Optics • Topic 3 of 3

Wave Optics

Ray optics fails when light meets obstacles or slits comparable to its wavelength. Wave optics treats light as an electromagnetic wave and explains interference, diffraction and polarisation. The foundation is Huygens' principle: every point on a wavefront acts as a source of secondary spherical wavelets, and the new wavefront is the forward envelope of these wavelets. This predicts both reflection and refraction (giving Snell's law) and shows that waves bend around edges.

Interference arises when two coherent waves (constant phase difference, same frequency) superpose. Where they arrive in phase the amplitude is maximum (constructive, bright fringe); where they arrive out of phase it is minimum (destructive, dark fringe). Energy is merely redistributed, never destroyed.

In Young's double-slit experiment, monochromatic light passes through two narrow slits separated by $d$, giving equally spaced fringes on a screen a distance $D$ away. The path difference at a point $y$ from the centre is $\Delta = \dfrac{y\,d}{D}$. Bright fringes occur for $\Delta = n\lambda$ (maxima) and dark fringes for $\Delta = (n+\tfrac{1}{2})\lambda$ (minima). The fringe width is $\beta = \dfrac{\lambda D}{d}$. Coherence is essential, which is why both slits are lit by the same source.

Diffraction is the bending and spreading of light around obstacles or through narrow openings. In single-slit diffraction of width $a$, the central maximum is bright and twice as wide as the side maxima; minima occur where $a \sin\theta = n\lambda$ ($n = 1, 2, 3, \dots$), and the angular half-width of the central maximum is $\theta \approx \dfrac{\lambda}{a}$. Diffraction sets the ultimate limit on the resolving power of optical instruments.

Polarisation demonstrates that light is a transverse wave — the electric field oscillates perpendicular to the direction of propagation. Ordinary light is unpolarised; a Polaroid transmits only one direction of vibration. When polarised light of intensity $I_0$ passes through an analyser whose axis makes angle $\theta$ with the polariser, the transmitted intensity follows Malus' law $I = I_0 \cos^2\theta$. Unpolarised light through a single Polaroid emerges with half its intensity, $I_0/2$.

Solved Examples

Worked Example

In a Young's double-slit experiment, two slits $0.5\,\text{mm}$ apart are illuminated by light of wavelength $600\,\text{nm}$. The screen is $1\,\text{m}$ away. Find the fringe width.

Solution

$\beta = \dfrac{\lambda D}{d}$ with $\lambda = 600\times10^{-9}\,\text{m}$, $D = 1\,\text{m}$, $d = 0.5\times10^{-3}\,\text{m}$.
$\beta = \dfrac{600\times10^{-9} \times 1}{0.5\times10^{-3}}$.
$\beta = \dfrac{6\times10^{-7}}{5\times10^{-4}} = 1.2\times10^{-3}\,\text{m}$.

Answer: The fringe width is $1.2\,\text{mm}$.

Worked Example

In a double-slit experiment the fringe width is $0.4\,\text{mm}$ for light of wavelength $500\,\text{nm}$ with slit separation $1\,\text{mm}$. Find the distance of the screen from the slits.

Solution

$\beta = \dfrac{\lambda D}{d} \Rightarrow D = \dfrac{\beta d}{\lambda}$.
$D = \dfrac{(0.4\times10^{-3})(1\times10^{-3})}{500\times10^{-9}}$.
$D = \dfrac{4\times10^{-7}}{5\times10^{-7}}$.
$D = 0.8\,\text{m}$.

Answer: The screen is $0.8\,\text{m}$ from the slits.

Worked Example

Plane-polarised light of intensity $I_0$ falls on an analyser whose axis is at $60^\circ$ to the plane of polarisation. Find the transmitted intensity.

Solution

Malus' law: $I = I_0 \cos^2\theta$ with $\theta = 60^\circ$.
$\cos 60^\circ = 0.5$, so $\cos^2 60^\circ = 0.25$.
$I = I_0 \times 0.25 = \dfrac{I_0}{4}$.

Answer: The transmitted intensity is $\dfrac{I_0}{4}$.

Worked Example

In a single-slit diffraction experiment, light of wavelength $600\,\text{nm}$ passes through a slit of width $0.2\,\text{mm}$. Find the angular position of the first minimum.

Solution

Minima condition: $a \sin\theta = n\lambda$; for the first minimum $n = 1$.
$\sin\theta = \dfrac{\lambda}{a} = \dfrac{600\times10^{-9}}{0.2\times10^{-3}}$.
$\sin\theta = \dfrac{6\times10^{-7}}{2\times10^{-4}} = 3\times10^{-3}$.
$\theta \approx 3\times10^{-3}\,\text{rad} \approx 0.17^\circ$.

Answer: The first minimum is at about $3\times10^{-3}\,\text{rad}$ ($\approx 0.17^\circ$) from the centre.

Worked Example

Unpolarised light of intensity $I_0$ passes through two Polaroids whose axes are at $30^\circ$ to each other. Find the final emergent intensity.

Solution

After the first Polaroid the intensity is halved: $I_1 = \dfrac{I_0}{2}$.
The second Polaroid applies Malus' law: $I_2 = I_1 \cos^2 30^\circ$.
$\cos^2 30^\circ = \left(\dfrac{\sqrt{3}}{2}\right)^2 = 0.75$.
$I_2 = \dfrac{I_0}{2} \times 0.75 = 0.375\,I_0 = \dfrac{3I_0}{8}$.

Answer: The emergent intensity is $\dfrac{3I_0}{8}$ ($0.375\,I_0$).

Worked Example

In a Young's double-slit experiment, the fringe width with light of wavelength $480\,\text{nm}$ is $0.96\,\text{mm}$. If the wavelength is changed to $600\,\text{nm}$, find the new fringe width (same geometry).

Solution

$\beta \propto \lambda$ since $D$ and $d$ are fixed: $\dfrac{\beta_2}{\beta_1} = \dfrac{\lambda_2}{\lambda_1}$.
$\beta_2 = \beta_1 \times \dfrac{\lambda_2}{\lambda_1} = 0.96 \times \dfrac{600}{480}$.
$\dfrac{600}{480} = 1.25$.
$\beta_2 = 0.96 \times 1.25 = 1.2\,\text{mm}$.

Answer: The new fringe width is $1.2\,\text{mm}$.

Key Points

Huygens' principle: each point on a wavefront is a source of secondary wavelets; their envelope is the new wavefront.
Young's double slit: maxima at $\Delta = n\lambda$, minima at $\Delta = (n+\tfrac{1}{2})\lambda$; fringe width $\beta = \dfrac{\lambda D}{d}$.
Coherent sources are essential for a stable interference pattern; interference redistributes energy.
Single-slit diffraction minima at $a\sin\theta = n\lambda$; central maximum is brightest and twice as wide.
Polarisation proves light is transverse; Malus' law $I = I_0\cos^2\theta$, and unpolarised light loses half its intensity through one Polaroid.

Quick Check — 4 MCQs

Tap an option to check your answer0 / 4

Q1.The fringe width in Young's double-slit experiment is given by:

Explanation: Consecutive bright fringes are separated by $\beta = \lambda D / d$.

Q2.Two sources can produce a sustained interference pattern only if they are:

Explanation: A stable pattern needs a constant phase difference, i.e. coherent sources of the same frequency.

Q3.When unpolarised light of intensity $I_0$ passes through a single ideal Polaroid, the emergent intensity is:

Explanation: Averaging $\cos^2\theta$ over all directions gives $1/2$, so the intensity becomes $I_0/2$.

Q4.In single-slit diffraction, the first minimum occurs when:

Explanation: Single-slit minima satisfy $a\sin\theta = n\lambda$; the first ($n=1$) gives $a\sin\theta = \lambda$.

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