Hello dear learner. Welcome to HSLC GURU. In this lesson we present the complete English-medium question-and-answer set, key concepts, derivations, formula sheet, fully worked numerical problems, hand-drawn SVG ray diagrams and additional important questions of the tenth chapter — Wave Optics — of the ASSEB (Assam State School Education Board) Class 12 Physics syllabus. Wave optics is the branch of physics that explains every observable phenomenon of light that ray optics cannot — the bright and dark fringes of Young’s experiment, the diffraction pattern at the edge of a razor blade, the rainbow colours on a soap bubble, the resolving power of telescopes and microscopes, the polarisation of sunlight reflected from a wet road, and the operation of LCD screens and Polaroid sun-glasses. Mastering this chapter is therefore essential for both your board examination and any future engineering / medical entrance test (NEET, JEE, CUET).

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Summary

The wave theory of light was put forward by Christian Huygens in 1678 and was firmly established by Thomas Young’s double-slit experiment (1801) and by Fresnel’s diffraction calculations. Light is treated as a transverse electromagnetic wave whose electric and magnetic field vectors oscillate perpendicular to each other and to the direction of propagation. A wavefront is the locus of all particles of the medium that are in the same phase of vibration; the perpendicular drawn on a wavefront in the direction of propagation is a ray. According to Huygens’s principle, every point of a wavefront acts as a fresh secondary source of disturbance, and the new wavefront at any later time is the forward envelope (tangent surface) of these secondary wavelets. Using this construction one can deduce the laws of reflection and refraction, and also Snell’s law $n_1\sin i = n_2\sin r$.

When two coherent waves overlap, their resultant intensity at any point depends on the path difference. Constructive interference (bright fringe) occurs when the path difference equals an integral multiple of the wavelength, and destructive interference (dark fringe) when it equals an odd multiple of half the wavelength. Young’s double-slit experiment realises this with two narrow slits illuminated by the same monochromatic source. The fringe spacing on a screen at distance $D$ is $\beta = \lambda D/d$, where $d$ is the slit separation. The chapter then turns to diffraction at a single slit of width $a$, where the central bright maximum has angular half-width $\lambda/a$ and linear width $2\lambda D/a$, with subsidiary minima at $a\sin\theta = n\lambda$. Diffraction limits the resolving power of optical instruments through the Rayleigh criterion. Finally polarisation — possible only for transverse waves — is described, together with Malus’s law $I = I_0\cos^2\theta$ and Brewster’s law $\tan\theta_B = n$, which gives the angle of incidence at which the reflected light is completely plane-polarised.

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Key Formulas

Quantity	Formula	SI Unit
Speed of light in medium	$v = c/n$	m s$^{-1}$
Wavelength in medium	$\lambda_n = \lambda/n$	metre (m)
Snell’s law (Huygens)	$n_1\sin i = n_2\sin r$	dimensionless
Path difference (YDSE)	$\Delta x = d\sin\theta \approx \dfrac{dy}{D}$	metre (m)
Bright fringe (YDSE)	$y_n = \dfrac{n\lambda D}{d}$, $n=0,1,2,\ldots$	metre (m)
Dark fringe (YDSE)	$\Delta x = (n+\tfrac{1}{2})\lambda$	metre (m)
Fringe width	$\beta = \dfrac{\lambda D}{d}$	metre (m)
Resultant intensity (two waves)	$I = I_1+I_2+2\sqrt{I_1I_2}\cos\phi$	W m$^{-2}$
Intensity (equal sources)	$I = 4I_0\cos^2(\phi/2)$	W m$^{-2}$
Single-slit minima	$a\sin\theta = n\lambda$, $n=\pm1,\pm2,\ldots$	dimensionless
Central maximum width	$W_0 = \dfrac{2\lambda D}{a}$	metre (m)
Resolving power (telescope)	$\theta_{\min} = \dfrac{1.22\lambda}{D}$	radian
Resolving power (microscope)	$d_{\min} = \dfrac{1.22\lambda}{2n\sin\beta}$	metre (m)
Malus’s law	$I = I_0\cos^2\theta$	W m$^{-2}$
Brewster’s law	$\tan\theta_B = n$	dimensionless
Brewster–reflection geometry	$\theta_B + \theta_r = 90^\circ$	dimensionless

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10.1 Huygens’s Principle

A wavefront is the locus of all points in the medium which oscillate in the same phase. From a point source the wavefronts are spherical; very far from a source (or at the focus of a converging lens) they become essentially plane. The direction of propagation is everywhere perpendicular to the wavefront and is what we ordinarily call a ray.

Huygens’s principle may be stated in two parts:

1. Every point on a given wavefront (called the primary wavefront) is itself a source of fresh disturbance — a secondary source — that emits secondary spherical wavelets in all forward directions with the same speed as that of the primary wave.
2. The new position of the wavefront after a time $t$ is given by the forward envelope (tangent surface) drawn to all these secondary wavelets, each of radius $vt$.

Huygens’s construction applied to a plane wavefront in a uniform medium just regenerates a plane wavefront moving forward — the rays remain parallel, in agreement with rectilinear propagation. The principle, however, becomes powerful when the wave encounters a boundary, an aperture, or an obstacle, because it then automatically gives the laws of reflection, of refraction and of diffraction.

Reflection by Huygens’s principle

Consider a plane wavefront $AB$ incident on a plane reflecting surface $MN$ at angle of incidence $i$. While the disturbance from $B$ travels to the surface at $C$ (covering distance $BC = vt$), the disturbance that arrived at $A$ at the start of the time interval has been emitting secondary wavelets that have grown to a sphere of radius $vt$. The new (reflected) wavefront $CE$ must therefore be tangent to this sphere from $C$. From the congruent right triangles $ABC$ and $AEC$ (common hypotenuse $AC$, $AE=BC=vt$) one finds $\angle BAC = \angle ECA$, that is

$$\sin i = \sin r,\qquad \therefore\; i = r.$$

This is the law of reflection. Moreover the incident ray, the reflected ray and the normal lie in the same plane.

Refraction by Huygens’s principle

Let a plane wavefront $AB$ travelling in medium 1 (speed $v_1$) be incident on the plane separation surface $XY$ between media 1 and 2 (speed $v_2$). While the disturbance from $B$ moves to $C$ in time $t$ — covering $BC = v_1 t$ in medium 1 — the disturbance from $A$ enters medium 2 and, in the same time, expands into a sphere of radius $AE = v_2 t$. The refracted wavefront is the tangent $CE$ from $C$ to that sphere. From the right triangles $ABC$ and $AEC$ with common hypotenuse $AC$:

$$\sin i = \dfrac{BC}{AC} = \dfrac{v_1 t}{AC},\qquad \sin r = \dfrac{AE}{AC} = \dfrac{v_2 t}{AC}.$$

$$\therefore\;\; \dfrac{\sin i}{\sin r} = \dfrac{v_1}{v_2} = \dfrac{n_2}{n_1} = n_{12}.$$

This is Snell’s law of refraction. The wavelength changes as $\lambda_2 = \lambda_1\,v_2/v_1$ but the frequency is unchanged. When light goes from a rarer to a denser medium ($v_2 < v_1$, $n_{21}>1$), the refracted ray bends toward the normal.

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10.2 Coherent Sources, Interference and Young’s Double-Slit Experiment

Two sources are said to be coherent if they emit light of the same frequency, with a phase difference between them that is constant in time. Independent ordinary light sources can never be coherent because each atom emits a wave-train of $\sim 10^{-9}$ s and then radiates again with a random phase; for sustained interference we therefore use a single source split into two by a slit-pair, biprism or other device. With a laser, the long coherence length makes coherence almost trivial.

If two coherent waves of amplitudes $a_1,a_2$ and phase difference $\phi$ superpose, the resultant intensity is

$$I = I_1 + I_2 + 2\sqrt{I_1 I_2}\,\cos\phi.$$

For equal intensities ($I_1=I_2=I_0$) this becomes $I = 4I_0\cos^2(\phi/2)$, with maxima $I_{\max}=4I_0$ when $\phi = 2n\pi$ and minima $I_{\min}=0$ when $\phi=(2n+1)\pi$. Energy is not destroyed at the dark fringes; it is merely redistributed from the dark fringes to the bright ones, so that the total energy is conserved.

Young’s double-slit experiment — geometry

Two narrow slits $S_1$ and $S_2$, separated by a small distance $d$, are illuminated by a single monochromatic source $S$ placed symmetrically behind them. Let $D$ be the distance of the screen from the plane of the slits, and let $P$ be a point on the screen at perpendicular distance $y$ from the central axis. The path difference for waves reaching $P$ is

$$\Delta x = S_2P – S_1P = d\sin\theta \approx d\tan\theta = \dfrac{dy}{D}\quad (\text{since } y,d\ll D).$$

For bright fringes (constructive interference) the path difference must be an integral multiple of $\lambda$:

$$\Delta x = n\lambda \;\;\Rightarrow\;\; y_n = \dfrac{n\lambda D}{d},\quad n = 0, \pm1, \pm2,\ldots$$

For dark fringes (destructive interference) the path difference must be an odd multiple of $\lambda/2$:

$$\Delta x = (n+\tfrac{1}{2})\lambda \;\;\Rightarrow\;\; y_n’ = \dfrac{(2n+1)\lambda D}{2d},\quad n = 0,1,2,\ldots$$

The distance between two consecutive bright (or two consecutive dark) fringes is the fringe width:

$$\boxed{\;\beta = y_{n+1}-y_n = \dfrac{\lambda D}{d}\;}$$

Conditions for clearly visible fringes: (i) the two sources must be coherent and monochromatic, (ii) they should have equal amplitudes for maximum contrast, (iii) the slit separation $d$ should be small (of the order of $1$ mm), (iv) the screen distance $D$ should be large (of the order of $1$ m), and (v) the slits should be narrow.

SVG: Young’s double-slit set-up

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10.3 Diffraction at a Single Slit

When a plane wavefront of monochromatic light is incident on a single slit of width $a$, the slit itself acts as a continuous distribution of secondary sources (Huygens). Waves from these sources superpose on a distant screen and produce a diffraction pattern: a bright central maximum flanked on both sides by alternate dark and bright fringes of decreasing intensity.

The path difference between waves from the two edges of the slit, observed at angle $\theta$ from the central axis, is $a\sin\theta$. We divide the slit into pairs of strips that destructively cancel:

$$\boxed{\;a\sin\theta = n\lambda,\quad n=\pm1,\pm2,\ldots\;\text{(minima)}\;}$$

$$a\sin\theta = (n+\tfrac{1}{2})\lambda,\quad n=\pm1,\pm2,\ldots\;\text{(secondary maxima, approx.)}$$

For small angles $\sin\theta\approx\theta\approx y/D$. The first minimum on each side of the centre is therefore at $y_1 = \pm\lambda D/a$, and the linear width of the central maximum is

$$\boxed{\;W_0 = 2y_1 = \dfrac{2\lambda D}{a}\;}$$

The angular width of the central maximum is $2\lambda/a$. The width of any non-central bright fringe is $\lambda D/a$, exactly half the width of the central one. As the slit becomes wider ($a \gg \lambda$) the diffraction pattern collapses to a sharp bright spot — recovering geometrical optics. As $a$ shrinks toward $\lambda$ the central maximum spreads out, eventually filling the geometric shadow.

Comparison: Interference vs Diffraction

Feature	Interference (YDSE)	Diffraction (single slit)
Sources	Two coherent point/line sources	Many secondary sources within one slit
Fringe widths	All bright fringes equally wide ($\beta=\lambda D/d$)	Central maximum is twice as wide as the rest
Intensity	All maxima equally bright (uniform)	Intensity falls off rapidly from centre outward
Minima	Perfectly dark for equal-amplitude sources	Not perfectly dark
Visibility	Fine, regular fringes	Broad, unequal fringes

SVG: Single-slit diffraction intensity

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10.4 Resolving Power (Rayleigh’s Criterion)

Because every aperture diffracts light, the image of a point source is never a geometric point but a small Airy disc surrounded by faint rings. Two point sources can be told apart only when their diffraction discs are sufficiently separated. Rayleigh’s criterion states that two equally bright point sources are just resolved when the central maximum of one coincides with the first minimum of the other.

Telescope. The minimum angular separation between two distant point sources just resolved by a telescope of objective diameter $D$ is

$$\theta_{\min} = \dfrac{1.22\lambda}{D}.$$

The reciprocal of this angle is the resolving power of the telescope, $R = D/(1.22\lambda)$. Larger objectives and shorter wavelengths give better resolution.

Microscope. The smallest distance between two object points that can be just resolved is

$$d_{\min} = \dfrac{1.22\lambda}{2n\sin\beta},$$

where $n\sin\beta$ is the numerical aperture and $2\beta$ is the angle subtended by the objective at the object. Because $d_{\min}\propto\lambda$, ultraviolet and electron microscopes resolve much finer detail than ordinary visible-light microscopes.

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10.5 Polarisation

Light is a transverse electromagnetic wave; the electric field vector $\vec{E}$ vibrates perpendicular to the direction of propagation. Natural (unpolarised) light contains $\vec{E}$ vibrations in all directions perpendicular to the ray with equal probability. Plane (linearly) polarised light is light in which $\vec{E}$ vibrates only along one fixed direction. The plane containing $\vec{E}$ and the direction of propagation is called the plane of vibration; the perpendicular plane is the plane of polarisation. Only transverse waves can be polarised — this is the experimental proof that light is transverse.

Methods of producing polarised light include selective absorption (Polaroid sheets), reflection at the Brewster angle, refraction (in birefringent crystals like calcite) and scattering (e.g. blue sky light viewed at 90$^\circ$ from the sun is partially polarised).

Malus’s law

If plane-polarised light of intensity $I_0$ is incident on a polariser (analyser) whose pass-axis makes angle $\theta$ with the polarisation direction of the incident light, the transmitted intensity is

$$\boxed{\;I = I_0\cos^2\theta\;}\quad(\text{Malus’s law}).$$

When $\theta=0$ the analyser passes the maximum intensity $I_0$; when $\theta=90^\circ$ the transmission is zero (the analyser is “crossed”). For unpolarised light incident on a polariser, the transmitted intensity is exactly $I_0/2$ (averaging $\cos^2\theta$ over all angles).

SVG: Malus’s law

Brewster’s law

When unpolarised light strikes a transparent dielectric (glass, water) at a particular angle of incidence $\theta_B$, called the Brewster angle or polarising angle, the reflected light is completely plane-polarised perpendicular to the plane of incidence. Brewster (1812) showed that

$$\boxed{\;\tan\theta_B = n\;}\qquad(\text{Brewster’s law}),$$

where $n$ is the refractive index of the second medium relative to the first. At this angle the reflected ray and the refracted ray are mutually perpendicular: $\theta_B + \theta_r = 90^\circ$. For air → glass ($n=1.5$), $\theta_B = \tan^{-1}(1.5) \approx 56.3^\circ$. Polaroid sun-glasses with their pass-axis vertical eliminate the strong horizontally-polarised glare reflected from horizontal surfaces (wet roads, water, car bonnets).

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NCERT/ASSEB Textbook Exercise — Solved

Q1. Monochromatic light of wavelength 589 nm is incident from air on a water surface ($n=1.33$). What are the wavelength, frequency and speed of (a) reflected and (b) refracted light?

Answer: Frequency depends only on the source and is the same in both reflected and refracted beams.

$$\nu = \dfrac{c}{\lambda} = \dfrac{3\times10^8}{589\times10^{-9}} = 5.09\times10^{14}\ \text{Hz}.$$

(a) Reflected light is in air, so $\lambda$ and $v$ are unchanged: $\lambda = 589$ nm, $v = 3\times10^8$ m s$^{-1}$.

(b) In water, $v = c/n = 3\times10^8/1.33 = 2.26\times10^8$ m s$^{-1}$ and $\lambda_w = \lambda/n = 589/1.33 = 442.9$ nm $\approx 443$ nm.

Q2. What is the shape of the wavefront in each of the following cases? (a) Light diverging from a point source. (b) Light emerging out of a convex lens when a point source is placed at its focus. (c) The portion of the wavefront from a distant star intercepted by the Earth.

Answer: (a) Spherical diverging wavefront — every point at the same distance from the source has the same phase. (b) Plane wavefront — a point source at the focus of a convex lens gives a parallel emerging beam. (c) Plane wavefront — the radius of the spherical wavefront from a distant star is so large that the small portion intercepted by the Earth is essentially flat.

Q3. (a) The refractive index of glass is 1.5. What is the speed of light in glass? Speed of light in vacuum is $3\times10^8$ m s$^{-1}$. (b) Is the speed of light in glass independent of the colour of light? If not, which colour travels slower, red or violet?

Answer: (a) $v = c/n = 3\times10^8/1.5 = 2.0\times10^8$ m s$^{-1}$.

(b) No, the refractive index varies with the wavelength of light (dispersion). Violet light has a higher refractive index than red light in glass ($n_V > n_R$), so violet light travels slower than red light in glass.

Q4. In a Young’s double-slit experiment, the slits are separated by $0.28$ mm and the screen is placed $1.4$ m away. The distance between the central bright fringe and the fourth bright fringe is measured to be $1.2$ cm. Determine the wavelength of light used.

Answer: $y_n = n\lambda D/d$. With $n=4$, $d=0.28\times10^{-3}$ m, $D=1.4$ m, $y_4=1.2\times10^{-2}$ m,

$$\lambda = \dfrac{y_n\,d}{nD} = \dfrac{1.2\times10^{-2}\times 0.28\times10^{-3}}{4\times1.4} = 6.0\times10^{-7}\ \text{m} = 600\ \text{nm}.$$

Q5. In Young’s double-slit experiment using monochromatic light of wavelength $\lambda$, the intensity of light at a point on the screen where path difference is $\lambda$ is $K$ units. What is the intensity of light at a point where path difference is $\lambda/3$?

Answer: Phase difference $\phi = (2\pi/\lambda)\Delta x$. For two equal sources $I = 4I_0\cos^2(\phi/2)$.

At $\Delta x = \lambda$: $\phi = 2\pi$, $I = 4I_0\cos^2\pi = 4I_0 = K$, so $I_0 = K/4$.

At $\Delta x = \lambda/3$: $\phi = 2\pi/3$, $I’ = 4I_0\cos^2(\pi/3) = 4I_0\,(1/2)^2 = I_0 = K/4$.

Q6. A beam of light consisting of two wavelengths, $650$ nm and $520$ nm, is used to obtain interference fringes in a Young’s double-slit experiment. (a) Find the distance of the third bright fringe on the screen from the central maximum for the wavelength $650$ nm. (b) What is the least distance from the central maximum where the bright fringes due to both the wavelengths coincide? Take $d = 2$ mm and $D = 1.2$ m.

Answer: (a) $y_3 = 3\lambda_1 D/d = 3\times650\times10^{-9}\times 1.2/(2\times10^{-3}) = 1.17\times10^{-3}$ m $= 1.17$ mm.

(b) The bright fringes coincide when $n_1\lambda_1 = n_2\lambda_2$, i.e. $n_1\times 650 = n_2\times 520$, giving $n_1/n_2 = 520/650 = 4/5$. The smallest integer pair is $n_1=4,\,n_2=5$. The corresponding position is

$$y = \dfrac{4\lambda_1 D}{d} = \dfrac{4\times650\times10^{-9}\times1.2}{2\times10^{-3}} = 1.56\times10^{-3}\ \text{m} = 1.56\ \text{mm}.$$

Q7. In a double-slit experiment using light of wavelength $600$ nm, the angular width of a fringe formed on a distant screen is $0.1^\circ$. What is the spacing between the two slits?

Answer: Angular fringe width $\theta = \beta/D = \lambda/d$. So

$$d = \dfrac{\lambda}{\theta} = \dfrac{600\times10^{-9}}{0.1\times\pi/180} = \dfrac{6\times10^{-7}}{1.745\times10^{-3}} \approx 3.44\times10^{-4}\ \text{m} = 0.344\ \text{mm}.$$

Q8. Light of wavelength $5000$ Å falls on a plane reflecting surface. What are the wavelength and frequency of the reflected light? For what angle of incidence is the reflected ray normal to the incident ray?

Answer: The wavelength and frequency of the reflected light are unchanged: $\lambda = 5000$ Å $=5\times10^{-7}$ m and $\nu = c/\lambda = 6\times10^{14}$ Hz. The reflected ray will be perpendicular to the incident ray when $i + r = 90^\circ$ and (since $i = r$) $i = 45^\circ$.

Q9. Estimate the distance for which ray optics is a good approximation for an aperture of $4$ mm and wavelength $400$ nm.

Answer: Ray optics holds within the Fresnel distance $z_F = a^2/\lambda$.

$$z_F = \dfrac{(4\times10^{-3})^2}{400\times10^{-9}} = \dfrac{1.6\times10^{-5}}{4\times10^{-7}} = 40\ \text{m}.$$

For distances much smaller than 40 m the geometric image is sharp; beyond this, diffraction begins to broaden it noticeably.

Q10. For what distance is ray optics a good approximation when the aperture is 3 mm wide and the wavelength is 500 nm?

Answer: $z_F = a^2/\lambda = (3\times10^{-3})^2/(500\times10^{-9}) = 9\times10^{-6}/5\times10^{-7} = 18$ m.

Q11. The 6563 Å H$_\alpha$ line emitted by hydrogen in a star is found to be red-shifted by $15$ Å. Estimate the speed with which the star is receding from the Earth.

Answer: Doppler shift $\Delta\lambda/\lambda = v/c$.

$$v = c\,\dfrac{\Delta\lambda}{\lambda} = 3\times10^8\times\dfrac{15}{6563} = 6.86\times10^5\ \text{m s}^{-1}.$$

The star is receding at about $6.9\times10^5$ m s$^{-1}$.

Q12. Show that the speed of light in a medium of refractive index $n$ is $v = c/n$ using Huygens’s construction for refraction.

Answer: See the derivation in §10.1. Huygens’s construction gives $\sin i/\sin r = v_1/v_2$. By definition $n_{12} = v_1/v_2$, and for a vacuum→medium pair $v_1=c$, $v_2=v$, so $n = c/v$ giving $v = c/n$.

Q13. Light of wavelength 5000 Å is used to obtain a single-slit diffraction pattern on a screen $1$ m away from a slit of width $0.2$ mm. Find the linear width of the central maximum.

Answer: $W_0 = 2\lambda D/a = 2\times5000\times10^{-10}\times1/(0.2\times10^{-3}) = 5\times10^{-3}$ m $= 5$ mm.

Q14. A polariser and an analyser are oriented so that the maximum amount of light is transmitted. To what fraction of its maximum value is the intensity of the transmitted light reduced when the analyser is rotated through (i) $30^\circ$, (ii) $45^\circ$, (iii) $60^\circ$?

Answer: By Malus’s law $I/I_0 = \cos^2\theta$.

• $\theta=30^\circ$: $\cos^230^\circ = 3/4 = 0.75$
• $\theta=45^\circ$: $\cos^245^\circ = 1/2 = 0.5$
• $\theta=60^\circ$: $\cos^260^\circ = 1/4 = 0.25$

Q15. At what angle of incidence should a light beam strike a glass slab of refractive index $\sqrt{3}$ so that the reflected and refracted rays are perpendicular to each other?

Answer: This is exactly the Brewster condition: $\tan\theta_B = n = \sqrt{3}$, hence $\theta_B = 60^\circ$.

Q16. Yellow light of wavelength $6000$ Å produces fringes of width $0.8$ mm in a Young’s double-slit experiment. If the source is replaced by another monochromatic source of wavelength $7500$ Å and the separation between the slits is doubled, what is the new fringe width?

Answer: $\beta = \lambda D/d$. Old: $\beta_1 = 0.8$ mm with $\lambda_1 = 6000$ Å, $d_1$. New: $\lambda_2 = 7500$ Å, $d_2 = 2d_1$, same $D$.

$$\dfrac{\beta_2}{\beta_1} = \dfrac{\lambda_2}{\lambda_1}\cdot\dfrac{d_1}{d_2} = \dfrac{7500}{6000}\times\dfrac{1}{2} = \dfrac{5}{8}.$$

$\therefore \beta_2 = 0.8\times 5/8 = 0.5$ mm.

Q17. Two slits in YDSE have widths in the ratio $4:1$. Find the ratio of intensity at the maxima to that at the minima.

Answer: Intensity is proportional to slit-width, hence amplitude $a \propto \sqrt{\text{width}}$. So $a_1/a_2 = 2/1$.

$$\dfrac{I_{\max}}{I_{\min}} = \left(\dfrac{a_1+a_2}{a_1-a_2}\right)^2 = \left(\dfrac{3}{1}\right)^2 = 9.$$

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Additional Important Questions

Very Short Answer (1 mark)

1. Define a wavefront.

Answer: A wavefront is the locus of all points in a medium that vibrate in the same phase.

2. What is the geometric relation between a ray of light and a wavefront?

Answer: A ray is the perpendicular drawn on the wavefront in the direction of propagation.

3. What type of wavefront is associated with (i) a point source, (ii) a parallel beam of light?

Answer: (i) Spherical wavefront, (ii) Plane wavefront.

4. Why are coherent sources required for sustained interference?

Answer: A constant phase difference is essential for the position of the maxima and minima to remain stationary on the screen; otherwise the fringe pattern would shift continuously and average out.

5. Two independent bulbs cannot produce an interference pattern. Why?

Answer: Because they are not coherent — atoms in two independent sources radiate with random and rapidly varying phase difference.

6. State the conditions for constructive and destructive interference.

Answer: Constructive: path difference $= n\lambda$ ($n=0,1,2,\ldots$); destructive: path difference $= (n+\tfrac{1}{2})\lambda$.

7. What is the effect of immersing the YDSE apparatus in water on the fringe width?

Answer: Inside water the wavelength reduces from $\lambda$ to $\lambda/n$, hence $\beta_w = \beta/n$. Fringe width decreases.

8. What happens to the fringe width if the entire YDSE set-up is moved farther from the screen?

Answer: Fringe width $\beta = \lambda D/d$ increases linearly with $D$.

9. State Brewster’s law.

Answer: When unpolarised light is incident on a transparent medium of refractive index $n$ at an angle $\theta_B$ such that $\tan\theta_B = n$, the reflected light is completely plane-polarised perpendicular to the plane of incidence.

10. What is the polarising angle for water ($n=1.33$)?

Answer: $\theta_B = \tan^{-1}(1.33) \approx 53.1^\circ$.

11. State Malus’s law.

Answer: When plane-polarised light of intensity $I_0$ passes through an analyser whose pass-axis makes angle $\theta$ with that of the polariser, the transmitted intensity is $I = I_0\cos^2\theta$.

12. Why can sound waves not be polarised whereas light waves can?

Answer: Polarisation is possible only for transverse waves. Sound waves are longitudinal; light waves are transverse, so only light can be polarised.

13. What is meant by resolving power of an optical instrument?

Answer: The reciprocal of the smallest distance (or angle) between two close points/objects which the instrument can image as separate.

14. How does the intensity of central maximum in single-slit diffraction change if the slit width is halved?

Answer: Halving $a$ doubles the angular width $\propto 1/a$ but quarters the central intensity ($\propto a^2$).

15. What is the path difference between two waves that are in opposite phase?

Answer: An odd multiple of $\lambda/2$.

Short Answer (2–3 marks)

16. State Huygens’s principle. How is it used to construct a new wavefront?

Answer: Huygens’s principle states that (i) every point on a primary wavefront acts as a fresh source emitting secondary spherical wavelets in the forward direction with the same speed as the primary wave, and (ii) the new wavefront after a time $t$ is given by the forward envelope of all these secondary wavelets, each of radius $vt$. By drawing the envelope tangent to all the wavelets we obtain the position of the wavefront at the later time.

17. Differentiate between interference and diffraction.

Answer: Interference involves superposition of waves from two (or a few) coherent sources; the fringes are equally spaced and equally bright. Diffraction is the bending of light around an obstacle/aperture and arises from the superposition of secondary wavelets from the same wavefront; the fringes have unequal widths and rapidly decreasing intensities.

18. Explain why no interference pattern is observed when two independent light sources are used.

Answer: Two independent ordinary light sources are not coherent — the phase difference between them changes rapidly and randomly with time. Hence the position of any bright/dark fringe shifts so quickly that the eye records only a uniform illumination.

19. What will be the effect on the YDSE pattern if the source is replaced by white light?

Answer: The central fringe (zero path difference) is white. On either side a few coloured fringes are visible — violet closest to the centre and red farthest, because $\beta\propto\lambda$. Beyond a few orders the fringes of different colours overlap and merge into general illumination.

20. What is meant by “diffraction limit” of an optical instrument?

Answer: Even an ideal aberration-free instrument cannot focus light to a true point — diffraction at the aperture creates an Airy disc of finite size, beyond which two close objects can no longer be distinguished. This fundamental limit on resolution is the diffraction limit.

21. Why does the Sun appear reddish at sunrise and sunset?

Answer: At sunrise/sunset sunlight has to travel a much greater thickness of atmosphere; blue and violet (short-wavelength) light is scattered out of the path more strongly than red (Rayleigh scattering, $\propto 1/\lambda^4$), so the directly transmitted light reaching us is predominantly red.

22. State and explain the conditions for sustained interference.

Answer: (i) The two sources must be coherent (constant phase difference). (ii) They must emit light of the same frequency and wavelength. (iii) Their amplitudes should be equal (or comparable) to give good contrast. (iv) The sources should be narrow and close together. (v) The screen should be sufficiently far from the slits.

23. Show that fringe width in YDSE is independent of the order of the fringe.

Answer: The position of the $n$th bright fringe is $y_n = n\lambda D/d$. Hence $\beta = y_{n+1}-y_n = \lambda D/d$, which contains no $n$. All consecutive bright (or dark) fringes are equally spaced.

24. What is the polarising angle of glass of refractive index $1.5$? At this angle, what is the angle of refraction inside the glass?

Answer: $\theta_B = \tan^{-1}(1.5) = 56.3^\circ$. Since $\theta_B + \theta_r = 90^\circ$, $\theta_r = 33.7^\circ$.

25. Two polaroids are placed in the path of an unpolarised beam of intensity $I_0$ such that no light is transmitted. A third polaroid is now introduced between them with its pass-axis making $45^\circ$ with each. Find the intensity emerging.

Answer: After P1 (unpolarised → polarised): $I_0/2$. After P3 at $45^\circ$: $(I_0/2)\cos^245^\circ = I_0/4$. After P2 (also at $45^\circ$ to P3, since P2$\perp$P1): $(I_0/4)\cos^245^\circ = I_0/8$.

26. The angular width of the central maximum in single-slit diffraction is twice that of any other maximum. Justify.

Answer: The first minima lie at $a\sin\theta = \pm\lambda$, i.e. at angles $\pm\lambda/a$. So the central maximum extends from $-\lambda/a$ to $+\lambda/a$, total angular width $2\lambda/a$. The next bright fringe lies between minima at $\lambda/a$ and $2\lambda/a$, of angular width $\lambda/a$, exactly half.

27. How does resolving power of a telescope depend on (i) wavelength, (ii) aperture diameter?

Answer: $R = D/(1.22\lambda)$. Resolving power is directly proportional to aperture diameter $D$ and inversely proportional to wavelength $\lambda$.

Long Answer (5 marks)

28. Use Huygens’s principle to derive the laws of refraction.

Answer: See §10.1. Briefly: a plane wavefront $AB$ in medium 1 (speed $v_1$) is incident on the boundary $XY$. While the disturbance from $B$ travels $BC = v_1 t$ to strike the boundary at $C$, the disturbance that arrived at $A$ enters medium 2 and grows into a sphere of radius $v_2 t$. The refracted wavefront $CE$ is the tangent from $C$. Right triangles $ABC$ and $AEC$ share the hypotenuse $AC$, giving $\sin i/\sin r = v_1/v_2 = n_2/n_1$, which is Snell’s law. The refracted ray and the normal lie in the plane of incidence, completing the laws.

29. Describe Young’s double-slit experiment. Derive the expression for fringe width on the screen.

Answer: A monochromatic source $S$ illuminates a single slit, behind which two narrow parallel slits $S_1$ and $S_2$ separated by $d$ act as coherent sources. The interference pattern is observed on a screen at distance $D$. For a point $P$ on the screen at distance $y$ from the central axis, the path difference is $\Delta x = d\sin\theta \approx dy/D$. Bright fringes occur where $\Delta x = n\lambda$, giving $y_n = n\lambda D/d$. The distance between consecutive bright (or consecutive dark) fringes is the fringe width $\beta = \lambda D/d$. The same expression also gives the spacing of the dark fringes; it is independent of the order $n$, so all fringes are equally spaced. Conditions for clear fringes: monochromatic source, narrow and equally illuminated slits, small $d$, large $D$.

30. Derive an expression for the intensity at any point on the screen due to two coherent waves of amplitudes $a_1$ and $a_2$.

Answer: Let the two waves be $y_1 = a_1\sin\omega t$ and $y_2 = a_2\sin(\omega t+\phi)$. By superposition $y = y_1+y_2 = A\sin(\omega t+\delta)$ where $A^2 = a_1^2+a_2^2+2a_1a_2\cos\phi$. Since $I\propto A^2$,

$$I = I_1 + I_2 + 2\sqrt{I_1I_2}\cos\phi.$$

For equal amplitudes ($I_1=I_2=I_0$) this reduces to $I = 4I_0\cos^2(\phi/2)$. Maximum: $I_{\max} = 4I_0$ when $\phi = 2n\pi$; minimum: $I_{\min}=0$ when $\phi=(2n+1)\pi$.

31. Discuss diffraction at a single slit and show that the linear width of the central maximum on a screen at distance $D$ is $W_0=2\lambda D/a$.

Answer: Consider a slit of width $a$ illuminated normally by a plane wave of wavelength $\lambda$. By Huygens, every point of the slit acts as a secondary source. Waves from two points separated by $a$ have path difference $a\sin\theta$ at angle $\theta$. Pair every point in the upper half of the slit with a point exactly $a/2$ below it. They produce destructive interference when their path difference $(a/2)\sin\theta = \lambda/2$, i.e. $a\sin\theta = \lambda$. Repeating with quarter-pairs gives $a\sin\theta=2\lambda$, and in general $a\sin\theta = n\lambda$ for $n=\pm1,\pm2,\ldots$ (minima). For small angles $\sin\theta\approx y/D$, so the first minima lie at $y=\pm\lambda D/a$. The linear width of the central maximum is therefore $W_0 = 2\lambda D/a$. Subsidiary maxima occur approximately at $a\sin\theta = (n+1/2)\lambda$ with rapidly decreasing intensity.

32. State and explain Brewster’s law. Show that when unpolarised light is incident at the polarising angle the reflected and refracted rays are perpendicular to each other.

Answer: Brewster’s law: $\tan\theta_B = n$, where $\theta_B$ is the polarising angle for the medium of refractive index $n$. By Snell’s law $\sin\theta_B/\sin\theta_r = n = \sin\theta_B/\cos\theta_B$, giving $\sin\theta_r = \cos\theta_B = \sin(90^\circ-\theta_B)$, hence $\theta_r = 90^\circ – \theta_B$, i.e. $\theta_B+\theta_r = 90^\circ$. The reflected and refracted rays are therefore mutually perpendicular at the polarising angle.

MCQs (1 mark each)

1. The shape of the wavefront from a point source is — (a) plane (b) cylindrical (c) spherical (d) elliptical.

2. In Young’s double-slit experiment, fringe width is — (a) $\lambda d/D$ (b) $\lambda D/d$ (c) $D/(\lambda d)$ (d) $d/(\lambda D)$.

3. The intensity at the central maximum of the interference pattern of two equal-amplitude coherent sources of intensity $I_0$ is — (a) $I_0$ (b) $2I_0$ (c) $4I_0$ (d) $I_0/4$.

4. Two coherent sources of light must have — (a) the same amplitude (b) a constant phase difference (c) different frequencies (d) different polarisations.

5. The angular width of the central maximum in a single-slit diffraction pattern of width $a$ is — (a) $\lambda/a$ (b) $2\lambda/a$ (c) $\lambda/(2a)$ (d) $a/\lambda$.

6. The first dark fringe in a single-slit diffraction occurs at — (a) $a\sin\theta = \lambda/2$ (b) $a\sin\theta = \lambda$ (c) $a\sin\theta = 2\lambda$ (d) $a\sin\theta = 3\lambda/2$.

7. Light reflected from a transparent dielectric is completely plane-polarised at — (a) $0^\circ$ (b) the critical angle (c) the Brewster angle (d) $90^\circ$.

8. The polarising angle for a glass of refractive index $1.732$ is — (a) $30^\circ$ (b) $45^\circ$ (c) $56^\circ$ (d) $60^\circ$.

9. Malus’s law: when angle between polariser and analyser axes is $60^\circ$, the transmitted intensity is — (a) $I_0$ (b) $I_0/2$ (c) $I_0/4$ (d) zero.

10. The resolving power of a telescope of objective diameter $D$ for light of wavelength $\lambda$ is proportional to — (a) $\lambda/D$ (b) $D/\lambda$ (c) $D\lambda$ (d) $1/(D\lambda)$.

11. If the slit separation $d$ in YDSE is doubled and screen distance $D$ is halved, the fringe width becomes — (a) twice (b) half (c) one-fourth (d) unchanged.

12. Polarisation establishes the — (a) longitudinal nature of light (b) wave nature of light (c) transverse nature of light (d) particle nature of light.

13. The fringe width of YDSE in air is $\beta$. When the apparatus is fully immersed in water ($n=4/3$), the fringe width becomes — (a) $4\beta/3$ (b) $3\beta/4$ (c) $\beta$ (d) $\beta/2$.

14. In Huygens’s wave theory of light, every point on a wavefront acts as — (a) a sink (b) a node (c) a source of secondary wavelets (d) an antinode.

15. Two unpolarised light beams of intensity $I_0$ each pass through a single polariser. The transmitted intensity is — (a) zero (b) $I_0$ (c) $I_0$ (since each becomes $I_0/2$, total $I_0$) (d) $2I_0$.

Fill in the blanks

• The fringe width in YDSE is given by $\beta = $ ____. ($\lambda D/d$)
• For destructive interference, path difference is ____. (an odd multiple of $\lambda/2$)
• Two sources are coherent if their ____ is constant. (phase difference)
• The angular width of the central maximum in single-slit diffraction equals ____. ($2\lambda/a$)
• Malus’s law states $I = I_0$ ____. ($\cos^2\theta$)
• Brewster’s law: $\tan\theta_B$ = ____. ($n$)
• The reflected and refracted rays at the Brewster angle are ____. (perpendicular)
• For light incident on glass ($n=1.5$) the polarising angle is ____. ($\approx 56.3^\circ$)
• The wavefront emitted by a distant star, intercepted by Earth, is essentially ____. (plane)
• An ordinary tube-light cannot give an interference pattern with another tube-light because the two are not ____. (coherent)

True / False

• Light from two independent sodium lamps can interfere. (False)
• Frequency of light remains unchanged on entering a denser medium. (True)
• The fringe width in YDSE depends on the order of the fringe. (False)
• Single-slit diffraction has a central maximum twice as wide as the other maxima. (True)
• Sound waves can be polarised. (False)
• At the polarising angle, the reflected light is plane-polarised parallel to the plane of incidence. (False — perpendicular)
• Resolving power of a microscope increases with the wavelength of light used. (False)
• Energy is conserved in interference; it is redistributed from dark fringes to bright fringes. (True)
• Polaroid sun-glasses cut horizontally polarised glare. (True)
• Huygens’s principle can derive the law of reflection but not Snell’s law. (False — it derives both)

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Worked Numerical Problems

N1. In a YDSE, $d=1$ mm, $D=1$ m, $\lambda=600$ nm. Find the fringe width and the position of the third bright fringe.

$$\beta = \dfrac{\lambda D}{d} = \dfrac{600\times10^{-9}\times 1}{1\times10^{-3}} = 6.0\times10^{-4}\ \text{m} = 0.6\ \text{mm}.$$

$$y_3 = 3\beta = 1.8\ \text{mm}.$$

N2. Light of wavelength $589$ nm is incident on a slit of width $0.1$ mm. The screen is $1$ m away. Find the linear width of the central maximum.

$$W_0 = \dfrac{2\lambda D}{a} = \dfrac{2\times 589\times10^{-9}\times 1}{0.1\times10^{-3}} = 1.178\times10^{-2}\ \text{m} = 11.78\ \text{mm}.$$

N3. The polarising angle of a medium is $60^\circ$. Find its refractive index and angle of refraction.

$$n = \tan 60^\circ = \sqrt{3} = 1.732,\qquad \theta_r = 90^\circ – 60^\circ = 30^\circ.$$

N4. Plane-polarised light of intensity $32$ W m$^{-2}$ is incident on a polariser oriented at $30^\circ$ to the polarisation direction. Find the transmitted intensity.

$$I = I_0\cos^2 30^\circ = 32\times \tfrac{3}{4} = 24\ \text{W m}^{-2}.$$

N5. Light of wavelength $500$ nm is used in a YDSE with $d=0.5$ mm. The first minimum is observed at $y=0.5$ mm from the centre. Find $D$.

$$y = \dfrac{\lambda D}{2d}\;\Rightarrow\; D = \dfrac{2yd}{\lambda} = \dfrac{2\times 0.5\times10^{-3}\times 0.5\times10^{-3}}{500\times10^{-9}} = 1.0\ \text{m}.$$

N6. A telescope has objective diameter $20$ cm. Find its angular resolving power for $\lambda=550$ nm.

$$\theta_{\min} = \dfrac{1.22\lambda}{D} = \dfrac{1.22\times 550\times10^{-9}}{0.20} = 3.36\times10^{-6}\ \text{rad}.$$

N7. Two coherent sources have intensities in the ratio $9:1$. Find the ratio of maximum to minimum intensities in the interference pattern.

$$\dfrac{I_{\max}}{I_{\min}} = \left(\dfrac{\sqrt{I_1}+\sqrt{I_2}}{\sqrt{I_1}-\sqrt{I_2}}\right)^2 = \left(\dfrac{3+1}{3-1}\right)^2 = 4.$$

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Glossary / Key Terms

Term	Meaning
Wavefront	Locus of points oscillating in the same phase.
Ray	Perpendicular drawn on a wavefront in the direction of propagation.
Huygens’s principle	Every point on a wavefront is a secondary source; the new wavefront is the forward envelope of the secondary wavelets.
Coherent sources	Sources that emit waves of the same frequency with a constant phase difference.
Interference	Modification of intensity from superposition of waves from two coherent sources.
Constructive interference	Path difference $= n\lambda$; intensity is maximum.
Destructive interference	Path difference $= (n+1/2)\lambda$; intensity is minimum (zero for equal amplitudes).
Fringe width ($\beta$)	Distance between consecutive bright (or dark) fringes; $\beta = \lambda D/d$.
Diffraction	Bending of light around obstacles/apertures and the resulting intensity distribution.
Single-slit minima	$a\sin\theta = n\lambda$, $n=\pm1,\pm2,\ldots$
Central maximum	The brightest, widest fringe in single-slit diffraction; angular width $2\lambda/a$.
Resolving power	Ability of an optical instrument to separate two close objects/wavelengths.
Rayleigh’s criterion	Two sources are just resolved when the centre of one Airy pattern falls on the first minimum of the other.
Polarisation	Restriction of $\vec{E}$ vibrations of a transverse wave to a single direction.
Polariser / analyser	An optical element that transmits light vibrations along one direction only.
Malus’s law	$I = I_0\cos^2\theta$ for plane-polarised light through an analyser.
Brewster angle ($\theta_B$)	Angle of incidence at which reflected light is fully plane-polarised; $\tan\theta_B = n$.
Plane of polarisation	Plane containing the direction of propagation and perpendicular to $\vec{E}$.
Doppler shift (light)	$\Delta\lambda/\lambda = v/c$ (for $v\ll c$, recession giving red shift).
Fresnel distance	$z_F = a^2/\lambda$, the distance up to which ray optics is a good approximation for an aperture $a$.

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Continue revising by re-deriving each result from a blank page — first Huygens’s construction for refraction, then YDSE fringe-width, then the single-slit minima condition. Make sure you can sketch a labelled YDSE and a single-slit intensity pattern from memory, and write down Malus’s and Brewster’s laws together with the Brewster geometry $\theta_B+\theta_r=90^\circ$. Best wishes from HSLC GURU for your ASSEB Class 12 Physics board examination!