell's Law. It also explains that light bends towards the normal when passing from a rarer to a denser medium because \(v_2 < v_1\).

PYQ — JEE Main 2022 (June 28 Shift 1)

Question: Light travels a distance \(x\) in time \(t_1\) in medium 1 and travels the same distance \(x\) in time \(t_2\) in medium 2. The ratio of refractive indices of medium 1 to medium 2 is:

(a) \(t_1 : t_2\)

(b) \(t_2 : t_1\)

(c) \(t_1 t_2 : 1\)

(d) \(1 : t_1 t_2\)

Solution:

Speed of light in medium 1: \(v_1 = x/t_1\).

Speed of light in medium 2: \(v_2 = x/t_2\).

Refractive index \(\mu \propto 1/v\). Therefore:

\frac{\mu_1}{\mu_2} = \frac{v_2}{v_1} = \frac{x/t_2}{x/t_1} = \frac{t_1}{t_2} = t_1 : t_2

\mu_1 : \mu_2 = t_1 : t_2

Answer: (a) \(t_1 : t_2\)

4

Principle of Superposition

When two or more light waves travel through the same region of space simultaneously, the resultant displacement at any point is the vector sum of the displacements due to the individual waves. This is the Principle of Superposition, and it is the mechanism behind interference and diffraction.

4.1 Mathematical Expression

If \(y_1, y_2, y_3, \ldots, y_n\) are the displacements of individual waves at a point, the resultant displacement is:

y = y_1 + y_2 + y_3 + \ldots + y_n

For two coherent waves arriving at a point with a phase difference \(\Delta \phi\):

y = y_1 + y_2 = a \sin(\omega t) + a \sin(\omega t + \Delta \phi)

Using trigonometric identity, this gives a resultant wave of amplitude:

A = \sqrt{a_1^2 + a_2^2 + 2a_1 a_2 \cos(\Delta \phi)}

4.2 Intensity of Resultant Wave

Most Repeated PYQ — Resultant Intensity

Since intensity \(I \propto A^2\), the resultant intensity is:

I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos(\Delta \phi)

Maximum Intensity (Constructive): \(\Delta \phi = 2n\pi\) (where \(n = 0, 1, 2, \ldots\)). \(I_{max} = (\sqrt{I_1} + \sqrt{I_2})^2\).
Minimum Intensity (Destructive): \(\Delta \phi = (2n-1)\pi\). \(I_{min} = (\sqrt{I_1} - \sqrt{I_2})^2\).
If \(I_1 = I_2 = I_0\), then \(I_{max} = 4I_0\) and \(I_{min} = 0\).

PYQ — JEE Main 2024 (Apr 5 Shift 1)

Question: Two light waves of intensities \(I\) and \(4I\) superimpose at a point. If the phase difference between them is \(\pi/2\), the resultant intensity at that point is:

(a) \(5I\)

(b) \(7I\)

(c) \(9I\)

(d) \(3I\)

Solution:

Using the intensity formula:

I_R = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos(\Delta \phi)

Given \(I_1 = I\), \(I_2 = 4I\), and \(\Delta \phi = \pi/2\). Since \(\cos(\pi/2) = 0\):

I_R = I + 4I + 2\sqrt{I \cdot 4I} \cdot 0 = 5I + 0 = 5I

\[ I_R = 5I \]

Answer: (a) \(5I\)

5

Interference of Light

Interference is the phenomenon of non-uniform redistribution of light energy when two or more coherent waves superpose. Bright and dark bands (fringes) are formed depending on whether the superposition is constructive or destructive.

5.1 Constructive Interference (Bright Fringe)

Most Repeated PYQ — Bright Fringe Condition

Occurs when the path difference is an integer multiple of wavelength:

\Delta x = n\lambda \quad (n = 0, 1, 2, \ldots)

The corresponding phase difference is \(\Delta \phi = 2n\pi\). The resultant intensity is maximum: \(I_{max} = (\sqrt{I_1} + \sqrt{I_2})^2\).

5.2 Destructive Interference (Dark Fringe)

Most Repeated PYQ — Dark Fringe Condition

Occurs when the path difference is an odd half-integer multiple of wavelength:

\Delta x = (2n-1)\frac{\lambda}{2} \quad (n = 1, 2, 3, \ldots)

The corresponding phase difference is \(\Delta \phi = (2n-1)\pi\). The resultant intensity is minimum: \(I_{min} = (\sqrt{I_1} - \sqrt{I_2})^2\).

5.3 Law of Conservation of Energy

During interference, light energy is not created or destroyed; it is merely redistributed. The total average intensity over a large region is the sum of the individual intensities.

I_{avg} = \frac{I_{max} + I_{min}}{2} = I_1 + I_2

PYQ — JEE Main 2023 (Jan 31 Shift 1)

Question: The intensity ratio of two interfering coherent sources is 9:1. The ratio of maximum and minimum intensities in the interference pattern is:

(a) 9:1

(b) 4:1

(c) 16:1

(d) 25:1

Solution:

Given \(I_1 / I_2 = 9 / 1\). Since intensity \(\propto \) amplitude\(^2\), let \(I_1 = 9I_0\) and \(I_2 = I_0\). Then \(a_1 = 3a\) and \(a_2 = a\).

\frac{I_{max}}{I_{min}} = \frac{(\sqrt{I_1} + \sqrt{I_2})^2}{(\sqrt{I_1} - \sqrt{I_2})^2} = \frac{(\sqrt{9} + \sqrt{1})^2}{(\sqrt{9} - \sqrt{1})^2} = \frac{(3+1)^2}{(3-1)^2} = \frac{16}{4} = \frac{4}{1}

I_{max} : I_{min} = 4 : 1

Answer: (b) 4:1

6

Coherent Sources

For a sustained and observable interference pattern, we need coherent sources. These are sources that emit light waves continuously with the same frequency (or wavelength) and a constant phase difference.

6.1 Conditions for Sustained Interference

Most Repeated PYQ — Conditions

Coherent Sources: Constant phase difference is mandatory.
Same Frequency/Wavelength: Essential for standing interference.
Same Amplitude (ideally): For good contrast between bright and dark fringes.
Small Separation: Sources should be close to each other to produce broad and distinct fringes.
Small Path Difference: To prevent overlapping of multiple orders.

6.2 Methods of Producing Coherent Sources

Two independent sources can never be coherent because light emission is a random process (spontaneous emission). We obtain coherent sources by dividing a single wave into two parts. The two main methods are:

Division of Wavefront: The wavefront is divided into parts using mirrors or slits (e.g., Young's Double Slit, Fresnel's Biprism, Lloyd's Mirror). Requires a point source.
Division of Amplitude: The amplitude of the incident beam is divided using partial reflection/refraction (e.g., thin films, Newton's rings, Michelson Interferometer). Can use an extended source.

PYQ — JEE Main 2021 (Aug 27 Shift 1)

Question: Two independent light sources cannot produce a sustained interference pattern because:

(a) They have different wavelengths

(b) They have different amplitudes

(c) Their phase difference changes randomly and rapidly

(d) They travel at different speeds

Solution:

Light emission from atoms occurs in short random bursts (10⁻⁸ s). The phase of light from independent sources changes millions of times per second, preventing any constant phase relationship. Thus, the interference pattern shifts so rapidly that the eye perceives only a uniform illumination.

\text{Phase difference changes randomly and rapidly}

Answer: (c) Their phase difference changes randomly and rapidly

7

Young's Double Slit Experiment (YDSE)

Thomas Young performed his famous double-slit experiment in 1801 to demonstrate the wave nature of light. It is the classic example of interference by division of wavefront.

7.1 Experimental Setup

A monochromatic source of light \(S\) illuminates a narrow slit \(S_0\).
Light from \(S_0\) falls on two extremely close and parallel slits \(S_1\) and \(S_2\).
According to Huygens' principle, \(S_1\) and \(S_2\) act as coherent secondary sources.
The waves from \(S_1\) and \(S_2\) superpose on a screen placed at a distance \(D\) from the slits.
Interference fringes (alternating bright and dark bands) are observed on the screen.

7.2 Path Difference in YDSE

Most Repeated PYQ — Path Difference

Let \(d\) be the distance between the slits \(S_1\) and \(S_2\), and \(D\) be the distance to the screen. For a point \(P\) at a distance \(y\) from the central maximum:

Path difference \(\Delta x = S_2P - S_1P\).

From geometry, applying the binomial approximation (\(D \gg d\) and \(D \gg y\)):

\Delta x = \frac{yd}{D}

Alternatively, if the point is at an angle \(\theta\) from the central axis:

\Delta x = d \sin \theta \approx d \tan \theta = \frac{yd}{D}

7.3 Positions of Bright and Dark Fringes

For Bright Fringes (Constructive): \(\Delta x = n\lambda\)

y_n = \frac{n\lambda D}{d} \quad (n = 0, 1, 2, \ldots)

For Dark Fringes (Destructive): \(\Delta x = (2n-1)\lambda/2\)

y_n = \frac{(2n-1)\lambda D}{2d} \quad (n = 1, 2, 3, \ldots)

PYQ — JEE Main 2024 (Jan 30 Shift 2)

Question: In YDSE, the slit separation is \(d = 0.5\) mm and the screen is at \(D = 1\) m. If the 3rd bright fringe is formed at a distance of 3 mm from the central maximum, the wavelength of light used is:

(a) 400 nm

(b) 500 nm

(c) 600 nm

(d) 700 nm

Solution:

Position of \(n\)-th bright fringe: \(y_n = \frac{n\lambda D}{d}\).

Convert to consistent units: \(d = 0.5 \times 10^{-3}\) m, \(D = 1\) m, \(y_3 = 3 \times 10^{-3}\) m.

3 \times 10^{-3} = \frac{3 \cdot \lambda \cdot 1}{0.5 \times 10^{-3}}

\lambda = \frac{3 \times 10^{-3} \times 0.5 \times 10^{-3}}{3} = 0.5 \times 10^{-6} \text{ m} = 500 \text{ nm}

\lambda = 500 \text{ nm}

Answer: (b) 500 nm

8

Fringe Width in YDSE

Fringe width is the distance between two successive bright or two successive dark fringes. It is a measure of the "width" of the interference pattern.

8.1 Fringe Width Formula

Most Repeated PYQ — Fringe Width

The distance between the \(n\)-th and \((n+1)\)-th bright fringe:

\beta = y_{n+1} - y_n = \frac{(n+1)\lambda D}{d} - \frac{n\lambda D}{d} = \frac{\lambda D}{d}

The same formula holds for dark fringes. Key observations:

Fringe width \(\beta \propto \lambda\): Longer wavelengths give wider fringes (red light gives broader fringes than blue).
\(\beta \propto D\): Moving the screen farther increases fringe width.
\(\beta \propto 1/d\): Bringing slits closer increases fringe width.
Frinspace is independent of the position \(n\) (all fringes have equal width).

8.2 Angular Fringe Width

The angular position of the \(n\)-th bright fringe is \(\theta_n = n\lambda/d\). The angular separation between consecutive fringes:

\theta = \frac{\lambda}{d} \text{ (in radians)}

8.3 Maximum Number of Fringes

Key Concept

The maximum order of interference fringes is limited by the geometry: \(|\sin \theta| \leq 1\).

n_{max} = \frac{d}{\lambda}

Total number of fringes on the screen = \(2n_{max} + 1\) (including central max).

PYQ — JEE Main 2022 (July 25 Shift 2)

Question: In YDSE, if the distance between the slits is doubled and the distance between the slits and the screen is halved, the fringe width becomes:

(a) 4 times

(b) 1/4 times

(c) 2 times

(d) 1/2 times

Solution:

Initial fringe width: \(\beta_1 = \frac{\lambda D}{d}\).

New parameters: \(d' = 2d\) and \(D' = D/2\).

\beta_2 = \frac{\lambda D'}{d'} = \frac{\lambda (D/2)}{2d} = \frac{1}{4} \frac{\lambda D}{d} = \frac{1}{4} \beta_1

\beta_2 = \frac{1}{4} \beta_1

Answer: (b) 1/4 times

9

Shape of Interference Fringes

The exact geometrical shape of the interference fringes depends on the path difference condition. While we often approximate them as straight lines near the center, the true shape is different.

9.1 Actual Shape (Hyperbola)

Most Repeated PYQ — Hyperbolic Fringes

A point of constant path difference \(\Delta x\) must satisfy \(S_2P - S_1P = \text{const}\). The locus of points where the difference of distances from two fixed points is constant is a hyperbola.

The fringes are hyperbolas with \(S_1\) and \(S_2\) as foci.
The central bright fringe (\(\Delta x = 0\)) is a straight line perpendicular to the line joining the slits, lying exactly midway.
Other fringes are curved, bending away from the central line.

9.2 Shape on the Screen

If the screen is held perpendicular to the line joining the midpoint of the slits to the central point on the screen:

For small \(y\) (near the axis), the hyperbolas intersect the screen in nearly straight lines. Thus, fringes are practically straight lines.
Far from the center, the curvature becomes noticeable.
If the screen is held parallel to the line joining the slits, the fringes are semi-circular/hyperbolic arcs.

PYQ — JEE Main 2023 (Apr 12 Shift 1)

Question: The shape of the interference fringes in Young's double slit experiment when the screen is held perpendicular to the central axis is:

(a) Concentric circles

(b) Straight lines

(c) Hyperbolic

(d) Parabolic

Solution:

The exact theoretical locus of constant path difference from two point sources \(S_1\) and \(S_2\) is a hyperbola in 3D space. When this intersects with a plane (the screen) perpendicular to the central axis, the resulting curve is mathematically a hyperbola. However, near the center, they appear as straight lines. The most accurate option is Hyperbolic.

\text{Hyperbolic}

Answer: (c) Hyperbolic

10

Intensity Distribution in YDSE

The intensity of light at any point on the screen in YDSE varies periodically. If the slits are identical, the maximum intensity is \(4I_0\) (where \(I_0\) is intensity from one slit), and the minimum is zero.

10.1 Intensity Variation with Phase/Path Difference

Most Repeated PYQ — Intensity Formula

If \(I_1 = I_2 = I_0\), the resultant intensity at phase difference \(\Delta \phi\) is:

I = 4I_0 \cos^2\left(\frac{\Delta \phi}{2}\right)

Since \(\Delta \phi = \frac{2\pi}{\lambda} \Delta x = \frac{2\pi y d}{\lambda D}\), intensity as a function of position \(y\):

I = 4I_0 \cos^2\left(\frac{\pi y d}{\lambda D}\right)

10.2 Ratio of Slit Widths

If the slits emit different intensities \(I_1\) and \(I_2\), the ratio of maximum to minimum intensity is given by:

\frac{I_{max}}{I_{min}} = \frac{(\sqrt{I_1} + \sqrt{I_2})^2}{(\sqrt{I_1} - \sqrt{I_2})^2}

Since intensity \(I \propto w\) (slit width), we have \(\frac{w_1}{w_2} = \frac{I_1}{I_2}\). Let \(\frac{I_{max}}{I_{min}} = R\). Then the slit width ratio is:

\frac{w_1}{w_2} = \left( \frac{\sqrt{R} + 1}{\sqrt{R} - 1} \right)^2

PYQ — JEE Main 2024 (Apr 6 Shift 2)

Question: In an interference experiment, the ratio of the intensities of the bright and dark fringes is 9. The ratio of the widths of the two slits is:

(a) 4:1

(b) 9:1

(c) 81:1

(d) 3:1

Solution:

Given \(\frac{I_{max}}{I_{min}} = 9 = R\).

Using the slit width ratio formula:

\frac{w_1}{w_2} = \left( \frac{\sqrt{R} + 1}{\sqrt{R} - 1} \right)^2 = \left( \frac{\sqrt{9} + 1}{\sqrt{9} - 1} \right)^2 = \left( \frac{3+1}{3-1} \right)^2 = \left( \frac{4}{2} \right)^2 = 2^2 = \frac{4}{1}

\[ w_1 : w_2 = 4 : 1 \]

Answer: (a) 4:1

11

YDSE in a Liquid Medium

When the entire YDSE apparatus is immersed in a transparent medium like water, the wavelength of light changes, altering the fringe pattern.

11.1 Effect on Wavelength and Fringe Width

Most Repeated PYQ — YDSE in Water

If the refractive index of the liquid is \(\mu\), the speed of light decreases to \(c/\mu\). Since frequency \(\nu\) remains unchanged, the new wavelength is:

\lambda' = \frac{\lambda}{\mu}

The new fringe width becomes:

\beta' = \frac{\lambda' D}{d} = \frac{\lambda D}{\mu d} = \frac{\beta}{\mu}

The fringes shrink by a factor of \(\mu\). The position of the central maximum remains the same (since \(\Delta x = 0\) there), but all other fringes shift closer to the center.

11.2 Key Observations

Important Points

Number of fringes in the same region increases by a factor of \(\mu\).
Angular fringe width \(\theta' = \lambda' / d = \lambda / (\mu d)\).
If 10 fringes were initially in a distance \(L\), now \(10\mu\) fringes will fit in the same \(L\).

PYQ — JEE Main 2024 (Jan 29 Shift 2)

Question: In YDSE, the fringe width is \(\beta\). If the entire apparatus is immersed in water of refractive index \(4/3\), the new fringe width will be:

(a) \(3\beta/4\)

(b) \(4\beta/3\)

(c) \(\beta/4\)

(d) \(4\beta\)

Solution:

When immersed in a medium of refractive index \(\mu\), the fringe width shrinks by a factor of \(\mu\).

\beta' = \frac{\beta}{\mu} = \frac{\beta}{4/3} = \frac{3\beta}{4}

\beta' = \frac{3\beta}{4}

Answer: (a) \(3\beta/4\)

12

YDSE with Transparent Slab

When a thin transparent slab of thickness \(t\) and refractive index \(\mu\) is introduced in front of one of the slits in YDSE, the entire interference pattern shifts towards the side of the slab. This is among the most frequently tested concepts in JEE.

12.1 Path Difference with Slab

Most Repeated PYQ — Slab Path Difference

Light travels distance \(t\) inside the slab, where its wavelength is \(\lambda/\mu\). If it had traveled the same distance \(t\) in air, it would have wavelength \(\lambda\). The extra path introduced by the slab:

\Delta x_{slab} = (\mu - 1)t

The total effective path difference at a point \(P\) at position \(y\) becomes:

\Delta x_{total} = \frac{yd}{D} + (\mu - 1)t

(Sign convention: Positive if the slab is placed on the upper slit, shifting the pattern downwards, and vice versa).

12.2 Fringe Shift

Most Repeated PYQ — Fringe Shift Formula

The central maximum (where \(\Delta x_{total} = 0\)) shifts to a new position \(y_0\):

0 = \frac{y_0 d}{D} + (\mu - 1)t \implies y_0 = -\frac{D(\mu - 1)t}{d}

The magnitude of the shift is:

y_0 = \frac{D(\mu - 1)t}{d} = \frac{(\mu - 1)t}{\lambda} \beta

Number of fringes shifted = \(\frac{(\mu - 1)t}{\lambda}\).

PYQ — JEE Main 2024 (Apr 8 Shift 2)

Question: In YDSE, a transparent slab of refractive index 1.5 is introduced in front of one of the slits. The wavelength of light used is 600 nm. If the interference pattern shifts by 10 fringe widths, the thickness of the slab is:

(a) 6 µm

(b) 12 µm

(c) 3 µm

(d) 4 µm

Solution:

Number of fringes shifted \(N = \frac{(\mu - 1)t}{\lambda}\).

Given \(N = 10\), \(\mu = 1.5\), \(\lambda = 600 \text{ nm} = 600 \times 10^{-9} \text{ m}\).

10 = \frac{(1.5 - 1) \cdot t}{600 \times 10^{-9}} \implies 10 = \frac{0.5 \cdot t}{600 \times 10^{-9}}

t = \frac{10 \times 600 \times 10^{-9}}{0.5} = 12000 \times 10^{-9} \text{ m} = 12 \times 10^{-6} \text{ m} = 12 \text{ µm}

t = 12 \text{ µm}

Answer: (b) 12 µm

13

Diffraction of Light (Single Slit)

Diffraction is the phenomenon of bending of light around the corners of an obstacle or aperture. In single slit diffraction, light passing through a narrow slit spreads out, and a pattern of alternating bright and dark bands is formed on a screen.

13.1 Condition for Minima (Dark Fringes)

Most Repeated PYQ — Diffraction Minima

For a slit of width \(a\), the path difference between waves coming from the top and bottom edges of the slit is \(a \sin \theta\). The condition for destructive interference (dark fringes) is:

a \sin \theta = n\lambda \quad (n = \pm 1, \pm 2, \pm 3, \ldots)

Note: \(n \neq 0\) for minima because \(n=0\) corresponds to the central maximum (the entire slit contributes constructively at the center).

13.2 Condition for Secondary Maxima (Bright Fringes)

The exact condition for secondary maxima involves calculus (derivative of the intensity function), giving the approximate condition:

a \sin \theta \approx (2n + 1)\frac{\lambda}{2} \quad (n = \pm 1, \pm 2, \ldots)

The intensity of these secondary maxima decreases rapidly. The first secondary maximum has about 4.5% of the intensity of the central maximum.

PYQ — JEE Main 2023 (Apr 11 Shift 1)

Question: Light of wavelength 500 nm is incident on a narrow slit of width 0.25 mm. The angle at which the first minimum of the diffraction pattern will occur is:

(a) \(2 \times 10^{-3}\) rad

(b) \(1.5 \times 10^{-3}\) rad

(c) \(1 \times 10^{-3}\) rad

(d) \(0.5 \times 10^{-3}\) rad

Solution:

For the first minimum, \(n = 1\):

a \sin \theta = \lambda \implies \sin \theta \approx \theta = \frac{\lambda}{a} \text{ (for small angles)}

Convert to standard units: \(\lambda = 500 \times 10^{-9}\) m, \(a = 0.25 \times 10^{-3}\) m.

\theta = \frac{500 \times 10^{-9}}{0.25 \times 10^{-3}} = 2000 \times 10^{-6} = 2 \times 10^{-3} \text{ rad}

\theta = 2 \times 10^{-3} \text{ rad}

Answer: (a) \(2 \times 10^{-3}\) rad

14

Width of Central Maximum

The diffraction pattern consists of a broad, bright central maximum flanked by much weaker secondary maxima. The width of the central maximum is a crucial parameter, differing from the uniform fringe widths seen in interference.

14.1 Angular and Linear Width

Most Repeated PYQ — Central Max Width

The central maximum extends from the first minimum on one side (\(n = -1\)) to the first minimum on the other side (\(n = +1\)).

First minimum occurs at: \(a \sin \theta = \lambda \implies \theta \approx \lambda/a\).

Angular width of central maximum:

2\theta = \frac{2\lambda}{a}

Linear width of central maximum (on screen at distance \(D\)):

Y = \frac{2\lambda D}{a}

Width of secondary maxima is half the width of the central maximum: \(\beta' = \frac{\lambda D}{a}\).

14.2 Key Differences from Interference

Common Misconception

In interference (YDSE), all bright fringes have the same intensity and width. In diffraction, the central maximum is twice as wide as the secondary maxima and is much brighter. Secondary maxima decrease in intensity as we move away from the center.

PYQ — JEE Main 2021 (Jul 27 Shift 1)

Question: In a single slit diffraction experiment, the width of the slit is made double. The width of the central maximum will become:

(a) Half

(b) Double

(c) Four times

(d) Remains same

Solution:

Width of central maximum: \(Y = \frac{2\lambda D}{a}\).

If slit width \(a\) is doubled to \(2a\), the new width is:

Y' = \frac{2\lambda D}{2a} = \frac{Y}{2}

Y' = \frac{Y}{2} \text{ (Half)}

Answer: (a) Half

15

Resolving Power & Rayleigh Criterion

The resolving power of an optical instrument is its ability to distinguish between two closely spaced objects. The limit of resolution is determined by diffraction.

15.1 Rayleigh's Criterion

Most Repeated PYQ — Rayleigh Criterion

According to Lord Rayleigh, two point sources are said to be "just resolved" when the central maximum of the diffraction pattern of one source falls exactly on the first minimum of the diffraction pattern of the other.

The minimum angular separation \(\theta_{min}\) between two objects that can be just resolved is:

\theta_{min} = \frac{1.22 \lambda}{a}

where \(\lambda\) is the wavelength of light and \(a\) is the aperture (diameter) of the objective lens.

15.2 Resolving Power of Telescope

For a telescope, the resolving power is the reciprocal of the limit of resolution:

R.P. = \frac{1}{\theta_{min}} = \frac{a}{1.22 \lambda}

To improve resolving power (see finer details):

Increase aperture \(a\) (e.g., large mirrors in observatories like James Webb Telescope).
Decrease wavelength \(\lambda\) (using blue or ultraviolet light).

15.3 Resolving Power of Microscope

For a microscope, the limit is the minimum distance \(d\) between two points that can be distinguished:

d = \frac{1.22 \lambda}{2 \mu \sin \theta}

where \(\mu\) is the refractive index of the medium between object and lens, and \(\theta\) is the semi-angle of the cone of light. The term \(\mu \sin \theta\) is the numerical aperture.

Resolving Power \(= 1/d \propto \mu \sin \theta\).

PYQ — JEE Main 2020 (Sep 5 Shift 1)

Question: The resolving power of a telescope can be increased by:

(a) Increasing the wavelength of light

(b) Decreasing the diameter of the objective lens

(c) Increasing the diameter of the objective lens

(d) Decreasing the focal length of the eyepiece

Solution:

Resolving power of a telescope is \(R.P. = \frac{a}{1.22 \lambda}\), where \(a\) is the aperture (diameter) of the objective lens.

It is directly proportional to the diameter \(a\). Hence, increasing the diameter of the objective lens increases the resolving power.

\text{Increasing the diameter of the objective lens}

Answer: (c) Increasing the diameter of the objective lens

16

Polarization of Light

Polarization is the phenomenon of restricting the vibrations of the electric vector of light to a specific plane. It is the definitive proof that light is a transverse wave (since longitudinal waves cannot be polarized).

16.1 Unpolarized vs Polarized Light

Unpolarized Light: The electric vector vibrates in all possible directions perpendicular to the direction of propagation (e.g., ordinary sunlight, light from a bulb).
Linearly (Plane) Polarized Light: The electric vector vibrates only in one single plane perpendicular to the direction of propagation.

16.2 Malus' Law

Most Repeated PYQ — Malus' Law

When plane polarized light of intensity \(I_0\) is incident on an analyzer (polaroid) whose axis is at an angle \(\theta\) to the vibration direction, the transmitted intensity \(I\) is given by Malus' Law:

I = I_0 \cos^2 \theta

If unpolarized light of intensity \(I_{in}\) is incident on a polarizer, the transmitted light is polarized with half the intensity: \(I_0 = I_{in} / 2\).

16.3 Brewster's Law (Polarization by Reflection)

Most Repeated PYQ — Brewster's Angle

When unpolarized light is incident on a transparent surface at a particular angle called Brewster's angle (or polarizing angle \(i_p\)), the reflected light is completely plane polarized.

\mu = \tan i_p

where \(\mu\) is the refractive index of the reflecting medium. At this angle, the reflected and refracted rays are perpendicular to each other (\(i_p + r = 90^\circ\)).

PYQ — JEE Main 2024 (Jan 27 Shift 1)

Question: Unpolarized light of intensity \(I_0\) is incident on a polarizer and an analyzer placed crossed to each other (angle \(90^\circ\)). If another polarizer is inserted between them at an angle of \(45^\circ\) to the first, the final intensity of light emerging will be:

(a) \(I_0 / 2\)

(b) \(I_0 / 4\)

(c) \(I_0 / 8\)

(d) Zero

Solution:

After first polarizer: \(I_1 = I_0 / 2\).

After inserted polarizer (at \(45^\circ\)): \(I_2 = I_1 \cos^2(45^\circ) = (I_0 / 2) \cdot (1/2) = I_0 / 4\).

After analyzer (at \(45^\circ\) to inserted polarizer): \(I_3 = I_2 \cos^2(45^\circ) = (I_0 / 4) \cdot (1/2) = I_0 / 8\).

I_{final} = \frac{I_0}{8}

Answer: (c) \(I_0 / 8\)

17

Interference vs Diffraction

While both interference and diffraction result from the superposition of waves and produce patterns of bright and dark bands, they arise from different physical setups and have distinct mathematical and visual properties.

17.1 Comparison Table

Property	Interference	Diffraction
Source of Superposition	Superposition of waves from two distinct coherent sources (e.g., two slits).	Superposition of waves coming from different parts of the same wavefront (e.g., single slit).
Fringe Width	All bright and dark fringes have equal width (\(\beta = \lambda D / d\)).	Central maximum is twice as wide as secondary maxima (\(\beta' = \lambda D / a\)).
Intensity Distribution	All bright fringes have the same intensity (ideally).	Intensity of bright fringes decreases rapidly away from the center.
Minima Condition	Path difference \(\Delta x = n\lambda\) for max, \((2n-1)\lambda/2\) for min.	Path difference \(a \sin \theta = n\lambda\) for minima (not maxima!).
Number of Sources	Discrete (2 or more point sources).	Continuous (infinite point sources across the slit).

PYQ — JEE Main 2022 (June 28 Shift 2)

Question: Which of the following statements is correct regarding diffraction and interference patterns?

(a) In interference, all maxima have different intensities.

(b) In diffraction, the central maximum is twice as wide as other maxima.

(c) In diffraction, all minima are perfectly zero intensity.

(d) Interference requires only one source.

Solution:

In diffraction from a single slit, the width of the central maximum is \(2\lambda D / a\), while the width of secondary maxima is \(\lambda D / a\). Thus, the central maximum is exactly twice as wide.

\text{Statement (b) is correct}

Answer: (b) In diffraction, the central maximum is twice as wide as other maxima.

18

Important Results & Formulae Summary

A consolidated list of important results and properties that frequently appear in JEE problems.

Most Repeated PYQ — Key Formulae

1. Phase Difference: \(\Delta \phi = \frac{2\pi}{\lambda} \Delta x\)

2. Resultant Intensity: \(I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos(\Delta \phi)\)

3. Max/Min Intensity Ratio: \(\frac{I_{max}}{I_{min}} = \left(\frac{\sqrt{I_1}+\sqrt{I_2}}{\sqrt{I_1}-\sqrt{I_2}}\right)^2\)

4. YDSE Path Diff: \(\Delta x = \frac{yd}{D} = d \sin \theta\)

5. YDSE Fringe Width: \(\beta = \frac{\lambda D}{d}\)

6. YDSE Angular Fringe Width: \(\theta = \frac{\lambda}{d}\)

7. YDSE in Liquid: \(\beta' = \frac{\beta}{\mu}\)

8. YDSE Slab Shift: \(y_0 = \frac{D(\mu-1)t}{d} = \frac{(\mu-1)t}{\lambda} \beta\)

9. Diffraction Minima: \(a \sin \theta = n\lambda\)

10. Central Max Width: \(Y = \frac{2\lambda D}{a}\)

11. Telescope Resolving Power: \(R.P. = \frac{a}{1.22\lambda}\)

12. Microscope Limit: \(d = \frac{1.22\lambda}{2\mu \sin \theta}\)

13. Malus' Law: \(I = I_0 \cos^2 \theta\)

14. Brewster's Law: \(\mu = \tan i_p\)

PYQ — JEE Main 2023 (Apr 10 Shift 1)

Question: Two polaroids are kept crossed to each other. If the angle of one polaroid is rotated by \(30^\circ\) from the crossed position, the fraction of unpolarized incident light that passes through the system is:

(a) 1/8

(b) 1/4

(c) 3/8

(d) 1/2

Solution:

Let unpolarized intensity be \(I_0\).

After first polaroid: \(I_1 = I_0 / 2\).

Initially they are crossed (\(90^\circ\)). If one is rotated by \(30^\circ\), the angle between them becomes \(90^\circ - 30^\circ = 60^\circ\).

I_2 = I_1 \cos^2(60^\circ) = \frac{I_0}{2} \times \left(\frac{1}{2}\right)^2 = \frac{I_0}{2} \times \frac{1}{4} = \frac{I_0}{8}

Fraction passing = \(I_2 / I_0 = 1/8\).

\text{Fraction} = \frac{1}{8}

Answer: (a) 1/8

Previous Year Questions Analysis

JEE Main (Last 10 Years)

✓ YDSE Fringe Width & Modifications: 25%
✓ YDSE with Glass Slab: 18%
✓ Intensity Distribution (Ratio): 15%
✓ Single Slit Diffraction: 12%
✓ Huygens' Principle & Wavefronts: 10%
✓ Resolving Power: 10%
✓ Polarization (Malus/Brewster): 10%

JEE Advanced (Last 10 Years)

✓ Multi-concept YDSE + Slab: 30%
✓ Diffraction combined with Interference: 25%
✓ Intensity Distributions with Phase: 20%
✓ Polarization with Multiple Polaroids: 15%
✓ Resolving Power limits: 10%

Weightage Analysis

JEE Main: 4-8 marks (1-2 questions)

JEE Advanced: 4-6 marks (1-2 questions)

Difficulty Level: Easy to Moderate

Time Required: 4-5 hours practice

Download Complete PYQ PDF (2015-2024)

Practice Problem Set

Level 1: Basic (JEE Main Standard)

Find the fringe width in YDSE if \(\lambda = 600\) nm, \(D = 1\) m, and \(d = 0.5\) mm.
What is the phase difference corresponding to a path difference of \( \lambda/4 \)?
Two coherent sources have intensity ratio 25:9. Find \( I_{max} / I_{min} \).
What is the angular width of the central maximum in single slit diffraction if \( a = 2 \lambda \)?
Find Brewster's angle for a glass of refractive index 1.5.
Unpolarized light of intensity \(32\) W/m\(^2\) passes through a polaroid. What is the transmitted intensity?

Level 2: Intermediate (JEE Main/Advanced)

In YDSE, a slab of \(\mu = 1.5\) and thickness \(t\) is introduced. If the pattern shifts by 5 fringes (\(\lambda = 500\) nm), find \(t\).
The entire YDSE setup is immersed in water (\(\mu = 4/3\)). If initial 10 fringes fit in a region, how many will fit now?
Three polaroids are arranged. The first and third are crossed. The second is at \(45^\circ\). If \(I_0\) is incident, find the final intensity.
In single slit diffraction, the first minimum for light of \(\lambda = 500\) nm coincides with the first maximum for another \(\lambda'\). Find \(\lambda'\).
Find the resolving power of a telescope with aperture 5 cm using light of wavelength 550 nm.
The ratio of slit widths in YDSE is 4:1. Find the ratio of maximum to minimum intensity.

Level 3: Advanced (JEE Advanced/Olympiad)

Derive the intensity distribution formula for single slit diffraction.
A double slit is illuminated by light of \(\lambda = 480\) nm. Slit separation is 0.2 mm. A slab of thickness \(1.44 \times 10^{-4}\) m is placed in front of one slit. Find the position of the central maximum.
Show that the secondary maxima in single slit diffraction do not exactly lie at \((2n+1)\lambda/2\).
An unpolarized beam is incident on a glass slab at Brewster's angle. Find the ratio of transmitted to reflected intensity if the glass has \(\mu = 1.5\).
In YDSE, if the source is moved slightly perpendicular to the line joining the slits, how does the fringe pattern shift? Calculate the shift.

Get Solutions PDF with Step-by-Step Working

> FAQ

Frequently Asked Questions

What is Huygens' Principle?

Huygens' principle states that every point on a given wavefront acts as a fresh source of new secondary wavelets, which travel in all directions with the speed of light in that medium. The forward envelope of these secondary wavelets gives the new position of the wavefront.

What is the formula for fringe width in YDSE?

The fringe width (β) in Young's Double Slit Experiment is given by β = λD/d, where λ is the wavelength of light, D is the distance from the slits to the screen, and d is the separation between the two slits.

What are coherent sources?

Coherent sources are sources of light that emit light waves of the same frequency (or wavelength) and have a constant phase difference between them. They are essential to produce a sustained interference pattern.

What is the difference between interference and diffraction?

Interference is the superposition of waves coming from two distinct coherent sources, resulting in uniform fringe widths. Diffraction is the superposition of waves coming from different parts of the same wavefront, resulting in non-uniform fringe widths where the central maximum is twice as wide as the others.

What is the resolving power of an optical instrument?

Resolving power is the ability of an optical instrument to distinguish between two closely spaced objects. According to Rayleigh's criterion, two objects are just resolved when the central maximum of one falls on the first minimum of the other.

How many questions come from Wave Optics in JEE Mains?

On average, 1-2 questions (4-8 marks) appear from Wave Optics in every JEE Main session. Most repeated topics are YDSE (fringe width, intensity, slab introduction), Huygens' principle, and single slit diffraction.

What happens to fringe width if YDSE is performed in water?

If YDSE is performed in a medium of refractive index μ, the wavelength of light decreases (λ' = λ/μ). Consequently, the fringe width decreases and becomes β' = β/μ.

Wave Optics notes, formulas & PYQs for JEE 2026

📑 Table of Contents

Introduction & Wavefronts

1.1 What is a Wavefront?

1.2 Types of Wavefronts

🔥 Most Repeated PYQ — Types of Wavefronts

1.3 Relation between Phase Difference and Path Difference

📝 PYQ — JEE Main 2024 (Jan 31 Shift 2)

Huygens' Principle

2.1 Postulates of Huygens' Principle

🔥 Most Repeated PYQ — Huygens' Postulates

2.2 Construction of Wavefronts

💡 Key Concept

2.3 Validity and Limitations

📝 PYQ — JEE Main 2023 (Apr 13 Shift 1)

Laws of Reflection & Refraction by Huygens

3.1 Law of Reflection

🔥 Most Repeated PYQ — Reflection Proof

3.2 Law of Refraction (Snell's Law)

🔥 Most Repeated PYQ — Snell's Law Proof

📝 PYQ — JEE Main 2022 (June 28 Shift 1)

Principle of Superposition

4.1 Mathematical Expression

4.2 Intensity of Resultant Wave

🔥 Most Repeated PYQ — Resultant Intensity

📝 PYQ — JEE Main 2024 (Apr 5 Shift 1)

Interference of Light

5.1 Constructive Interference (Bright Fringe)

🔥 Most Repeated PYQ — Bright Fringe Condition

5.2 Destructive Interference (Dark Fringe)

🔥 Most Repeated PYQ — Dark Fringe Condition

5.3 Law of Conservation of Energy

📝 PYQ — JEE Main 2023 (Jan 31 Shift 1)

Coherent Sources

6.1 Conditions for Sustained Interference

🔥 Most Repeated PYQ — Conditions

6.2 Methods of Producing Coherent Sources

📝 PYQ — JEE Main 2021 (Aug 27 Shift 1)

Young's Double Slit Experiment (YDSE)

7.1 Experimental Setup

7.2 Path Difference in YDSE

🔥 Most Repeated PYQ — Path Difference

7.3 Positions of Bright and Dark Fringes

📝 PYQ — JEE Main 2024 (Jan 30 Shift 2)

Fringe Width in YDSE

8.1 Fringe Width Formula

🔥 Most Repeated PYQ — Fringe Width

8.2 Angular Fringe Width

8.3 Maximum Number of Fringes

💡 Key Concept

📝 PYQ — JEE Main 2022 (July 25 Shift 2)

Shape of Interference Fringes

9.1 Actual Shape (Hyperbola)

🔥 Most Repeated PYQ — Hyperbolic Fringes

9.2 Shape on the Screen

📝 PYQ — JEE Main 2023 (Apr 12 Shift 1)

Intensity Distribution in YDSE

10.1 Intensity Variation with Phase/Path Difference

🔥 Most Repeated PYQ — Intensity Formula

10.2 Ratio of Slit Widths

📝 PYQ — JEE Main 2024 (Apr 6 Shift 2)

YDSE in a Liquid Medium

11.1 Effect on Wavelength and Fringe Width

🔥 Most Repeated PYQ — YDSE in Water

11.2 Key Observations

💡 Important Points

📝 PYQ — JEE Main 2024 (Jan 29 Shift 2)

YDSE with Transparent Slab

12.1 Path Difference with Slab

🔥 Most Repeated PYQ — Slab Path Difference

12.2 Fringe Shift

🔥 Most Repeated PYQ — Fringe Shift Formula

📝 PYQ — JEE Main 2024 (Apr 8 Shift 2)

Diffraction of Light (Single Slit)

13.1 Condition for Minima (Dark Fringes)

🔥 Most Repeated PYQ — Diffraction Minima

13.2 Condition for Secondary Maxima (Bright Fringes)

📝 PYQ — JEE Main 2023 (Apr 11 Shift 1)

Width of Central Maximum

14.1 Angular and Linear Width

Table of Contents

Most Repeated PYQ — Types of Wavefronts

PYQ — JEE Main 2024 (Jan 31 Shift 2)

Most Repeated PYQ — Huygens' Postulates

Key Concept

PYQ — JEE Main 2023 (Apr 13 Shift 1)

Most Repeated PYQ — Reflection Proof

Most Repeated PYQ — Snell's Law Proof

PYQ — JEE Main 2022 (June 28 Shift 1)

Most Repeated PYQ — Resultant Intensity

PYQ — JEE Main 2024 (Apr 5 Shift 1)

Most Repeated PYQ — Bright Fringe Condition

Most Repeated PYQ — Dark Fringe Condition

PYQ — JEE Main 2023 (Jan 31 Shift 1)

Most Repeated PYQ — Conditions

PYQ — JEE Main 2021 (Aug 27 Shift 1)

Most Repeated PYQ — Path Difference

PYQ — JEE Main 2024 (Jan 30 Shift 2)

Most Repeated PYQ — Fringe Width

Key Concept

PYQ — JEE Main 2022 (July 25 Shift 2)

Most Repeated PYQ — Hyperbolic Fringes

PYQ — JEE Main 2023 (Apr 12 Shift 1)

Most Repeated PYQ — Intensity Formula

PYQ — JEE Main 2024 (Apr 6 Shift 2)

Most Repeated PYQ — YDSE in Water

Important Points

PYQ — JEE Main 2024 (Jan 29 Shift 2)

Most Repeated PYQ — Slab Path Difference

Most Repeated PYQ — Fringe Shift Formula

PYQ — JEE Main 2024 (Apr 8 Shift 2)

Most Repeated PYQ — Diffraction Minima

PYQ — JEE Main 2023 (Apr 11 Shift 1)

Most Repeated PYQ — Central Max Width

Common Misconception

PYQ — JEE Main 2021 (Jul 27 Shift 1)

Most Repeated PYQ — Rayleigh Criterion

PYQ — JEE Main 2020 (Sep 5 Shift 1)

Most Repeated PYQ — Malus' Law

Most Repeated PYQ — Brewster's Angle

PYQ — JEE Main 2024 (Jan 27 Shift 1)

PYQ — JEE Main 2022 (June 28 Shift 2)

Most Repeated PYQ — Key Formulae

PYQ — JEE Main 2023 (Apr 10 Shift 1)

Previous Year Questions Analysis

Practice Problem Set

Why This Chapter is Important

Must-Remember Formulas & Values

5-Step Exam Strategy