Wave Optics | ISC Class 12 Physics Notes

This note covers wavefronts, Huygens’ principle, wave explanations of reflection and refraction, phase and path differences, coherent sources, interference, Young’s double-slit experiment, fringe width, single-slit Fraunhofer diffraction, intensity patterns and the width of the central maximum.

What are phase, wavefronts and rays?

Phase specifies the stage of an oscillation, or repeating variation, at a particular position and time. Points oscillating in phase reach corresponding stages together. A wavefront is a surface of constant phase: all its points oscillate in phase at a given instant.

A wavelength, represented by λ, is the distance between successive points in the same phase along the direction of propagation. The frequency, represented by ν, is the number of complete oscillations per second. The wave speed v is the distance travelled by the disturbance per unit time.

v = νλ

The SI unit of wavelength is the metre, symbol m. The SI unit of frequency is the hertz, symbol Hz. The SI unit of wave speed is the metre per second, written m s⁻¹. SI means the International System of Units.

How does the source determine the wavefront?

A ray indicates the direction of energy propagation. In a medium where the speed of light is independent of direction, rays are perpendicular to the wavefront. The ray and the wavefront therefore describe different geometrical features of the same wave.

Part (a) shows concentric circular sections of spherical wavefronts with outward arrows. Part (b) shows parallel plane surfaces and arrows indicating propagation approximately perpendicular to them.

See Fig. 10.1 in your NCERT textbook

Geometrical optics neglects the finite wavelength of light. Its straight-ray description is an approximation when the wavelength is very small compared with relevant obstacle or aperture dimensions. An aperture is an opening through which light passes. Wave optics retains wavelength and explains interference and diffraction.

How does Huygens’ principle construct a new wavefront?

Definition: Huygens’ principle treats each point of a wavefront as a source of secondary wavelets. The forward envelope of these wavelets gives the position of the wavefront at a later instant.

A secondary wavelet is the disturbance spreading from one point of the original wavefront. An envelope is a surface tangent to the wavelets. Where wave speed is independent of direction, each secondary wavelet is spherical.

What steps are used in the construction?

1. Mark the known wavefront at the initial instant and choose several points on it as secondary sources.
2. Choose an elapsed time τ, the Greek letter tau, during which the disturbance advances.
3. Draw a wavelet from each point with radius vτ, where v is the wave speed in the medium.
4. Draw the common forward tangent or envelope. It represents the new wavefront after time τ.

Writing R for the secondary wavelet radius gives the propagation relation below. The SI unit of elapsed time is the second, symbol s. The SI unit of wavelet radius is the metre.

R = vτ

For an initially spherical wavefront in a uniform medium, the forward envelope is another sphere with the same centre. For an initially plane wavefront, the envelope is another plane parallel to the first. A two-dimensional drawing represents spherical wavelets by circular arcs.

Two vertical lines represent the initial and later plane wavefronts. Small curved wavelets are drawn from points on the first line; horizontal arrows connect corresponding positions on the two lines.

See Fig. 10.3 in your NCERT textbook

What is the limitation of the elementary construction?

The geometrical construction also suggests a backward envelope. Huygens assumed that wavelet amplitude was greatest forwards and zero backwards. Amplitude means the maximum magnitude of the oscillating disturbance. This assumption is not satisfactory by itself; a more rigorous wave theory justifies the absence of the backwave.

The construction is particularly useful at a boundary. Different portions of an incident wavefront reach the boundary at different times. Following their wavelets for the same elapsed time establishes the orientation of the reflected or refracted wavefront.

How does Huygens’ principle prove the laws of reflection?

Reflection is the return of light into the medium from which it strikes a surface. The normal is a line perpendicular to that surface at the point considered. Let i be the angle of incidence and r the angle of reflection, both measured from the normal.

Consider a plane reflecting surface MN and an incident plane wavefront AB. These pairs of capital letters label the surface and wavefront in the construction. Point A reaches the reflector first. While point B advances to point C on the reflector, a reflected secondary wavelet spreads from A.

Derivation: The reflection relation

1. Let τ be the time taken for B to reach C. Since the wave speed in the incident medium is v, the distance BC equals vτ.
2. Draw a reflected secondary wavelet centred at A with radius AE equal to vτ. E labels its point of contact with the tangent drawn from C.
3. The tangent CE is the reflected wavefront. AE is perpendicular to CE, and BC is perpendicular to the incident wavefront AB.
4. Right triangles AEC and CBA share the hypotenuse AC, the side opposite each right angle, and have AE = BC. They are congruent, meaning identical in size and shape, so the corresponding wavefront angles, and hence the ray angles, are equal.

i = r

The incident ray, reflected ray and normal lie in the same plane. This coplanarity, together with equality of the two angles, states the laws of reflection. The proof uses equal propagation speeds before and after reflection because both paths are in the same medium.

The horizontal reflector is labelled MN. AB is the sloping incident wavefront and CE is the reflected wavefront. The construction joins A to E and B to C and marks the angles i and r.

See Fig. 10.6 in your NCERT textbook

Which geometrical distinction matters?

The angle between a ray and the normal equals the corresponding angle between its wavefront and the reflecting surface. This follows because ray and wavefront are perpendicular, as are normal and surface. Mixing a ray-to-surface angle with a ray-to-normal angle gives an incorrect comparison.

The incident and reflected waves have the same frequency. Atoms driven by the incident light oscillate at its frequency and produce the reflected disturbance at that frequency.

How does Huygens’ principle explain refraction?

Refraction changes the direction of propagation when a wave crosses obliquely between media with different wave speeds. Let v₁ and v₂ be the speeds in the first and second media. Here i is the angle of incidence and r is the angle of refraction, measured from the boundary normal.

Derivation: Snell’s law from wavefronts

1. Let AB be an incident plane wavefront. A reaches the interface first; in time τ, B travels to C along the boundary-reaching ray, giving BC = v₁τ.
2. In the same time, the secondary wavelet from A in the second medium has radius AE = v₂τ. Draw the tangent CE from C to form the refracted wavefront.
3. The right triangles give sin i = BC/AC and sin r = AE/AC. The function sin denotes the sine of an angle.
4. Divide these relations to obtain sin i/sin r = v₁/v₂. Define the absolute refractive indices n₁ = c/v₁ and n₂ = c/v₂, where c is the speed of light in vacuum.

n₁ sin i = n₂ sin r

Refractive index is a speed ratio and has no unit. When v₂ is smaller than v₁, the ray bends towards the normal at oblique incidence. When v₂ is greater, it bends away. The incident ray, refracted ray and normal lie in the same plane.

The boundary separates medium 1 above from medium 2 below. AB is the incident wavefront; CE is the refracted wavefront. The drawn second-medium wavelet has radius v₂τ, smaller than the first-medium travel distance v₁τ.

See Fig. 10.4 in your NCERT textbook

Let λ₁ and λ₂ be the wavelengths in the two media. Frequency remains the same on refraction, so both speed and wavelength change in the same ratio.

v₁/v₂ = λ₁/λ₂

How are speed, frequency and wavelength calculated?

Worked example 1. Light of wavelength 589 nm in air reflects from water. Take the air speed as c = 3.0 × 10⁸ m s⁻¹. Find the reflected frequency and wavelength; nm means nanometre, equal to 10⁻⁹ m.

Formula: ν = c/λ; λ = c/ν. Substitute: ν = (3.0 × 10⁸)/(589 × 10⁻⁹). Answer: ν ≈ 5.09 × 10¹⁴ Hz; reflected wavelength = 589 nm = 0.000000589 m, because reflection leaves the light in air.

Worked example 2. The same incident wavelength, 589 nm in air, enters water of refractive index n = 1.33. Take c = 3.0 × 10⁸ m s⁻¹. Find the water speed and wavelength.

Formula: v = c/n; λ₂ = λ₁/n, taking the air index as unity. Substitute: v = (3.0 × 10⁸)/1.33; λ₂ = 589/1.33 nm. Answer: v ≈ 2.26 × 10⁸ m s⁻¹ and λ₂ ≈ 443 nm = 0.000000443 m; frequency remains approximately 5.09 × 10¹⁴ Hz.

Worked example 3. Glass has refractive index n = 1.5. Calculate its light speed using c = 3.0 × 10⁸ m s⁻¹.

Formula: v = c/n. Substitute: v = (3.0 × 10⁸)/1.5. Answer: v = 2.0 × 10⁸ m s⁻¹, or 200000000 m s⁻¹. The refractive index divides the vacuum speed; it does not multiply it.

What makes two sources coherent and interference sustained?

Superposition means that the resultant disturbance at a point is the vector sum of the separate disturbances, adding them with their directions taken into account. Interference is the resulting redistribution of intensity when waves overlap. Intensity is the energy crossing unit area per unit time.

Coherent sources have the same frequency and maintain a constant phase difference. Constant does not necessarily mean zero. Incoherent sources have no stable phase relationship. A steady interference pattern requires the relative phase at each observation point to remain constant during observation.

Why are two ordinary lamps unsuitable?

Light from an ordinary source undergoes abrupt phase changes. Two independent sodium lamps do not maintain a fixed phase relationship. Their bright and dark interference positions change so rapidly that observation gives a time-averaged intensity, meaning an average over many rapid changes.

If each lamp separately produces intensity I₀ at the point, the combined time-averaged intensity is I. The subscript zero in I₀ labels the intensity of one contribution.

I = 2I₀

Young’s arrangement derives both interfering contributions from a single original source. A phase change in that source is shared by both contributions, so their relative phase remains fixed. Splitting a common wavefront therefore provides the required coherence.

Which conditions serve different purposes?

Monochromatic light has a single frequency. It gives one wavelength in a specified medium. Coherence fixes the pattern in time; equal or nearly equal amplitudes give strong contrast between bright and dark bands. Equal amplitudes are required for complete cancellation in the ideal two-wave case.

For the simple interference calculations below, the two disturbances have matching vibration directions, equal amplitudes at the screen and a fixed source phase difference of zero. These assumptions make the bright and dark conditions unambiguous; equal frequency alone does not ensure a steady pattern.

How do path and phase differences produce bright and dark fringes?

The path difference, written Δx, is the difference between the distances travelled by two waves to the same observation point. The phase difference, written φ, compares their stages of oscillation there. For sources initially in phase and paths through the same uniform medium, the two are related.

φ = 2πΔx/λ

The constant π is the circle constant. Phase is expressed in radians, the angular measure in which one complete cycle is 2π. A path difference of one wavelength therefore corresponds to one whole cycle; half a wavelength corresponds to half a cycle.

When do the waves reinforce or cancel?

Constructive interference occurs when the waves arrive in phase and reinforce each other. Destructive interference occurs when they arrive in opposite phase. For equal amplitudes, opposite disturbances cancel completely. Let m be an integer, meaning zero or a positive or negative whole number, specifying the interference order.

Intensity is proportional to the square of wave amplitude. Reinforcement doubles the amplitude of equal contributions, producing four times the intensity of either contribution alone. Destructive interference reduces the resultant amplitude to zero under the equal-amplitude assumption.

How does intensity vary between the extremes?

For two coherent contributions of equal individual intensity I₀, the resultant intensity I depends on φ through the following expression. The function cos denotes the cosine of an angle.

I = 4I₀ cos²(φ/2)

At intermediate phase differences, intensity lies between the two extreme values. If φ varies rapidly and irregularly, averaging the pattern gives 2I₀ instead. This is why adding intensities directly describes incoherent illumination but misses a stationary interference pattern.

Note: A dark fringe does not imply destruction of light energy. In interference and diffraction, intensity decreases in some regions and increases in others. Energy is redistributed, consistently with conservation of energy.

A path difference of 2λ gives phase difference 4π and reinforcement. A path difference of magnitude 2.5λ gives an odd half-cycle difference and cancellation for equal amplitudes. Whole cycles restore the same phase; an added half-cycle reverses it.

How is Young’s double-slit experiment arranged?

In Young’s double-slit experiment, a narrow source opening S illuminates two nearby openings S₁ and S₂. A slit is a narrow, elongated opening. These capital letters label the original source and the two secondary sources. The secondary waves overlap on a screen and produce alternating bright and dark bands called fringes.

Let d be the separation of the two slits and D their distance from the screen. Let O be the central screen point equidistant from both slits, and P another observation point. The displacement OP along the screen is y. The observation angle from the central axis is θ.

With S placed symmetrically relative to the slits, S₁ and S₂ are illuminated in phase. At O their distances are equal, so the path difference is zero and a central bright fringe forms. At other points, one wave generally travels farther than the other.

Part (a) shows a source opening, a pair of openings and a screen. Part (b) joins S₁ and S₂ to point P on screen GG′, with O on the central axis; d and D mark slit separation and screen distance. The figure labels the screen displacement x, corresponding to y here.

See Fig. 10.12 in your NCERT textbook

How is the path difference obtained geometrically?

1. Take the screen sufficiently far from the two slits that the paths from S₁ and S₂ towards P can be treated as parallel.
2. Draw a perpendicular from one slit to the path from the other. The short extra segment on the longer path represents the path difference.
3. Project the slit separation d onto the propagation direction. The extra path has magnitude d sin θ.
4. For small θ near the centre, sin θ ≈ tan θ ≈ y/D. The function tan denotes the tangent of an angle; hence the path difference is approximately dy/D.

Δx = d sin θ

Δx = dy/D

The second expression uses the small-angle approximation, while the first uses the far-screen, nearly parallel-ray geometry. Here D is much greater than d, and the relevant screen displacement is small compared with D. These conditions accompany the familiar fringe-position formulae.

The symbol Δx describes path difference, not a screen coordinate. Keeping it separate from y prevents confusing the distance travelled by a wave with the position at which the two waves meet.

How are fringe positions and fringe width calculated?

The fringe width β, the Greek letter beta, is the distance between successive bright fringes or successive dark fringes. It is not the distance from one bright fringe to the next dark fringe. The SI unit of fringe width is the metre.

Derivation: Fringe width in Young’s experiment

1. Use the small-angle path difference Δx = dy/D. For a bright fringe of integer order m, substitute the constructive condition Δx = mλ.
2. Solving gives yₘ = mλD/d, where yₘ is the position of the mth bright fringe relative to the central bright fringe.
3. The next bright fringe is at yₘ₊₁ = (m + 1)λD/d. Subtract yₘ from this position.
4. The difference is β = λD/d, independent of m. Substituting the destructive condition similarly gives dark fringes at y = (m + ½)λD/d.

β = λD/d

Thus, the ideal bright fringes are equally spaced near the centre. The dark fringes have the same spacing and lie halfway between adjacent bright fringes. At fixed d and D, a longer wavelength gives a larger fringe width.

For a fixed wavelength, moving the screen farther away increases β. Increasing slit separation reduces β. These comparisons assume that the coherence and small-angle conditions continue to hold.

Worked example 4. Slit separation is 0.28 mm, screen distance is 1.4 m and the fourth bright fringe is 1.2 cm from the central bright fringe. Find the wavelength. Here mm means millimetre and cm means centimetre.

Formula: λ = yₘd/(mD). Substitute: λ = (1.2 × 10⁻² × 0.28 × 10⁻³)/(4 × 1.4) m. Answer: λ = 6.0 × 10⁻⁷ m, or 0.000600 mm (600 nm). The central fringe has order zero, so the fourth bright fringe has m = 4.

Worked example 5. For slit separation d = 0.025 mm, screen distance D = 5 cm and wavelength λ = 5 × 10⁻⁵ cm, calculate the fringe width.

Formula: β = λD/d. Substitute: β = [(5 × 10⁻⁷)(5 × 10⁻²)]/(2.5 × 10⁻⁵) m. Answer: β = 1.0 × 10⁻³ m, or 1 mm. All three lengths were converted to metres before substitution.

Worked example 6. Use d = 0.025 mm, D = 5 cm and λ = 5 × 10⁻⁵ cm to find the distance from the central bright fringe to the nearest dark fringe.

Formula: fringe width = λD/d; y = (fringe width)/2. Substitute: β = 1.0 × 10⁻³ m; y = (1.0 × 10⁻³)/2 m. Answer: y = 5.0 × 10⁻⁴ m, or 0.5 mm, on either side of the centre.

What does the intensity graph show?

Label the horizontal axis θ and the vertical axis I. Draw equally spaced maxima of height 4I₀, separated by zero-intensity minima, in the small-angle region. Put a maximum at θ = 0 and successive maxima at θ ≈ mλ/d.

This ideal graph assumes equal amplitudes without appreciable variation across the region shown. Real slits have finite width, so their individual diffraction patterns modify the heights of interference fringes.

How does a single slit produce Fraunhofer diffraction?

Diffraction is the spreading of waves through openings or around obstacles into regions where geometrical optics would predict a shadow. It is exhibited by light, sound, water and matter waves. Light’s wavelength is much smaller than most everyday obstacles, so its diffraction is not usually apparent in everyday observations.

Fraunhofer diffraction describes the far-field pattern, where incoming illumination and the rays associated with one observation direction may be treated as parallel. A converging lens can bring rays with the same direction to a point in its focal plane, the plane containing the focused image of a distant source. The pattern can be observed there.

What is the arrangement and observed pattern?

A parallel monochromatic beam falls normally on a narrow slit of width a. Different parts of the wavefront within the slit act as secondary sources. For a point on the central axis, their paths are equal and their contributions arrive in phase.

The pattern has a broad central maximum, the central region of greatest intensity. On either side lie dark minima and weaker bright secondary maxima. The secondary maxima become weaker farther from the centre. A minimum is an intensity low point; a maximum is an intensity high point.

Figure 10.14 labels the slit ends L and N and midpoint M, with rays directed towards P at angle θ. Figure 10.15 shows incoming light, the slit, a viewing screen and a fringe photograph with a broad central bright band and weaker side bands.

See Figs. 10.14 and 10.15 in your NCERT textbook

How are the dark directions obtained?

1. For an observation direction θ, measured from the slit normal, the path difference between contributions from the two edges is a sin θ.
2. If a sin θ = λ, divide the slit into two equal halves. Corresponding points in the two halves have path difference λ/2.
3. Each such pair contributes equal disturbances in opposite phase under uniform illumination, meaning equal amplitude across the slit, producing cancellation and the first minimum.
4. For a sin θ = mλ, divide the slit into 2m equal strips and pair neighbouring strips. This gives further minima for positive integers m.

a sin θₘ = mλ

Here θₘ is the angle of the mth minimum on one side, with m = 1, 2, 3 and so on; matching minima occur on the other side. The zero order is excluded because θ = 0 is the central maximum, not a minimum.

For small angles measured in radians, θₘ ≈ mλ/a. If the screen is a distance D away in the far field, the corresponding screen positions are approximately yₘ = mλD/a. Here yₘ now labels diffraction minima rather than interference maxima.

What determines the central maximum and secondary maxima?

The first diffraction minima occur on opposite sides of the centre at sin θ₁ = λ/a. The central maximum lies between them. Its angular width is the full angular separation of those two first minima, rather than the centre-to-first-minimum angle.

Let Wθ denote this full angular width in radians. For λ/a small enough to use the small-angle approximation, the width is twice λ/a.

Wθ = 2λ/a

Let W be the full linear width on a distant screen at distance D. If a lens forms the pattern in its focal plane, let f be the focal length, meaning the distance from the lens to that plane for incoming parallel rays.

W = 2Dλ/a

For the focal-plane arrangement, replace D by f to obtain W = 2fλ/a. Both linear-width expressions use the same small-angle approximation. Decreasing slit width broadens the central maximum; increasing wavelength also broadens it.

Where are the weaker maxima?

Secondary maxima lie between successive dark minima. Their approximate positions satisfy a sin θ ≈ (m + ½)λ for m = 1, 2, 3 and so on on either side. This gives an elementary estimate, not an exact condition for peak intensity.

Label angle θ horizontally and intensity I vertically. Draw the highest peak at zero, first zeros at approximately ±λ/a, and further zeros at ±2λ/a and ±3λ/a. Draw progressively weaker side peaks between the zeros.

The central bright region is twice the width of the regions between successive minima on either side in the small-angle pattern. All these widths refer to intervals bounded by dark minima, not distances between intensity peak positions.

How do the two patterns compare?

Both effects involve superposition. The distinction describes the arrangement of contributing sources, rather than a separate physical principle. In a real double-slit experiment, single-slit diffraction from each opening and interference between the two openings appear together.

Glossary

• Phase — The stage reached by an oscillation at a specified position and time.
• Wavefront — A surface whose points are at the same phase at a given instant.
• Ray — A line indicating energy propagation, perpendicular to the wavefront when wave speed is independent of direction.
• Secondary wavelet — A disturbance spreading from an individual point on a wavefront in Huygens’ construction.
• Envelope — The common tangent surface to secondary wavelets, giving the new forward wavefront.
• Coherent sources — Sources with equal frequency and a phase difference that remains constant during observation.
• Monochromatic light — Light of a single frequency, corresponding to one wavelength in a specified medium.
• Path difference — The difference between the distances travelled by two waves reaching the same observation point.
• Constructive interference — Reinforcement when disturbances arrive in phase, producing an intensity maximum.
• Destructive interference — Cancellation when disturbances arrive in opposite phase, complete for equal amplitudes.
• Fringe width — The distance between adjacent bright fringes or adjacent dark fringes in an interference pattern.
• Diffraction — The spreading of waves through openings or around obstacles beyond geometrical shadow boundaries.
• Fraunhofer diffraction — Far-field diffraction described using parallel incident rays and parallel outgoing rays for each observation direction.
• Central maximum — The central bright region of the single-slit diffraction pattern, bounded by its first minima.

Common errors and misconceptions

• Misconception: A ray lies along a wavefront. Correct: The ray is perpendicular to the wavefront when speed is independent of propagation direction.
• Misconception: Frequency decreases when light enters water. Correct: Frequency remains unchanged; speed and wavelength decrease together.
• Misconception: Any two sources of the same colour are coherent. Correct: A constant phase difference is also necessary for sustained interference.
• Misconception: Complete darkness follows from opposite phase regardless of amplitude. Correct: Complete cancellation requires equal amplitudes as well as opposite phase.
• Misconception: Fringe width is the bright-to-nearest-dark distance. Correct: It is the bright-to-next-bright or dark-to-next-dark distance; the former distance is half as large.
• Misconception: Zero order satisfies the diffraction minimum condition. Correct: Zero angle gives the central maximum; minima start at the first nonzero order.
• Misconception: The central diffraction width is λD/a. Correct: That is its approximate half-width; the full width is 2λD/a.
• Misconception: Diffraction secondary maxima occur exactly at the halfway path-difference rule. Correct: The rule a sin θ ≈ (m + ½)λ estimates their positions.

Exam-style questions with model answers

Q1. Define a wavefront and state its relation to a ray when wave speed is independent of direction. [2 marks]

1. A wavefront is a surface joining points that have the same phase at a particular instant.
2. A ray indicates energy propagation and is perpendicular to the wavefront when wave speed is independent of direction.

Q2. Why can two slits illuminated symmetrically by one monochromatic source produce sustained interference, whereas two independent ordinary lamps do not? [3 marks]

1. The two illuminated slits derive their light from a common source, so phase changes in that source are shared by both contributions.
2. The slit contributions consequently have the same frequency and a stable relative phase, allowing stationary bright and dark fringes to form.
3. Independent lamps undergo unrelated phase changes. Their fringe positions vary rapidly, so observation gives the sum of their time-averaged intensities instead of a stable fringe pattern.

Q3. Use Huygens’ construction to derive Snell’s law for a plane wave crossing a plane boundary. Define the quantities used and state the coplanarity law. [5 marks]

1. Let AB be the incident wavefront, with A first reaching the boundary. Let i and r be the incidence and refraction angles to the normal, and v₁ and v₂ the speeds in the two media.
2. In elapsed time τ, B reaches boundary point C, so BC = v₁τ. The wavelet from A in the second medium has radius AE = v₂τ.
3. Draw the tangent CE to that wavelet. It is the refracted wavefront. Right-triangle geometry gives sin i = BC/AC and sin r = AE/AC.
4. Division gives sin i/sin r = v₁/v₂. With refractive indices n₁ = c/v₁ and n₂ = c/v₂, where c is vacuum light speed, this becomes n₁ sin i = n₂ sin r.
5. The incident ray, refracted ray and boundary normal lie in the same plane, completing the laws of refraction.

Q4. Light of wavelength 589 nm in air enters water of refractive index 1.33. Take the light speed in air as 3.0 × 10⁸ m s⁻¹ and the air index as unity. Calculate its frequency, speed in water and wavelength in water, stating what remains unchanged. [4 marks]

1. Use ν for frequency and λ₁ for the air wavelength. Since 589 nm = 589 × 10⁻⁹ m, ν = (3.0 × 10⁸)/(589 × 10⁻⁹) ≈ 5.09 × 10¹⁴ Hz.
2. With water index n = 1.33, its light speed is v = c/n = (3.0 × 10⁸)/1.33 ≈ 2.26 × 10⁸ m s⁻¹.
3. The wavelength in water is λ₂ = λ₁/n = 589/1.33 ≈ 443 nm.
4. Frequency remains unchanged on refraction. The reductions in speed and wavelength have the same ratio, so the relation v = νλ remains satisfied.

Q5. In Young’s experiment, two in-phase slits are separated by 0.28 mm and the screen is 1.4 m away. The fourth bright fringe lies 1.2 cm from the central bright fringe. Using the small-angle approximation, find the wavelength and fringe width. [3 marks]

1. Let d be slit separation, D screen distance and y₄ the fourth-fringe displacement. Convert the lengths: d = 0.28 × 10⁻³ m, D = 1.4 m and y₄ = 1.2 × 10⁻² m.
2. For bright order m = 4, λ = y₄d/(4D) = 6.0 × 10⁻⁷ m, or 600 nm.
3. The fringe width is β = y₄/4 = 3.0 × 10⁻³ m, or 3.0 mm. Four equal bright-fringe intervals separate the centre from the fourth bright fringe.

Q6. Derive the double-slit fringe-width expression for two coherent in-phase slits separated by d, with a screen distance D and wavelength λ. Assume D is much greater than d and screen displacements are small compared with D. [5 marks]

1. Let y be displacement from the central bright fringe and θ the angle to the central axis. Nearly parallel paths from the slits give path difference Δx = d sin θ.
2. For small angles, sin θ ≈ tan θ ≈ y/D. Therefore Δx ≈ dy/D in the region of the screen being considered.
3. At a bright fringe, Δx = mλ, where integer m is the fringe order. Combining the two relations gives the bright position yₘ = mλD/d.
4. The next bright fringe is at yₘ₊₁ = (m + 1)λD/d. Subtracting yₘ defines the fringe width β and gives β = λD/d.
5. The expression is independent of order, so successive bright fringes are equally spaced within these approximations. Applying the dark condition Δx = (m + ½)λ gives the same spacing for dark fringes.

Q7. A uniformly illuminated slit of width a receives normally incident monochromatic light of wavelength λ. Explain its first Fraunhofer minimum, obtain the small-angle full angular width of its central maximum, and describe the secondary peaks. Assume λ/a is small. [5 marks]

1. Let θ be the observation angle to the slit normal. Contributions from the two edges have path difference a sin θ.
2. At a sin θ = λ, divide the slit into equal halves. Corresponding points in the halves have path difference λ/2 and hence opposite phase.
3. The paired equal-amplitude contributions cancel. This gives a first dark minimum on each side at angles whose magnitudes satisfy sin θ₁ = λ/a.
4. For small angles in radians, θ₁ ≈ λ/a. The central maximum extends between the two first minima, so its full angular width is Wθ ≈ 2λ/a.
5. The side maxima are weaker and become weaker farther from the centre. Their positions are approximately a sin θ ≈ (m + ½)λ, with positive integer m, rather than exactly given by this estimate.

Q8. For two coherent, in-phase slits giving equal amplitudes at the screen, let the intensity at path difference λ be K units. Here λ is the wavelength. Find the intensity at path difference λ/3 using I = 4I₀ cos²(φ/2), where I₀ is either individual intensity and φ is the phase difference. [3 marks]

1. At path difference λ, the phase difference is 2π, where π is the circle constant. The waves reinforce, so the given maximum intensity is K = 4I₀.
2. At path difference λ/3, the phase difference is φ = 2π/3. The half-angle in the intensity expression is therefore π/3.
3. Since cos(π/3) = ½, the intensity is I = K(½)² = K/4 units. Thus this point has one quarter of the stated maximum intensity.

Key takeaways

• A wavefront joins points of equal phase; Huygens’ principle constructs its later position from the forward envelope of secondary wavelets.
• Equal propagation distances in the reflecting medium lead to equal incidence and reflection angles in Huygens’ construction.
• Refraction changes speed and wavelength together while preserving frequency; wavefront geometry gives the relation n₁ sin i = n₂ sin r.
• Coherence requires equal frequencies and a constant phase difference; equal amplitudes additionally allow complete destructive interference.
• Young’s experiment gives fringe width β = λD/d under the far-screen and small-angle assumptions used in its derivation.
• Single-slit diffraction minima obey a sin θₘ = mλ for nonzero orders; the central direction is a maximum.
• The central diffraction maximum has full angular width approximately 2λ/a, and the secondary maxima become weaker away from the centre.
• Interference and diffraction redistribute light energy into brighter and darker regions without violating conservation of energy.

Test yourself

What wavefront does a uniformly emitting point source produce?

It produces spherical wavefronts centred on the point source; a small distant portion can be approximated as plane.

What fixes a secondary wavelet’s radius after an elapsed time?

Its radius equals the wave speed in the medium multiplied by the elapsed time.

Does a coherent pair have to remain exactly in phase?

No. Coherence requires a constant phase difference, which need not be zero.

What phase difference corresponds to a path difference of half a wavelength for initially in-phase waves?

It corresponds to π radians, so the disturbances arrive in opposite phase.

What happens to Young’s fringe width if slit separation increases while other quantities remain fixed?

It decreases because fringe width is inversely proportional to slit separation under the stated approximations.

Why is the zero order excluded from the single-slit minimum condition?

At zero angle, the contributions arrive in phase and form the central bright maximum.

What happens to the central diffraction width when the slit becomes narrower?

The central maximum broadens because its angular width varies inversely with slit width.

Where does the energy missing from dark fringes appear?

It is redistributed into brighter regions of the pattern, so total energy is conserved.