Wave Optics

1. Check this before you revise anything

The "Additional Exercises" section has been removed from this chapter, as from all 14 chapters of the current Class 12 Physics book. Only 6 questions remain, making this the smallest exercise set in the book.

Exercise 10.6 is printed without the data it needs. It asks for "the distance of the third bright fringe on the screen from the central maximum" but supplies only the two wavelengths — no slit separation and no screen distance. We rendered the page at 300 dpi to confirm this is genuinely how it is printed, not a text-extraction fault.

Earlier editions stated $d=2$ mm and $D=1.2$ m. Part (b), which asks which fringes coincide, is fully answerable as printed, because it depends only on the ratio of the wavelengths. Our solution gives the symbolic answer first and then evaluates it with the older data.

Four topics have been removed:

• Brewster's law — zero hits. Malus' law, $I=I_0{\cos }^2\theta$, is retained.
• Resolving power of optical instruments — zero hits.
• Fresnel distance and the validity of ray optics — zero hits.
• Polarisation by scattering — zero hits. Combined with the removal of scattering from Chapter 9, why the sky is blue is now explained nowhere in the book.

The fringe width is never named. The chapter gives the fringe positions, $x_n=n\lambda D/d$ for bright fringes and $\left(n+\frac{1}{2}\right)\lambda D/d$ for dark ones, and notes only that they are "equally spaced". The separation $\beta =\lambda D/d$ — the single most-used quantity in every double-slit numerical — is never written as a formula of its own. Derive it once from Eq. 10.13 and keep it.

2. Huygens Principle and Wavefronts (Textbook 10.2 to 10.3)

A wavefront is the surface joining all points that are in the same phase. Its shape follows the symmetry of the source: spherical near a point source, and plane at a large distance, since a small patch of a huge sphere is indistinguishable from a flat surface.

Huygens principle states that every point on a wavefront acts as a secondary source of spherical wavelets, and the new wavefront is the surface tangent to all of them.

What this explains. Applying the construction at a boundary reproduces both laws of refraction and reflection. In particular, requiring the wavefronts to stay continuous across the surface gives:

$\frac{\sin i}{\sin r}=\frac{v_1}{v_2}=n_{21}$

This settled a two-century argument. The corpuscular model predicted that light bends towards the normal because it speeds up in the denser medium. The wave model predicts it bends towards the normal because it slows down. Measurement confirmed the wave model.

Frequency is the invariant. On crossing a boundary the frequency is fixed by the source and cannot change. The speed changes to $c/n$, so the wavelength must change in the same ratio, ${\lambda }^{\prime }=\lambda /n$. This single fact answers Exercise 10.1 completely, and it is why an object does not change colour underwater.

3. Coherence and Young's Experiment (Textbook 10.4 to 10.5)

Two sources are coherent if they maintain a constant phase difference over time. Two independent lamps never manage this: an ordinary source emits in bursts with abrupt phase changes about every $10^{-10}$ s, so any interference pattern would be washed out long before the eye could register it.

Young's solution was to derive both beams from a single source by passing light through two closely spaced slits. Whatever the phase of the parent wave does, both slits follow it together, so their phase difference stays locked.

Path difference and the fringe positions. For slits separated by $d$ with a screen at distance $D$, the path difference to a point $x$ from the centre is $xd/D$. Constructive interference needs this to be a whole number of wavelengths:

$\text{Bright: }{x}_n=\frac{n\lambda D}{d}\text{Dark: }{x}_n=\left(n+\frac{1}{2}\right)\frac{\lambda D}{d}$

The fringe separation follows immediately by subtracting consecutive bright positions:

$\beta =\frac{\lambda D}{d}$

Fringes are therefore equally spaced, wider for longer wavelengths and for a larger screen distance, and narrower for more widely separated slits.

Intensity. For two waves of equal amplitude:

$I=4I_0{\cos }^2\left(\frac{\varphi }{2}\right),\varphi =\frac{2\pi }{\lambda }\times \text{path difference}$

The maximum is $4I_0$, not $2I_0$: amplitudes add, not intensities. Exercise 10.5 turns on converting a path difference of $\lambda /3$ into a phase of $120^{\circ }$, which drops the intensity to a quarter of the maximum.

Coincidence of two wavelengths. Fringes from two colours overlap where $n_1{\lambda }_1=n_2{\lambda }_2$, which needs no geometry at all — only the ratio of the wavelengths, as Exercise 10.6(b) shows.

4. Diffraction at a Single Slit (Textbook 10.6)

Diffraction is the bending of light into the geometrical shadow, and a single slit of width $a$ produces a pattern with a broad central maximum flanked by much weaker secondary maxima.

The minima, counter-intuitively, are given by:

$a\sin \theta =n\lambda ,n=\pm 1,\pm 2,\dots$

Note that this condition marks dark fringes, whereas the corresponding relation in the double-slit case marks bright ones. Mixing the two up is the standard error.

The central maximum is twice as wide as the others, spanning from $-\lambda /a$ to $+\lambda /a$ in angle, with half-angular width $\lambda /a$ and linear width $2\lambda D/a$ on a screen.

Narrower slits spread the light more. Since the width goes as $1/a$, closing the slit widens the pattern. In the limit of a very wide slit the pattern shrinks to the geometrical image, which is why ray optics works at all for ordinary apertures.

Interference against diffraction:

5. Polarisation (Textbook 10.7)

Interference and diffraction show that light is a wave; polarisation shows further that it is a transverse one. Longitudinal waves such as sound cannot be polarised at all.

Unpolarised light has its electric field vibrating in every direction perpendicular to propagation. A polaroid transmits only the component along its pass axis, so the emerging light is plane polarised and its intensity is halved.

Malus' law governs what happens at a second polaroid:

$I=I_0{\cos }^2\theta$

where $\theta$ is the angle between the two pass axes and $I_0$ is the intensity after the first polaroid.

Reading the two extremes. With the axes parallel, $\theta =0$ and all the light passes. With them crossed at $90^{\circ }$, ${\cos }^{2}90^{\circ }=0$ and the transmitted intensity falls to zero. Rotating one polaroid through a full turn therefore produces two maxima and two complete extinctions.

Note on what is no longer here. Brewster's law, which gives the polarising angle through $\tan i_B=n$, has been removed from this edition, as has polarisation by scattering. Both appear in older guides and in many practice papers.

Summary

• A wavefront joins points of equal phase: spherical near a point source, plane from a distant one.
• Huygens principle: every point on a wavefront is a source of secondary wavelets, and their tangent surface is the new wavefront.
• The construction gives $\frac{\sin i}{\sin r}=\frac{v_1}{v_2}=n_{21}$, and predicts light slows in a denser medium — confirming the wave model over the corpuscular one.
• Frequency never changes at a boundary; the speed becomes $c/n$ and the wavelength $\lambda /n$.
• Coherent sources keep a constant phase difference; two independent lamps cannot, because phase jumps every $10^{-10}$ s.
• Young derived both beams from one source, locking their phase difference.
• Bright fringes at $x_n=n\lambda D/d$, dark at $\left(n+\frac{1}{2}\right)\lambda D/d$.
• Fringe separation $\beta =\lambda D/d$ — the chapter never states this, so derive it from the positions.
• $I=4I_0{\cos }^2(\varphi /2)$ with $\varphi =\frac{2\pi }{\lambda }\times$ path difference; the maximum is $4I_0$ because amplitudes add, not intensities.
• Two wavelengths coincide where $n_1{\lambda }_1=n_2{\lambda }_2$, which needs no geometry.
• Single-slit minima at $a\sin \theta =n\lambda$ — this marks dark fringes, the opposite of the double-slit condition.
• The central diffraction maximum is twice as wide as the others, of angular half-width $\lambda /a$.
• Narrower slits spread the pattern more, since the width varies as $1/a$.
• Interference gives equally spaced fringes of equal intensity; diffraction gives a dominant central maximum with rapidly weakening side maxima.
• Polarisation proves light is transverse; a single polaroid halves the intensity of unpolarised light.
• Malus' law $I=I_0{\cos }^2\theta$, giving zero through crossed polaroids.
• Brewster's law, resolving power, Fresnel distance and polarisation by scattering have been removed; with scattering also gone from Chapter 9, the blue sky is explained nowhere in the book.
• Exercise 10.6 is printed without the slit separation or screen distance that part (a) requires.