JEE Physics Wave Optics Notes Set 03

Wavefronts and Huygens' Principle

A wavefront is a surface on which all points oscillate in the same phase. A point source gives spherical wavefronts, a distant source gives plane wavefronts, and the rays are perpendicular to the wavefronts.
Huygens' principle: every point on a wavefront acts as a source of secondary wavelets that spread in all directions with the speed of the wave. The envelope of these wavelets after a time t is the new wavefront. It explains reflection and refraction and gives \( \frac{\sin i}{\sin r}=\frac{v_1}{v_2} \), with speed and wavelength smaller in the denser medium while the frequency remains unchanged.
The principle is a geometrical construction. It does not account for polarisation or for the origin of spectra.

Superposition and Interference

When two waves overlap, the resultant displacement is the sum of the individual displacements. For two sources of intensities \( I_1 \) and \( I_2 \) with phase difference \( \phi \): \( I=I_1+I_2+2\sqrt{I_1I_2}\cos\phi \).
Maximum intensity \( (\sqrt{I_1}+\sqrt{I_2})^2 \) when \( \phi=2n\pi \) (path difference \( n\lambda \)); minimum intensity \( (\sqrt{I_1}-\sqrt{I_2})^2 \) when \( \phi=(2n+1)\pi \) (path difference \( (2n+1)\frac\lambda2 \)). Phase difference and path difference are related by \( \phi=\frac{2\pi}{\lambda}\Delta x \).
For a stable pattern the sources must be coherent: same frequency and a constant phase difference. Two independent ordinary sources are not coherent, so coherent sources are obtained by dividing a single wavefront (Young's slits, Fresnel's biprism) or a single beam's amplitude (thin films).
If the two intensities are equal, \( I=4I_0\cos^2\frac\phi2 \), with minimum zero and maximum \( 4I_0 \); covering one slit leaves a quarter of the central intensity.

Young's Double Slit Experiment

Two narrow slits a distance d apart, illuminated by a single source, act as coherent sources. On a screen at a distance D (with D much larger than d), the path difference at a point at height y from the centre is \( \Delta x=\frac{yd}{D} \).
Bright fringes: \( y_n=\frac{n\lambda D}{d} \). Dark fringes: \( y_n=\left(n-\tfrac12\right)\frac{\lambda D}{d} \). Fringe width \( \beta=\frac{\lambda D}{d} \), the same for bright and dark fringes.
The fringe width grows with D and \( \lambda \) and falls as d grows. In a medium of index \( \mu \) it reduces to \( \frac{\beta}{\mu} \), but the central fringe stays on the perpendicular bisector. The number of orders that can be seen is limited by \( \sin\theta\le1 \), so \( |n|\le\frac d\lambda \).
A thin transparent sheet of thickness t and index \( \mu \) over one slit adds an optical path \( (\mu-1)t \) and shifts the whole pattern towards that slit by \( \frac{(\mu-1)tD}{d} \), which is \( \frac{(\mu-1)t}{\lambda} \) fringes. Moving the source upwards shifts the pattern downwards.
With white light the central fringe is white (zero path difference for every colour), with violet nearest to the centre in each order and red farthest, and the fringes soon wash out. Intensity at a point with phase difference \( \phi \) is \( I_0\cos^2\frac\phi2 \), where \( I_0 \) is the central intensity.

Thin Film Interference

In reflected light from a film of thickness t and index \( \mu \), the path difference between the rays reflected from the two surfaces is \( 2\mu t\cos r \). A wave reflected from a denser medium undergoes a phase change of \( \pi \) (an extra \( \frac\lambda2 \)).
When only one of the two reflections involves a phase change (a film in air, for example), bright fringes need \( 2\mu t\cos r=\left(n+\tfrac12\right)\lambda \) and dark fringes need \( 2\mu t\cos r=n\lambda \). A film that is very thin compared with the wavelength therefore looks dark in reflection. If both reflections or neither give a phase change (a coating of intermediate index on glass), the conditions swap.
Reflected and transmitted patterns are complementary. Newton's rings (an air film between a convex lens and a flat plate) are circular, with a dark centre in reflected light; ring radii vary as \( \sqrt n \), so the rings crowd together outwards.

Fresnel's Biprism

A biprism is two thin prisms joined base to base. It refracts the wavefront from a narrow slit into two beams that appear to come from two virtual coherent sources. This is division of wavefront.
If the slit is at distance a from the biprism and the screen at distance b, the prism angle is \( \alpha \) and its index is \( \mu \): the source separation is \( d=2a(\mu-1)\alpha \), the fringe width is \( \beta=\frac{\lambda(a+b)}{d} \), and fringes are seen only in the overlap region of width \( 2b(\mu-1)\alpha \).
A larger prism angle increases d and makes the fringes narrower; moving the screen away widens them. Using a convex lens between biprism and screen, the two lens positions give image separations \( d_1 \) and \( d_2 \), and \( d=\sqrt{d_1d_2} \).

Diffraction

Diffraction is the bending of light around obstacles and the spreading of light through apertures comparable in size with the wavelength. Unlike interference, the fringes of a single slit are unequal: the central maximum is brightest and twice as wide as the others.
For a single slit of width a, minima occur at \( a\sin\theta=n\lambda \) (n = 1, 2, 3, ...). On a screen at D, or in the focal plane of a lens of focal length f, the first minimum is at \( y=\frac{\lambda D}{a} \) and the central maximum has width \( \frac{2\lambda D}{a} \).
Resolution: two point sources are just resolved when the angular separation is \( \theta=\frac{1.22\lambda}{a} \) (Rayleigh criterion), where a is the aperture diameter. A larger aperture and a shorter wavelength resolve finer detail.

Polarisation

Light is a transverse wave, so its electric vector can be confined to one plane (plane polarised light). Longitudinal waves such as sound cannot be polarised. Natural light is an unpolarised mixture of all transverse directions.
Malus' law: a polaroid transmits \( I=I_0\cos^2\theta \) of plane polarised light whose plane makes an angle \( \theta \) with its axis. An ideal polaroid transmits half the intensity of unpolarised light. Crossed polaroids transmit nothing.
Brewster's law: reflected light is completely plane polarised when light is incident at the angle \( i_B \) with \( \tan i_B=\mu \). The reflected and refracted rays are then at right angles, \( i_B+r=90^\circ \), and the refracted ray is only partially polarised.
Other methods: scattering of sunlight by the atmosphere (the light scattered at 90 degrees is polarised), double refraction in crystals such as calcite, and selective absorption in a polaroid.

Worked Examples

Worked Example 1. In a Young's double slit experiment, d = 1 mm, D = 1.5 m and \( \lambda=600\ \text{nm} \). Find the fringe width and the position of the third bright and second dark fringes.
Answer:
\( \beta=\frac{\lambda D}{d}=\frac{6\times10^{-7}\times1.5}{10^{-3}}=0.9\ \text{mm} \).
Third bright: \( 3\beta=2.7\ \text{mm} \). Second dark: \( 1.5\beta=1.35\ \text{mm} \).

Worked Example 2. A thin sheet of \( \mu=1.5 \) and thickness 2 micron is placed over one slit of a double slit set-up using light of wavelength 500 nm. By how many fringes does the pattern shift?
Answer:
Extra path \( =(\mu-1)t=0.5\times2\times10^{-6}=10^{-6}\ \text{m} \).
Shift \( =\frac{10^{-6}}{5\times10^{-7}}=2 \) fringes, towards the slit covered by the sheet.

Worked Example 3. Two coherent sources have intensities I and 9I. Find the maximum and minimum intensities.
Answer:
Amplitudes are in the ratio 1 : 3, so \( I_{max}=(1+3)^2I=16I \) and \( I_{min}=(3-1)^2I=4I \).

Worked Example 4. A slit of width 0.2 mm is illuminated by light of wavelength 500 nm. Find the width of the central maximum on a screen 1 m away.
Answer:
Width \( =\frac{2\lambda D}{a}=\frac{2\times5\times10^{-7}\times1}{2\times10^{-4}}=5\ \text{mm} \).

Worked Example 5. Unpolarised light of intensity \( I_0 \) passes through two polaroids whose axes are at 60° to each other. Find the transmitted intensity. At what angle of incidence is light reflected from glass of index \( \sqrt3 \) completely polarised?
Answer:
After the first polaroid, \( \frac{I_0}{2} \). After the second, \( \frac{I_0}{2}\cos^260^\circ=\frac{I_0}{8} \).
Brewster's angle: \( \tan i_B=\sqrt3\Rightarrow i_B=60^\circ \).