Class 12Physics · OpticsFull chapter

Wave Optics

The whole chapter in one place — read it, then test yourself. Clear notes, formula sheet, a practice quiz, and worked NCERT solutions & PYQs.

Wavefronts and Huygens' Principle

Quick answer Wave optics treats light as a wave described by wavefronts; Huygens' Principle gives a geometric method to construct how a wavefront moves forward in time.

In wave optics, light is treated as a wave spreading out from its source. A wavefront is the locus of all points in a medium that are vibrating in the same phase at a given instant. The direction of propagation of light (a ray) at any point is always perpendicular to the wavefront at that point.

The shape of a wavefront depends on the source. A point source in a homogeneous medium produces spherical wavefronts, since every point at the same distance from the source has travelled the same optical path and is in the same phase. A line source produces cylindrical wavefronts. When the source is extremely far away, a small patch of the (nearly spherical) wavefront looks essentially flat, and we call it a plane wavefront — light reaching the Earth from a distant star is the standard example.

Huygens' Principle is a geometric construction used to find the position of a wavefront at a later instant from its position at an earlier instant. It has two parts: (a) every point on an existing wavefront becomes a source of secondary spherical wavelets, which spread out in the forward direction with the speed of light in that medium; (b) the new wavefront at a later time is the forward envelope, i.e. the common tangent surface, touching all these secondary wavelets.

Worked example (construction): A plane wavefront AB travels with speed v in a medium. To locate the wavefront after time τ, draw secondary wavelets of radius vτ centred on several points along AB. The new wavefront A′B′ is the plane tangent to all these wavelets, and simple geometry shows A′B′ is parallel to AB and displaced forward by exactly vτ — consistent with straight-line propagation of a plane wave at speed v. This same construction, applied to a curved (e.g. spherical) wavefront, correctly predicts that it keeps expanding while remaining centred on the source.

Huygens originally could not explain why the wavelets do not also produce a backward-travelling wave; this was later resolved by Fresnel using the idea that the amplitude of secondary wavelets is not the same in all directions (it is zero in the backward direction), a result that follows properly only from a more complete diffraction theory. For CBSE Class 12 purposes, it is enough to know that Huygens' Principle, despite this limitation, correctly predicts reflection, refraction and the general forward motion of light waves.

Speed of light in a medium v = c / n m s⁻¹ · c = speed of light in vacuum, n = (absolute) refractive index of the medium
Forward displacement of a wavefront distance moved = v × τ m · τ = time interval; secondary wavelets in Huygens' construction have this radius
Remember
  • A wavefront is a surface of constant phase; rays are always perpendicular to wavefronts.
  • Point sources give spherical wavefronts; line sources give cylindrical wavefronts; a very distant source gives (locally) plane wavefronts.
  • Huygens' Principle: every point on a wavefront is a source of secondary wavelets travelling forward at the wave speed of the medium.
  • The new wavefront is the forward tangential envelope of all the secondary wavelets.
  • Huygens' construction correctly predicts straight-line propagation, reflection and refraction of light.

Reflection and Refraction of Plane Waves (Huygens' Construction)

Quick answer Huygens' construction can be used to derive the laws of reflection and Snell's law of refraction from first principles, showing how speed, wavelength and frequency change (or don't) across a boundary.

Law of reflection. Let a plane wavefront AB be incident obliquely on a plane reflecting surface MN, touching it first at A. While the secondary wavelet from B travels a distance BC = v·t to reach the surface at C, the wavelet from A (which started earlier) has already spread out into the medium as a hemisphere of radius AE = v·t = BC. The reflected wavefront is the tangent EC. Since the right triangles EAC and BAC share the hypotenuse AC and have AE = BC, they are congruent, so the angle between the incident wavefront and AC equals the angle between AC and the reflected wavefront. Converting to rays (perpendicular to the wavefronts) gives the familiar law: angle of incidence = angle of reflection, and the incident ray, reflected ray and normal all lie in the same plane.

Law of refraction (Snell's law). Now let the wavefront cross from medium 1 (speed v₁) into medium 2 (speed v₂). In the same time t that the wavelet from B travels BC = v₁t to reach the second medium, the wavelet from A spreads into medium 2 with radius AE = v₂t. From the geometry, BC = AC sin i and AE = AC sin r, where i is the angle of incidence and r the angle of refraction. Hence sin i / sin r = v₁/v₂, a constant for the two media, called the refractive index of medium 2 with respect to medium 1, n₂₁. Since n = c/v for each medium, n₂₁ = v₁/v₂ = n₂/n₁. Because frequency depends only on the source and stays unchanged as the wave crosses the boundary, while speed changes, the wavelength must also change: v = fλ gives λ₁/λ₂ = v₁/v₂. This immediately explains why light slows down and its wavelength shrinks on entering a denser medium (bending towards the normal), while its frequency and colour remain the same.

Worked example: A ray of light travelling in air (n₁ = 1) strikes a glass surface (n₂ = 1.5) at an angle of incidence of 30°. Find the angle of refraction.
Using sin i / sin r = n₂/n₁: sin r = sin i × (n₁/n₂) = sin 30° × (1/1.5) = 0.5/1.5 = 0.333.
r = sin⁻¹(0.333) ≈ 19.5°. As expected, the ray bends towards the normal on entering the denser medium.

Law of reflection ∠i = ∠r Angle of incidence equals angle of reflection, measured from the normal
Snell's law (Huygens form) sin i / sin r = v₁ / v₂ = n₂₁ = n₂ / n₁ n₂₁ = refractive index of medium 2 w.r.t. medium 1
Wavelength ratio across a boundary λ₁ / λ₂ = v₁ / v₂ Frequency f is unchanged on refraction; only v and λ change together
Remember
  • Huygens' construction geometrically derives both the law of reflection and Snell's law of refraction.
  • Law of reflection: angle of incidence = angle of reflection; incident ray, reflected ray, normal are coplanar.
  • Snell's law from Huygens: sin i / sin r = v₁/v₂ = n₂₁ = n₂/n₁.
  • On refraction, frequency stays constant; speed and wavelength both change, and they change together (λ₁/λ₂ = v₁/v₂).
  • Light bends towards the normal when it slows down entering a denser (optically) medium.

Interference of Light: Coherent Sources and Young's Double-Slit Experiment

Quick answer When two coherent light waves overlap, they produce a stable pattern of bright and dark fringes; Young's double-slit experiment is the classic demonstration and gives a simple formula for fringe width.

By the principle of superposition, when two light waves overlap at a point, their displacements add. If the two waves maintain a constant phase difference at every point (i.e. they are coherent), the addition produces a stable pattern of alternately bright (constructive interference) and dark (destructive interference) regions — this phenomenon is called interference. Two independent sources are almost never coherent because their phases fluctuate randomly and independently; in practice coherent sources are made by splitting light from a single source into two parts, as Thomas Young did by illuminating two narrow, closely spaced slits with light from a single slit.

In Young's double-slit experiment (YDSE), two slits S₁ and S₂, separated by a small distance d, are illuminated by a common monochromatic source, and the resulting pattern is observed on a screen at distance D (with D >> d). At a point P on the screen at a small angle θ from the centre, the path difference between the two waves arriving at P is Δx = d sinθ ≈ d y/D, where y is the distance of P from the centre of the pattern (valid for small angles). Constructive interference (bright fringe) occurs when Δx is an integer multiple of the wavelength; destructive interference (dark fringe) occurs when Δx is an odd multiple of half the wavelength.

The spacing between successive bright (or dark) fringes, called the fringe width β, works out to β = λD/d, and is uniform across the pattern (for small angles). Using the phase difference φ = (2π/λ)Δx, the resultant intensity at any point is I = I₁ + I₂ + 2√(I₁I₂) cosφ. For two sources of equal intensity I₀ each, this simplifies to I = 4I₀cos²(φ/2), giving a maximum of 4I₀ and a minimum of zero. Notice that the average intensity over the whole pattern is still 2I₀ (= I₁+I₂), so energy is only redistributed by interference, not created or destroyed — energy conservation is not violated.

Worked example: In a YDSE set-up, d = 1 mm, D = 1 m and λ = 600 nm. Find the fringe width and the position of the 5th bright fringe.
β = λD/d = (600 × 10⁻⁹ m)(1 m) / (1 × 10⁻³ m) = 6 × 10⁻⁴ m = 0.6 mm.
Position of the 5th bright fringe: y₅ = nβ = 5 × 0.6 mm = 3 mm from the centre.

Path difference Δx = d sinθ ≈ d y / D m · d = slit separation, D = screen distance, y = distance from central fringe
Bright fringe (constructive interference) Δx = nλ, n = 0, ±1, ±2, …
Dark fringe (destructive interference) Δx = (n + ½)λ
Fringe width β = λD / d m · Uniform spacing between consecutive bright (or dark) fringes
Resultant intensity (general) I = I₁ + I₂ + 2√(I₁I₂) cosφ φ = (2π/λ)Δx is the phase difference
Resultant intensity (equal sources) I = 4I₀ cos²(φ/2) Valid when I₁ = I₂ = I₀; Imax = 4I₀, Imin = 0
Remember
  • Interference requires coherent sources: a constant (not necessarily zero) phase difference and the same frequency.
  • Path difference at a point on the screen: Δx = d sinθ ≈ dy/D.
  • Bright fringe: Δx = nλ; dark fringe: Δx = (n + ½)λ, n = 0, ±1, ±2, ...
  • Fringe width β = λD/d is the same for all fringes (equal spacing).
  • Resultant intensity I = 4I₀cos²(φ/2) for equal-amplitude coherent sources; interference redistributes energy without violating conservation.

Diffraction of Light: Single Slit

Quick answer Diffraction is the bending of light around obstacles or through narrow openings comparable in size to its wavelength; a single narrow slit produces a characteristic pattern with a wide bright centre and much fainter side bands.

Diffraction is the spreading of light as it passes through a narrow opening or around an edge whose size is comparable to the wavelength of light. Unlike Young's double-slit interference, which combines light from two separate, widely-spaced coherent point sources, single-slit diffraction arises from the superposition of secondary wavelets originating from every point across the same slit — effectively, interference among infinitely many coherent sources spread continuously over the slit width.

When a plane wavefront of wavelength λ falls normally on a narrow slit of width a, the diffraction pattern observed on a distant screen consists of a broad, intense central maximum flanked symmetrically by much fainter secondary maxima, separated by minima. To locate the minima, the slit is imagined as divided into pairs of strips; the condition for the m-th minimum is a sinθ = mλ (m = ±1, ±2, …; m ≠ 0). Secondary maxima occur roughly midway between successive minima, near a sinθ = (m + ½)λ, and are far weaker than the central maximum — the first secondary maximum has only about 1/22 of the central maximum's intensity.

Because the first minima occur at sinθ = ±λ/a, the central maximum is twice as wide (in angle) as any secondary maximum: its angular width is Δθ = 2λ/a, and its linear width on a screen at distance D is w = 2λD/a. A narrower slit (smaller a) produces a wider, more spread-out diffraction pattern — showing that diffraction effects become more pronounced as the opening size approaches the wavelength of light.

Worked example: A parallel beam of light of wavelength 600 nm falls normally on a slit of width 0.3 mm, and the pattern is observed on a screen 1.5 m away. Find the distance of the first minimum from the centre, and the width of the central maximum.
sinθ₁ = λ/a = (600 × 10⁻⁹)/(0.3 × 10⁻³) = 2 × 10⁻³ (small angle, so θ₁ ≈ sinθ₁ in radians).
y₁ = Dθ₁ = 1.5 × 2 × 10⁻³ = 3 × 10⁻³ m = 3 mm.
Width of central maximum = 2y₁ = 6 mm.

Minima condition (single slit) a sinθ = mλ, m = ±1, ±2, … a = slit width; central maximum lies at θ = 0
Angular width of central maximum Δθ = 2λ / a rad · Twice the angular position of the first minimum
Linear width of central maximum on screen w = 2λD / a m · D = distance of screen from the slit
Remember
  • Diffraction is significant when the slit/obstacle size is comparable to the wavelength of light.
  • Single-slit diffraction comes from interference of wavelets from many points across the same slit, unlike two-source interference in YDSE.
  • Minima condition: a sinθ = mλ (m = ±1, ±2, …); the centre (θ = 0) is always a maximum.
  • The central maximum is twice as wide as, and far more intense than, the secondary maxima.
  • Angular width of central maximum = 2λ/a; linear width on screen = 2λD/a.
  • A narrower slit produces a more spread-out (wider) diffraction pattern.

Polarisation of Light

Quick answer Polarisation shows that light is a transverse wave: it can be restricted to vibrate in a single plane, using devices like Polaroids, and following Malus's law and Brewster's law.

Ordinary light from sources such as the sun or a bulb is unpolarised: its electric field vibrates randomly in all directions perpendicular to the direction of travel. That light can be polarised at all is direct evidence that light is a transverse wave (a longitudinal wave, like sound, cannot be polarised, since its vibrations are already confined to the direction of propagation).

A Polaroid is a synthetic material containing long-chain molecules aligned in one direction, which absorbs the component of the electric field parallel to the chains and freely transmits the perpendicular component. It thus has a special direction called its pass axis. When unpolarised light passes through a single Polaroid, exactly half its intensity is transmitted (I = I₀/2), regardless of the Polaroid's orientation, and the transmitted light is now plane polarised, vibrating only along the pass axis.

If this plane-polarised light of intensity I₀ is then passed through a second Polaroid (called the analyser) whose pass axis makes an angle θ with the first Polaroid's pass axis, only the component of the electric field along the analyser's axis is transmitted. Since intensity is proportional to the square of amplitude, the transmitted intensity follows Malus's Law: I = I₀cos²θ. When the two axes are parallel (θ = 0), all the polarised light passes through; when they are crossed (θ = 90°), no light is transmitted.

Light can also become polarised on reflection. At a special angle of incidence called the polarising angle or Brewster's angle θB, the reflected ray is entirely plane polarised (with vibrations perpendicular to the plane of incidence), and the reflected and refracted rays are exactly perpendicular to each other. Brewster's Law states tanθB = n₂₁, where n₂₁ is the refractive index of the reflecting medium with respect to the medium of incidence.

Worked example: Unpolarised light of intensity 20 W/m² is incident on a Polaroid, and the transmitted light then passes through a second Polaroid whose axis is at 30° to the first. Find the final transmitted intensity.
After the first Polaroid: I₁ = I₀/2 = 20/2 = 10 W/m² (now plane polarised).
After the second Polaroid (Malus's Law): I₂ = I₁cos²30° = 10 × (0.75) = 7.5 W/m².

Polarisation has many practical applications in everyday devices: polarised sunglasses cut out glare from horizontally polarised light reflected off water or roads, LCD screens use polarisers to control transmitted light, and polarising filters are used in photography and 3D cinema glasses.

Malus's Law I = I₀ cos²θ θ = angle between polariser and analyser pass axes
Intensity after first polariser (from unpolarised light) I = I₀ / 2 Independent of the Polaroid's orientation
Brewster's Law tanθB = n₂₁ ; θB + r = 90° θB = polarising angle, r = angle of refraction at that incidence
Remember
  • Polarisation of light is direct evidence that light waves are transverse.
  • Unpolarised light through one Polaroid: transmitted intensity is always I₀/2, and the light becomes plane polarised.
  • Malus's Law: I = I₀cos²θ gives transmitted intensity through an analyser at angle θ to a polariser.
  • Crossed Polaroids (θ = 90°) transmit no light; parallel Polaroids (θ = 0°) transmit all the polarised light.
  • Brewster's Law: tanθB = n₂₁; at this angle, reflected and refracted rays are mutually perpendicular and the reflected ray is fully polarised.
  • Applications include polarised sunglasses, LCD displays, camera filters and 3D glasses.

The formula sheet

Every formula in this chapter, in one place — screenshot it before your exam.

v = c / n
Speed of light in a mediumm s⁻¹
distance moved = v × τ
Forward displacement of a wavefrontm
∠i = ∠r
Law of reflection
sin i / sin r = v₁ / v₂ = n₂₁ = n₂ / n₁
Snell's law (Huygens form)
λ₁ / λ₂ = v₁ / v₂
Wavelength ratio across a boundary
Δx = d sinθ ≈ d y / D
Path differencem
Δx = nλ, n = 0, ±1, ±2, …
Bright fringe (constructive interference)
Δx = (n + ½)λ
Dark fringe (destructive interference)
β = λD / d
Fringe widthm
I = I₁ + I₂ + 2√(I₁I₂) cosφ
Resultant intensity (general)
I = 4I₀ cos²(φ/2)
Resultant intensity (equal sources)
a sinθ = mλ, m = ±1, ±2, …
Minima condition (single slit)
Δθ = 2λ / a
Angular width of central maximumrad
w = 2λD / a
Linear width of central maximum on screenm
I = I₀ cos²θ
Malus's Law
I = I₀ / 2
Intensity after first polariser (from unpolarised light)
tanθB = n₂₁ ; θB + r = 90°
Brewster's Law

Test yourself

Tap an answer to check it instantly — you'll see why it's right, and what to revise if it isn't.

0 correct · 0/12 answered
Q1 Huygens' Principle easy

According to Huygens' Principle, a wavefront is best described as:

Q2 Wavefronts easy

Light reaching the Earth from a very distant star has a wavefront that is essentially:

Q3 Interference easy

Two light sources are said to be coherent if they have:

Q4 Young's Double Slit Experiment medium

In Young's double-slit experiment, if the slit separation is halved and the screen distance is doubled, the fringe width becomes:

Q5 Young's Double Slit Experiment medium

In a YDSE set-up, the slit separation is 0.5 mm, the screen is 1 m away, and light of wavelength 5000 Å is used. The fringe width is:

Q6 Polarisation medium

Which of these phenomena provides the most direct evidence that light is a transverse wave?

Q7 Interference medium

Two coherent sources, each of intensity I₀, superpose at a point where their phase difference is 60°. The resultant intensity at that point is:

Q8 Refraction (Huygens) medium

When light travels from a rarer medium into a denser medium, Huygens' construction shows that:

Q9 Diffraction hard

In a single-slit diffraction experiment, a = 0.2 mm, λ = 600 nm, and the screen is 2 m away. The width of the central maximum is:

Q10 Diffraction hard

The angular width of the central maximum in single-slit diffraction does NOT depend on:

Q11 Polarisation (Brewster's Law) hard

Light is incident on a glass surface (refractive index √3) exactly at the polarising angle. The angle of refraction is:

Q12 Interference hard

In a YDSE pattern, the intensity at the central bright fringe is I₀. The intensity at a point on the screen where the path difference is λ/4 is:

NCERT solutions & previous-year questions

Step-by-step model answers — tap a question to reveal the full solution.

NCERT questions 6

1 Using Huygens' wave theory, show that the angle of incidence is equal to the angle of reflection when a plane wavefront is reflected by a plane mirror.Huygens' Principle — Law of Reflection

Let a plane wavefront AB be incident obliquely on a plane reflecting surface MN, touching the surface first at point A while B is still in the medium. Let the time taken for the wavelet from B to reach the surface at C be t, so BC = v·t (v = speed of light in the medium).

During this same time t, the secondary wavelet originating from A (which reached the surface earlier) spreads out as a hemisphere into the medium, with radius AE = v·t. Since AE = v·t = BC, and the right-angled triangles EAC and BAC share the common hypotenuse AC, the two triangles are congruent (RHS criterion: right angle, hypotenuse AC common, side AE = side BC).

From this congruence, ∠EAC = ∠BCA. But ∠BCA is the angle between the incident wavefront direction and the surface, related to the angle of incidence i, and ∠EAC is similarly related to the angle of reflection r for the reflected wavefront EC. Converting these wavefront angles to ray angles (rays are perpendicular to their wavefronts) gives:

∠i = ∠r

Thus the angle of incidence equals the angle of reflection, and it can further be shown that the incident ray, the reflected ray and the normal at the point of incidence all lie in the same plane — establishing the laws of reflection from Huygens' Principle.

2 Monochromatic light of wavelength 589 nm is incident from air onto a water surface (refractive index of water = 1.33). Find the wavelength, frequency and speed of (a) the reflected light, and (b) the refracted light.Refraction — Speed, Wavelength, Frequency

(a) Reflected light: Reflection occurs back into the same medium (air), so none of the wave properties change.
Wavelength = 589 nm (unchanged).
Speed = c = 3 × 10⁸ m/s.
Frequency f = c/λ = (3 × 10⁸)/(589 × 10⁻⁹) = 5.09 × 10¹⁴ Hz.

(b) Refracted light: The frequency of light never changes on refraction (it is fixed by the source), so:
Frequency f = 5.09 × 10¹⁴ Hz (same as incident light).

Speed in water: v = c/n = (3 × 10⁸)/1.33 = 2.26 × 10⁸ m/s.

Wavelength in water: λ′ = v/f = λ₀/n = 589/1.33 = 442.9 nm ≈ 443 nm.

So the refracted light has the same frequency (5.09 × 10¹⁴ Hz) but a reduced speed (2.26 × 10⁸ m/s) and shorter wavelength (≈ 443 nm) compared to the incident light in air.

3 What is the shape of the wavefront in each of the following cases: (a) light diverging from a point source, (b) light emerging from a convex lens when a point source is placed at its focus, (c) the wavefront of light from a distant star intercepted by the Earth?Wavefronts

(a) Light diverging from a point source produces spherical wavefronts — concentric spheres centred on the source, since all points at equal radial distance are reached at the same time and hence share the same phase.

(b) When a point source is placed exactly at the focus of a convex lens, each ray leaving the lens emerges parallel to the principal axis (this is simply the reverse of the lens focusing a parallel beam to its focus). Parallel rays correspond to a plane wavefront, so the lens converts the diverging spherical wavefront from the source into a plane wavefront on the far side.

(c) A star is so far away that, over the small size of the Earth, the enormous spherical wavefront emitted by it is indistinguishable from a flat surface. Hence the wavefront intercepted by the Earth is (very nearly) a plane wavefront.

4 In Young's double-slit experiment using monochromatic light of wavelength λ, the intensity of light at a point on the screen where the path difference is λ is K units. What is the intensity of light at a point where the path difference is λ/3?Interference — Intensity

For two coherent sources of equal intensity I₀, the resultant intensity is I = 2I₀(1 + cosφ), where φ = (2π/λ) × (path difference).

At path difference = λ: φ = (2π/λ)(λ) = 2π, so cosφ = cos(2π) = 1.
I = 2I₀(1 + 1) = 4I₀ = K ⇒ I₀ = K/4.

At path difference = λ/3: φ = (2π/λ)(λ/3) = 2π/3 = 120°, so cosφ = cos(120°) = −0.5.
I = 2I₀(1 − 0.5) = 2I₀(0.5) = I₀ = K/4.

So the intensity at the point with path difference λ/3 is K/4.

5 A parallel beam of light of wavelength 500 nm falls on a narrow slit, and the resulting diffraction pattern is observed on a screen 1 m away. It is found that the first minimum lies 2.5 mm from the centre of the central maximum. Calculate the width of the slit.Diffraction — Single Slit

For the first minimum in single-slit diffraction: a sinθ = λ, so a = λ / sinθ.

Since the angle is small, sinθ ≈ tanθ = y/D, where y = 2.5 mm = 2.5 × 10⁻³ m and D = 1 m.

sinθ ≈ (2.5 × 10⁻³)/1 = 2.5 × 10⁻³.

a = λ / sinθ = (500 × 10⁻⁹) / (2.5 × 10⁻³) = 2 × 10⁻⁴ m = 0.2 mm.

6 Estimate the polarising (Brewster's) angle for a glass slab of refractive index 1.5 placed in air, and verify that the angle of refraction at this incidence is consistent with Brewster's law.Polarisation — Brewster's Law

By Brewster's Law, tanθB = n₂₁ = 1.5 (refractive index of glass with respect to air).

θB = tan⁻¹(1.5) ≈ 56.3°.

Verification: At Brewster's angle, the refracted ray makes an angle r = 90° − θB = 90° − 56.3° = 33.7° with the normal. Checking with Snell's law: n = sinθB / sin r = sin(56.3°)/sin(33.7°) = 0.832/0.555 ≈ 1.50, which matches the given refractive index, confirming the result. At this angle, the reflected ray is completely plane polarised, with its vibrations perpendicular to the plane of incidence.

Previous-year board questions 4

Q1 Using Huygens' construction, derive Snell's law of refraction for a plane wavefront passing from a rarer medium (speed v₁) to a denser medium (speed v₂). 2022 3 marks

Let a plane wavefront AB be incident on the plane boundary XY separating medium 1 (speed v₁) and medium 2 (speed v₂), touching the boundary first at A. Let t be the time taken for the secondary wavelet from B to travel through medium 1 and reach the boundary at C, so BC = v₁t.

In this same time t, the wavelet from A, now travelling in medium 2, spreads out to a hemisphere of radius AE = v₂t. The refracted wavefront is the tangent EC.

From the right triangles ABC and AEC (both share hypotenuse AC): BC = AC sin i and AE = AC sin r, where i is the angle of incidence and r is the angle of refraction (measured from the normal, using ray directions perpendicular to the respective wavefronts).

Dividing: sin i / sin r = BC/AE = (v₁t)/(v₂t) = v₁/v₂.

Since sin i / sin r is constant for a given pair of media, this is Snell's law, with the refractive index n₂₁ = v₁/v₂ = n₂/n₁ (using n = c/v). This shows that when light enters an optically denser medium (v₂ < v₁), sin r < sin i, so the ray bends towards the normal, consistent with observation.

Q2 In a Young's double-slit experiment, the two slits are 1 mm apart and the screen is placed 1 m away. Monochromatic light of wavelength 6000 Å is used. Calculate (a) the fringe width, and (b) the distance of the 10th bright fringe from the central maximum. 2023 3 marks

Given: d = 1 mm = 1 × 10⁻³ m, D = 1 m, λ = 6000 Å = 6 × 10⁻⁷ m.

(a) Fringe width:
β = λD/d = (6 × 10⁻⁷ × 1) / (1 × 10⁻³) = 6 × 10⁻⁴ m = 0.6 mm.

(b) Position of the 10th bright fringe:
y₁₀ = nβ = 10 × 0.6 mm = 6 mm from the central maximum.

Q3 Light of wavelength 6000 Å falls normally on a single slit of width 0.3 mm, producing a diffraction pattern on a screen placed 2 m away. Calculate the distance between the first minima on either side of the central maximum (i.e. the width of the central maximum). 2022 4 marks

Given: λ = 6000 Å = 6 × 10⁻⁷ m, a = 0.3 mm = 3 × 10⁻⁴ m, D = 2 m.

For the first minimum: a sinθ = λ ⇒ sinθ = λ/a = (6 × 10⁻⁷)/(3 × 10⁻⁴) = 2 × 10⁻³.

Since θ is small, sinθ ≈ θ ≈ y/D, so the distance of the first minimum from the centre:
y = Dθ = 2 × (2 × 10⁻³) = 4 × 10⁻³ m = 4 mm.

By symmetry, the first minimum on the other side is also 4 mm from the centre, so the total distance between the first minima on either side (= width of the central maximum) is:
2 × 4 mm = 8 mm.

Q4 Unpolarised light of intensity I₀ is first passed through a Polaroid P₁. The transmitted light is then passed through a second Polaroid P₂ whose pass axis makes an angle of 60° with that of P₁. Find the intensity of light emerging from P₂, in terms of I₀. 2024 2 marks

When unpolarised light of intensity I₀ passes through the first Polaroid P₁, exactly half the intensity is transmitted (regardless of orientation), and the light becomes plane polarised:
I₁ = I₀/2.

This plane-polarised light now passes through the second Polaroid P₂, whose axis is at θ = 60° to P₁'s axis. By Malus's Law:
I₂ = I₁ cos²θ = (I₀/2) × cos²(60°) = (I₀/2) × (1/2)² = (I₀/2) × (1/4) = I₀/8.

So only one-eighth of the original unpolarised intensity emerges from the second Polaroid.

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