Engineering Physics (BT-201) - Unit 2 Short Notes

How unit 2 is examined

This unit covers interference (superposition, Young, Newton's rings, Michelson, Mach-Zehnder) and diffraction (single slit, Rayleigh criterion, grating); Newton's rings and superposition carry the most marks.

Huygens' principle

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Definition. <mark>Every point on a wavefront acts as a source of secondary wavelets, and the new wavefront after time $t$ is the forward envelope of these wavelets.</mark>

Key points.

1. A wavefront is the locus of points vibrating in the same phase.
2. Each secondary wavelet spreads with the speed of the wave, so its radius after time $t$ is $vt$.
3. The tangent drawn to the wavelets in the forward direction gives the new wavefront.
4. The principle explains reflection, refraction, interference and diffraction, and it ignores the backward wave.

Superposition of waves and interference by wavefront and amplitude splitting

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Definition. ==Superposition principle: when two or more waves overlap in a medium, the resultant displacement is the algebraic sum of the individual displacements, $y = y_1 + y_2 + \dots + y_n$.== It holds for linear media. Interference is the redistribution of light energy when two coherent waves superpose, giving maxima and minima of intensity.

Formula. For $y_1 = a_1\sin\omega t$ and $y_2 = a_2\sin(\omega t + \phi)$:

$$y = R\sin(\omega t + \theta),\quad R^2 = a_1^2 + a_2^2 + 2a_1a_2\cos\phi,\quad I \propto R^2$$

Key points.

1. Constructive interference occurs when $\phi = 2n\pi$, i.e. path difference $\Delta = n\lambda$, giving $R_{max} = a_1 + a_2$ and $I_{max} \propto (a_1+a_2)^2$.
2. Destructive interference occurs when $\phi = (2n+1)\pi$, i.e. $\Delta = (2n+1)\lambda/2$, giving $R_{min} = |a_1 - a_2|$ and $I_{min} \propto (a_1-a_2)^2$.
3. Sustained fringes need coherent sources: the same frequency and a constant phase difference, so they are obtained from one source.
4. In wavefront splitting, a point source or slit is divided into two parts of the same wavefront, as in Young's slits and the Fresnel biprism.
5. In amplitude splitting, one beam is divided by partial reflection and transmission into two beams of lower amplitude, as in thin films, Newton's rings and the Michelson interferometer.
6. Energy is conserved in interference: it is redistributed from minima to maxima, not destroyed.
7. Superposition also underlies diffraction, beats and standing waves.

Answer frame. Open with the superposition principle and $y=y_1+y_2$; draw two waves and the resultant; derive $R$ then the two conditions in phase and path difference; for the difference question give the 6-row table with examples; close with "energy is redistributed, not lost".

Asked: [7 marks] (Jun 2022) What is superposition of waves? Explain constructive and destructive interference. Asked: [5 marks] (Dec 2023) Differentiate between division of amplitude and division of wavefront. Asked: [5 marks] (Dec 2023) Define superposition of waves. Asked: [7 marks] (Jun 2025) What is superposition of wave and interference of light by amplitude splitting?

Young's double slit experiment

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Definition. <mark>Light from a narrow slit $S$ falls on two close slits $S_1, S_2$ (separation $d$), which act as coherent sources and give alternate bright and dark fringes on a screen at distance $D$.</mark>

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Formula. Path difference at $P$: $\Delta = S_2P - S_1P = \dfrac{yd}{D}$.

$$\text{Bright: } y_n = \frac{n\lambda D}{d},\qquad \text{Dark: } y_n = \frac{(2n-1)\lambda D}{2d},\qquad \beta = \frac{\lambda D}{d}$$

Key points.

1. Bright fringes need $\Delta = n\lambda$ and dark fringes need $\Delta = (2n-1)\lambda/2$.
2. All fringes have the same width $\beta = \lambda D/d$, so they are equally spaced, straight and parallel to the slits.
3. Fringe width grows with $\lambda$ and $D$ and falls with $d$; white light gives a white centre with coloured edges.
4. The experiment proved the wave nature of light and gave a way to measure $\lambda$.

Answer frame. Open with the definition of interference and the need for coherent sources; draw the labelled figure; derive $\Delta = yd/D$ using $S_1P^2$ and $S_2P^2$ (so $S_2P^2 - S_1P^2 = 2yd$ and $S_1P + S_2P \approx 2D$); write maxima, minima, then $\beta$; close with significance.

Asked: [7 marks] (Dec 2023, Jun 2023) Discuss Young's double slit experiment on the basis of labelled diagram, formula used and significance; What is interference of light? Describe Young's experiment and derive the expression of fringe width.

Newton's rings

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Definition. <mark>Newton's rings are concentric circular interference fringes formed in the wedge-shaped air film between a plano-convex lens of large radius of curvature $R$ and a plane glass plate, seen in reflected light.</mark>

<figure class="ds-fig" style="margin:1.4rem 0;overflow-x:auto"><svg xmlns="http://www.w3.org/2000/svg" id="dsfig-u2-02" viewBox="0 0 424 252" width="424" height="252" role="img" aria-label="Newton's rings setup: Src monochromatic source, M glass plate at 45 degrees, L plano-convex lens on plane plate Gp, Mic microscope"><style>#dsfig-u2-02 .e{stroke:#454C5A;stroke-width:1.4;fill:none}#dsfig-u2-02 .e.hi{stroke:#2340B8;stroke-width:2.6}#dsfig-u2-02 .n{fill:#FFFFFF;stroke:#16181D;stroke-width:1.4}#dsfig-u2-02 .n.hi{fill:#E3E9FC;stroke:#2340B8;stroke-width:2.2}#dsfig-u2-02 .n.rb-b{fill:#16181D;stroke:#16181D}#dsfig-u2-02 .n.rb-r{fill:#BD3227;stroke:#BD3227}#dsfig-u2-02 text{font-family:"JetBrains Mono",ui-monospace,Menlo,Consolas,monospace;font-size:13px}#dsfig-u2-02 .t{fill:#16181D;font-weight:500}#dsfig-u2-02 .t.inv{fill:#FFFFFF;font-weight:700}#dsfig-u2-02 .kd{stroke:#16181D;stroke-width:1.2}#dsfig-u2-02 .dot{fill:#16181D}#dsfig-u2-02 .ann{fill:#2340B8;font-size:11px;font-weight:700}#dsfig-u2-02 .lbl{fill:#6F7787;font-family:system-ui,-apple-system,sans-serif;font-size:12px;font-weight:700}#dsfig-u2-02 .ptr{fill:#2340B8;font-size:12px;font-weight:700}#dsfig-u2-02 .ah{fill:#454C5A}#dsfig-u2-02 .ah.hi{fill:#2340B8}#dsfig-u2-02 .wl rect{fill:#FFFFFF;stroke:#DCE0E7}#dsfig-u2-02 .wl .t{font-size:12px;font-weight:700}#dsfig-u2-02 .wl.hi rect{fill:#2340B8;stroke:#2340B8}#dsfig-u2-02 .wl.hi .t{fill:#FFFFFF}html.dark #dsfig-u2-02 .e{stroke:#B1B7C3}html.dark #dsfig-u2-02 .e.hi{stroke:#8FA3FF}html.dark #dsfig-u2-02 .n{fill:#161920;stroke:#E6E8ED}html.dark #dsfig-u2-02 .n.hi{fill:#1E2748;stroke:#8FA3FF}html.dark #dsfig-u2-02 .n.rb-b{fill:#E6E8ED;stroke:#E6E8ED}html.dark #dsfig-u2-02 .n.rb-r{fill:#FF7E71;stroke:#FF7E71}html.dark #dsfig-u2-02 .t{fill:#E6E8ED}html.dark #dsfig-u2-02 .t.inv{fill:#0F1115}html.dark #dsfig-u2-02 .kd{stroke:#E6E8ED}html.dark #dsfig-u2-02 .dot{fill:#E6E8ED}html.dark #dsfig-u2-02 .ann{fill:#8FA3FF}html.dark #dsfig-u2-02 .lbl{fill:#858D9C}html.dark #dsfig-u2-02 .ptr{fill:#8FA3FF}html.dark #dsfig-u2-02 .ah{fill:#B1B7C3}html.dark #dsfig-u2-02 .ah.hi{fill:#8FA3FF}html.dark #dsfig-u2-02 .wl rect{fill:#161920;stroke:#2A2E37}html.dark #dsfig-u2-02 .wl.hi rect{fill:#8FA3FF;stroke:#8FA3FF}html.dark #dsfig-u2-02 .wl.hi .t{fill:#0F1115}</style><defs><marker id="ah2" viewBox="0 0 10 10" refX="9" refY="5" markerWidth="7" markerHeight="7" orient="auto-start-reverse"><path class="ah" d="M0,1 L9,5 L0,9 z"/></marker><marker id="ahh2" viewBox="0 0 10 10" refX="9" refY="5" markerWidth="7" markerHeight="7" orient="auto-start-reverse"><path class="ah hi" d="M0,1 L9,5 L0,9 z"/></marker></defs><path class="e" d="M40,59 L40,107"/><path class="e" d="M365.6,44.6 L232.4,77.9" marker-end="url(#ah2)"/><path class="e" d="M193.6,87.6 L60.4,120.9" marker-end="url(#ah2)"/><path class="e" d="M40,145 L40,191" marker-end="url(#ah2)"/><g class="wl"><rect x="-11.9" y="74" width="103.8" height="18" rx="9"/><text class="t" x="40" y="83" dy=".35em" text-anchor="middle">lens_on_plate</text></g><circle class="n" cx="40" cy="40" r="18"/><text class="t" x="40" y="40" dy=".35em" text-anchor="middle">L</text><circle class="n" cx="40" cy="126" r="18"/><text class="t" x="40" y="126" dy=".35em" text-anchor="middle">Gp</text><circle class="n" cx="212" cy="83" r="18"/><text class="t" x="212" y="83" dy=".35em" text-anchor="middle">M</text><circle class="n" cx="384" cy="40" r="18"/><text class="t" x="384" y="40" dy=".35em" text-anchor="middle">Src</text><circle class="n" cx="40" cy="212" r="18"/><text class="t" x="40" y="212" dy=".35em" text-anchor="middle">Mic</text></svg><figcaption style="font-size:.82em;opacity:.72;margin-top:.45rem">Newton's rings setup: Src monochromatic source, M glass plate at 45 degrees, L plano-convex lens on plane plate Gp, Mic microscope</figcaption></figure>

Formation. Light from the source is reflected by the plate at $45^\circ$ down onto the lens. Rays reflected from the top and bottom surfaces of the air film interfere. The film thickness is constant on a circle about the contact point, so the fringes are circles; the centre is dark because of the $\pi$ phase change on reflection at the plate.

Derivation. For a ring of radius $r$ at film thickness $t$: $r^2 = 2Rt - t^2 \approx 2Rt$, so $t = r^2/2R$. In reflected light (normal incidence) the condition is $2t = n\lambda$ for dark and $2t = (2n-1)\lambda/2$ for bright. Hence $r_n^2 = n\lambda R$, so for dark rings $D_n^2 = 4n\lambda R$, and for bright rings $D_n^2 = 2(2n-1)\lambda R$. Between ring $n$ and $n+p$:

$$\lambda = \frac{D_{n+p}^2 - D_n^2}{4pR}$$

Key points.

1. Dark ring diameter is proportional to $\sqrt n$, so the rings crowd together outward.
2. The centre is dark because of the $\pi$ phase change at the glass surface.
3. Using $D_{n+p}^2 - D_n^2$ removes the error in locating the centre.
4. The refractive index of a liquid is found from $\mu = (D_n^2)_{air}/(D_n^2)_{liquid}$.

Example. Given $D_n = 0.42$ cm, $D_{n+14} = 0.70$ cm, $R = 100$ cm, $p = 14$.

$\lambda = 5600\ \text{\AA}$.

Answer frame. Open with the definition and draw the setup; explain formation by the air film; derive $t = r^2/2R$, then $D_n^2 = 4n\lambda R$, then $\lambda$ using $n$ and $n+p$; close with the dark centre and $\sqrt n$ spacing. For the numerical, write formula, substitute, box the answer.

Asked: [7 marks] (Jun 2022, Nov 2022, Dec 2024) Explain the formation of Newton's rings. Obtain the expression for the wavelength of light; How are Newton's rings formed? Deduce the expression for diameter of dark and bright fringes. Asked: [5 marks] (Dec 2023) In Newton's ring method the diameters of the $n^{th}$ and $(n+14)^{th}$ rings are 0.42 cm and 0.70 cm. If $R$ = 100 cm, calculate the wavelength of light.

Michelson interferometer

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Definition. <mark>Michelson interferometer is an amplitude-splitting instrument in which a beam is divided by a beam splitter into two perpendicular beams that are reflected back by two mirrors and recombined to give interference fringes.</mark>

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Key points.

1. Light from an extended source falls on $G_1$, a half-silvered glass plate at $45^\circ$, which splits it into a reflected beam going to $M_2$ and a transmitted beam going to $M_1$.
2. Both beams return, recombine at $G_1$ and enter the telescope, where they interfere because they come from the same source.
3. $G_2$ has the same thickness as $G_1$ and is placed in the path of the transmitted beam so that both beams pass through equal glass thickness, which makes the path difference independent of wavelength.
4. $M_1$ moves on a micrometer screw, so the path difference $2d$ can be varied, where $d$ is the distance between $M_1$ and the image of $M_2$.
5. If $M_1$ and $M_2$ are exactly perpendicular the fringes are circular (equal inclination); if slightly tilted the fringes are straight (equal thickness); with white light a few coloured fringes appear at $d = 0$.
6. When $M_1$ moves by $\lambda/2$, one fringe crosses the field, so $\lambda = 2x/N$ for a displacement $x$ giving $N$ fringes.

Answer frame. Open with the definition; draw the labelled figure; describe construction, then working, then fringe types; close with the formula $\lambda = 2x/N$.

Asked: [7 marks] (Jun 2022, Dec 2023, Jun 2025) Discuss the construction and working of Michelson interferometer; explain it on the basis of labelled diagram and types of fringes.

Mach-Zehnder interferometer

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Definition. <mark>Mach-Zehnder interferometer is an amplitude-splitting device in which a collimated beam is split by a first beam splitter into two separate paths that are recombined by a second beam splitter.</mark>

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Key points.

1. The set-up has a collimated source, two beam splitters $BS_1$ and $BS_2$, two fully reflecting mirrors $M_1$ and $M_2$, and detectors or a screen.
2. $BS_1$ divides the beam into a reference arm and a test arm, each of which reflects only once from its mirror.
3. The two beams recombine at $BS_2$, and the phase difference between the arms decides whether the output is bright or dark.
4. A sample of refractive index $n$ and thickness $t$ placed in one arm adds a path difference $(n-1)t$.
5. The two beams travel separate paths, so a sample can be placed in one arm without disturbing the other; this is the main advantage over the Michelson.
6. It gives two complementary outputs and is used to measure refractive index changes, in flow visualisation in aerodynamics and in optical modulators.

Answer frame. Open with the definition; draw the schematic with labelled arms; describe construction, then working, then path difference; close with applications.

Asked: [7 marks] (Nov 2022, Jun 2023, Jun 2025) Describe the construction and working of Mach-Zehnder interferometer; explain its working principle with suitable ray diagram.

Fraunhofer diffraction from a single slit and a circular aperture

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Definition. <mark>Fraunhofer diffraction is the bending of light at a slit of width $a$ when the source and screen are at infinite distance, using parallel incident and diffracted beams.</mark>

Key points.

1. Wavelets from the two edges have path difference $a\sin\theta$, so the phase difference is $2\beta = \dfrac{2\pi}{\lambda}a\sin\theta$ with $\beta = \dfrac{\pi a\sin\theta}{\lambda}$.
2. Adding the wavelet contributions by integration or vector polygon gives amplitude $R = A\dfrac{\sin\beta}{\beta}$, so the intensity is $I = I_0\left(\dfrac{\sin\beta}{\beta}\right)^2$.
3. The central maximum is at $\theta = 0$ where $I = I_0$; minima occur at $a\sin\theta = \pm m\lambda$ ($m = 1, 2, \dots$).
4. Secondary maxima lie near $\beta = \pm\dfrac{3\pi}{2}, \pm\dfrac{5\pi}{2}$ and have intensities about $4.5\%$ and $1.6\%$ of $I_0$.
5. For a circular aperture of diameter $D$ the first minimum is at $\sin\theta = 1.22\lambda/D$.

Asked: [7 marks] (Dec 2024) Explain the Fraunhofer diffraction due to single slit with necessary analysis.

Rayleigh criterion and its application to vision

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Definition. <mark>Rayleigh criterion: two point objects or spectral lines are just resolved when the central maximum of the diffraction pattern of one falls on the first minimum of the pattern of the other.</mark>

Key points.

1. Every image is a diffraction pattern, so two close objects give overlapping patterns.
2. In the just-resolved case the resultant intensity at the mid-point is about $81\%$ of either maximum, and the eye can detect this dip.
3. For a circular aperture of diameter $D$, the smallest resolvable angle is $\theta_{min} = 1.22\lambda/D$.
4. A larger aperture or a shorter wavelength gives better resolution; this is why telescopes are large and electron microscopes resolve finely.
5. For the human eye, with pupil $D \approx 2$ mm and $\lambda = 550$ nm, $\theta_{min} \approx 3.4\times10^{-4}$ rad, about $1'$ of arc.
6. For a grating the criterion fixes the resolving power $\lambda/d\lambda$.

Answer frame. Open with the statement of the criterion; draw the three intensity curves (well resolved, just resolved, unresolved) with the dip marked; then the $0.81$ point and $1.22\lambda/D$; close with the application to the eye and instruments.

Asked: [7 marks] (Dec 2023, Jun 2025) Explain the Rayleigh's criteria for resolving power; what is the importance of the Rayleigh criteria; explain the Rayleigh criterion for limit of resolution.

Diffraction gratings and their resolving power

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Definition. <mark>A diffraction grating is an optical component with a large number of equally spaced parallel slits ($a$ transparent, $b$ opaque) that splits light into spectra of different orders.</mark> The grating element is $a + b = 1/N'$, where $N'$ is the number of lines per unit length.

Formula. Grating equation: $(a+b)\sin\theta = n\lambda$ for the $n^{th}$ principal maximum.

Resolving power derivation. Resolving power is $\lambda/d\lambda$, where $\lambda$ and $\lambda + d\lambda$ are two lines just resolved. The $n^{th}$ maximum of $\lambda + d\lambda$ is at angle $\theta$: $(a+b)\sin\theta = n(\lambda + d\lambda)$. By the Rayleigh criterion, this must fall on the first minimum next to the $n^{th}$ maximum of $\lambda$, which for $N$ lines is at $(a+b)\sin\theta = \left(n + \dfrac{1}{N}\right)\lambda$. Equating, $n\,d\lambda = \lambda/N$, so

$$\frac{\lambda}{d\lambda} = nN$$

Key points.

1. The resolving power is proportional to the order $n$ and to the total number of lines $N$ illuminated.
2. A wider grating with more lines resolves closer lines, and higher orders resolve better but are fainter.
3. Resolving power does not depend directly on the grating element.

Example. $N' = 4250$ lines/cm, $n = 2$, $\theta = 30^\circ$.

$\lambda = 588\ \text{nm}$ ($5882\ \text{\AA}$).

Answer frame. Open with the definition and grating equation; state the Rayleigh criterion; derive $\lambda/d\lambda = nN$ using the $(n+1/N)$ minimum; add the factors (order and lines); for the numerical, write $a+b$ from $N'$, substitute and box.

Asked: [7 marks] (Nov 2022) Explain about the diffraction grating. A parallel beam of sodium light is incident on a plane transmission grating with 4250 lines per cm and a second-order line is seen at $30^\circ$. Find the wavelength. Asked: [7 marks] (Jun 2023) Explain Rayleigh's criterion of resolution. Derive an expression for the resolving power of a grating.

Last-minute revision

• Superposition: $y = y_1 + y_2$; $R^2 = a_1^2 + a_2^2 + 2a_1a_2\cos\phi$.
• Constructive: $\Delta = n\lambda$ ($\phi = 2n\pi$); destructive: $\Delta = (2n+1)\lambda/2$.
• Wavefront splitting: Young, biprism; amplitude splitting: Newton's rings, thin films, Michelson.
• Young: $\Delta = yd/D$, $\beta = \lambda D/d$.
• Newton's rings: $D_n^2 = 4n\lambda R$ (dark), $\lambda = (D_{n+p}^2 - D_n^2)/4pR$; centre dark.
• Newton's numerical answer: $5600$ \AA (0.42, 0.70 cm, $p=14$, $R=100$ cm).
• Michelson: division of amplitude, compensating plate $G_2$, $\lambda = 2x/N$.
• Mach-Zehnder: two beam splitters, two mirrors, separate arms, sample adds $(n-1)t$.
• Single slit: $I = I_0(\sin\beta/\beta)^2$, minima $a\sin\theta = m\lambda$.
• Rayleigh: peak on first minimum, dip $0.81$; $\theta_{min} = 1.22\lambda/D$.
• Grating: $(a+b)\sin\theta = n\lambda$; $\lambda/d\lambda = nN$; sodium answer $588$ nm.

Memory hooks

• Wavefront splitting = Young; Amplitude splitting = Newton and Michelson (thin films, "A for Aggregate of reflections").
• "Dark centre, root-n rings": Newton's rings have a dark centre and diameter $\propto \sqrt n$.
• Michelson has one compensating plate, Mach-Zehnder has two beam splitters and separate arms.
• Rayleigh: "peak on trough", the dip is 81%.
• Grating power: order times lines, $nN$.

Coverage checklist

• Huygens' principle: definition, secondary wavelets, envelope (no past question).
• superposition of waves and interference of light by wave front splitting and amplitude splitting: covers Jun 2022 superposition, Dec 2023 difference, Dec 2023 definition, Jun 2025 amplitude splitting.
• Young's double slit experiment: covers Jun 2023 and Dec 2023.
• Newton's rings: covers Jun 2022, Nov 2022, Dec 2024 derivation and Dec 2023 numerical.
• Michelson interferometer: covers Jun 2022, Dec 2023, Jun 2025.
• Mach-Zehnder interferometer: covers Nov 2022, Jun 2023, Jun 2025.
• Farunhofer diffraction from a single slit and a circular aperture: covers Dec 2024.
• the Rayleigh criterion for limit of resolution and its application to vision: covers Dec 2023 and Jun 2025.
• Diffraction gratings and their resolving power: covers Nov 2022 and Jun 2023.