JEE Physics Wave Optics Notes Set 04

Superposition, Intensity and Visibility

Two waves of intensities \( I_1,I_2 \) with phase difference \( \phi \) give \( I=I_1+I_2+2\sqrt{I_1I_2}\cos\phi \). Maximum \( (\sqrt{I_1}+\sqrt{I_2})^2 \) at \( \phi=2n\pi \); minimum \( (\sqrt{I_1}-\sqrt{I_2})^2 \) at \( \phi=(2n+1)\pi \).
Path difference \( \Delta \) and phase difference are linked by \( \phi=\frac{2\pi}{\lambda}\Delta \). For equal slits \( I=I_{0}\cos^2\frac{\phi}{2} \), so a point with half the maximum intensity has \( \Delta=(2n+1)\frac{\lambda}{4} \) and one with three quarters has \( \Delta=(n\pm\tfrac16)\lambda \).
Visibility \( V=\frac{I_{max}-I_{min}}{I_{max}+I_{min}}=\frac{2\sqrt{I_1I_2}}{I_1+I_2} \) depends only on the intensity ratio and equals 1 for equal slits. Only coherent sources (constant phase difference) give stable fringes; for incoherent sources the intensities simply add.

Young's Double Slit Experiment: Standard Results

Bright fringes at \( y_n=\frac{n\lambda D}{d} \), dark fringes at \( y_n=\left(n-\tfrac12\right)\frac{\lambda D}{d} \), fringe width \( \beta=\frac{\lambda D}{d} \). These small-angle forms need \( d\ll D \); for a large angle use \( d\sin\theta=n\lambda \) and \( y=D\tan\theta \).
Immersing the set-up in a liquid of index \( \mu \) reduces \( \beta \) to \( \beta/\mu \). With white light the central fringe is white, the first maximum on each side is violet at the inside edge and red at the outside edge, and no point is completely dark.
Two wavelengths: the fringes coincide where \( n_1\lambda_1=n_2\lambda_2 \); reduce the ratio of the wavelengths to the smallest integers to find the nearest coincidence.
The maximum order is limited by \( n\le d/\lambda \): if \( d=\lambda \) only the central maximum is seen.

Slabs, Oblique Incidence and Source Position

A thin sheet of index \( \mu \) and thickness t in front of one slit adds an optical path \( (\mu-1)t \) and shifts the whole pattern, without changing the fringe width, by \( \frac{(\mu-1)tD}{d} \) toward the slit that carries the sheet. The number of fringes crossing a point is \( (\mu-1)t/\lambda \).
Oblique incidence at angle \( \theta \): the extra path before the slits is \( d\sin\theta \), so the central maximum shifts to the opposite side where \( d\sin\theta=d\sin\alpha \) for the screen direction \( \alpha \). The condition for the n-th maximum is \( d(\sin\theta+\sin\alpha)=n\lambda \).
If the whole arrangement is placed in a medium, the wavelength in it is used. If a point source is displaced sideways by s at distance L from the slits, the pattern shifts the opposite way by \( \frac{sD}{L} \).
Lloyd's mirror: the slit and its mirror image act as two sources; reflection adds a phase change of \( \pi \), so the region at the mirror plane is dark and the bright and dark fringes are interchanged.

Thin Films

Reflected light from a film of index \( \mu \) and thickness t at normal incidence has path difference \( 2\mu t \). When exactly one of the two reflections has a phase change of \( \pi \) (for example a soap film in air), reflection is strong for \( 2\mu t=\left(n+\tfrac12\right)\lambda \) and weak for \( 2\mu t=n\lambda \). The transmitted light has the complementary pattern.
A coating whose index is higher than that of the glass (for example 1.8 on glass of 1.5, in air) has a phase change at the top surface only, so reflection is strong for \( 2\mu t=\left(n+\tfrac12\right)\lambda \). A coating of index between that of air and glass (an anti-reflection layer) has a phase change at both surfaces, so reflection is weak for \( 2\mu t=\left(n+\tfrac12\right)\lambda \).

Diffraction: Fresnel and Fraunhofer

Diffraction is the bending of waves around the edge of an obstacle or aperture into the geometrical shadow. It is noticeable only when the size of the obstacle is comparable to the wavelength; sound (wavelength from mm to metres) diffracts easily and light (about \( 10^{-7} \) m) does not in daily life.
Fresnel diffraction: source or screen (or both) at finite distance; wavefronts are spherical or cylindrical; no lenses needed; the centre may be bright or dark. Fraunhofer diffraction: source and screen effectively at infinity (lenses used); plane wavefronts; the centre of the pattern is always bright.
Single slit of width a: minima at \( a\sin\theta=n\lambda \), secondary maxima near \( a\sin\theta=\left(n+\tfrac12\right)\lambda \); the central maximum has width \( \frac{2\lambda D}{a} \). Narrowing the slit or increasing the wavelength widens the pattern.
Resolution: the limit for a circular aperture of diameter D is \( \Delta\theta=\frac{1.22\lambda}{D} \). A telescope resolves better with a larger objective; a microscope resolves better with shorter wavelength.

Polarisation

Light is a transverse wave, so its electric vector can be confined to a single plane (plane polarised light). Unpolarised light has electric vectors in all directions in the plane perpendicular to the ray. An ideal polaroid transmits half of the intensity of unpolarised light, \( I=\frac{I_0}{2} \).
Law of Malus: plane polarised light of intensity \( I_0 \) passing through an analyser at angle \( \theta \) emerges with \( I=I_0\cos^2\theta \). Crossed polaroids transmit nothing; a third polaroid inserted between two crossed ones at \( 45^\circ \) restores \( I_0/8 \) of the initial plane polarised intensity.
Polarisation by reflection: at the Brewster angle \( \tan\theta_B=\mu \) the reflected light is completely polarised with its electric vector perpendicular to the plane of incidence, and the reflected and refracted rays are at \( 90^\circ \). Scattering of sunlight by air molecules also polarises the light from the clear blue sky.
Uses: polaroid sun glasses to cut glare, 3-D projection, stress analysis in transparent solids, LCD displays and the determination of the optical activity of sugar solutions.

Worked Examples

Example 1. In a YDSE d = 1 mm, D = 1 m and \( \lambda=600 \) nm. A sheet of index 1.5 and thickness 12 \( \mu \)m is placed over one slit. Find the shift of the central maximum and the number of fringes by which the pattern shifts.
Answer:
Extra optical path \( (1.5-1)\times12\ \mu\text{m}=6\ \mu\text{m} \).
Shift \( =\frac{6\times10^{-6}\times1}{10^{-3}}=6\ \text{mm} \); fringe width \( \beta=0.6\ \text{mm} \), so the pattern shifts by 10 fringes.

Example 2. Light of intensity I falls on a polaroid A, then on a polaroid C at 30 degrees to A, then on a polaroid B at 90 degrees to A. Find the final intensity if the incident light is unpolarised.
Answer:
After A: \( \frac I2 \). After C: \( \frac I2\cos^230^\circ=\frac{3I}{8} \).
B is at 60 degrees to C: \( \frac{3I}{8}\cos^260^\circ=\frac{3I}{32} \).

Example 3. The Brewster angle for a transparent liquid is 53 degrees. Find the refractive index and the angle of refraction at that incidence.
Answer:
\( \mu=\tan53^\circ=1.33 \).
At the Brewster angle \( r=90^\circ-53^\circ=37^\circ \).