Physics Lab

Unit 12: Optics, Modern Physics & Electronics

This is the broadest unit of JEE, combining ray optics, wave optics, photoelectric effect, atomic & nuclear physics, and semiconductor electronics — together worth 15–20 % of the paper. Expect:

  • JEE-Main: 1–2 ray-optics (lens, prism, total internal reflection), 1 wave-optics (Young's slits, single-slit diffraction), 1 photoelectric, 1 Bohr-model, 1 nuclear/radioactivity, 1 semiconductor/diode/logic gate.
  • JEE-Advanced: Multi-step optics (lens + mirror combo); photoelectric + de Broglie crossover; Bohr model with isotope/recoil correction.

This unit splits neatly into five blocks:

  1. Ray optics (geometric, classical).
  2. Wave optics (interference, diffraction, polarization).
  3. Photoelectric + de Broglie (the wave–particle bridge).
  4. Atoms, nuclei, radioactivity (Bohr → modern).
  5. Semiconductors and basic logic (the JEE electronics core).

Concept Map

RAY OPTICS
├── Mirrors: 1/v + 1/u = 1/f, m = -v/u
├── Refraction at plane / spherical surface
├── Lenses: lens-maker, 1/v − 1/u = 1/f
├── Combinations of thin lenses, power P = 1/f
├── Prism: A + D = i₁ + i₂; thin prism D = (n−1)A
├── TIR: sin θ_c = 1/n
└── Dispersion, scattering

WAVE OPTICS
├── Huygens construction
├── Young's: β = λD/d; intensity I = I_max cos²(πd sinθ/λ)
├── Single-slit diffraction: a sinθ = mλ
├── Resolving power
├── Thin films
└── Polarization: Malus I = I₀cos²θ, Brewster tan θ_B = n

MODERN
├── Photoelectric: hν = φ + KE_max
├── de Broglie: λ = h/p
├── Bohr model: E_n = −13.6 Z²/n² eV
├── Spectral series (Lyman, Balmer, …)
├── X-rays (Moseley): √f = a(Z − b)
├── Nuclear: ΔE = Δm c²
├── BE/A curve, fission, fusion
└── Radioactivity: N = N₀e^(−λt), t½, mean life

ELECTRONICS
├── Band theory: insulator / conductor / semiconductor
├── Intrinsic vs extrinsic (n, p)
├── pn-junction, forward / reverse I–V
├── Half-wave, full-wave rectifier
├── Zener (regulator)
└── Logic gates (NOT, AND, OR, NAND, NOR, XOR)

Topic 1: Ray Optics

Sub-topic A: Reflection at a Spherical Mirror — Mirror Formula

Sign convention (Cartesian): take pole as origin, optic axis along xx, light incident from the left as the positive direction. Distances measured from pole; positive in the direction of incident light.

Mirror formula:

1v+1u=2R=1f\boxed{\frac{1}{v} + \frac{1}{u} = \frac{2}{R} = \frac{1}{f}}

where f=R/2f = R/2 (concave mirror f>0f > 0 in many texts; sign depends on convention).

Lateral magnification: m=v/um = -v/u (negative when image is inverted).

Sub-topic B: Refraction at a Plane Surface (Snell's Law)

n1sinθ1=n2sinθ2n_1\sin\theta_1 = n_2\sin\theta_2

Apparent depth (object in medium 1, viewer in medium 2):

Apparent depth=Real depth×n2n1\text{Apparent depth} = \text{Real depth}\times \frac{n_2}{n_1}

Sub-topic C: Refraction at a Spherical Surface

Convention same as above. For a single refracting surface of radius RR:

n2vn1u=n2n1R\boxed{\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}}

Derivation sketch: Use paraxial geometry. Consider an object at distance uu in medium n1n_1 in front of a spherical surface of radius RR separating media n1,n2n_1, n_2. A small-angle ray traces from object to a point on the axis; applying Snell's law in linear approximation and using sign convention yields the result.

Sub-topic D: Thin Lens — Lens-maker Formula and Thin-Lens Equation

Apply the spherical-surface formula twice, once at each surface (radii R1,R2R_1, R_2). For the first surface light goes n1n2n_1\to n_2; for the second n2n1n_2\to n_1. Adding the two:

n1vn1u=(n2n1)(1R11R2)\frac{n_1}{v} - \frac{n_1}{u} = (n_2 - n_1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)

For n1=1n_1 = 1 (air), n2=nn_2 = n:

1v1u=(n1)(1R11R2)1f\boxed{\frac{1}{v} - \frac{1}{u} = (n - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) \equiv \frac{1}{f}}

This defines the focal length ff via the lens-maker formula:

1f=(n1)(1R11R2)\frac{1}{f} = (n - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)

For a biconvex lens both RR's contribute the same sign; for biconcave, the opposite.

Thin-lens equation:

1v1u=1f\frac{1}{v} - \frac{1}{u} = \frac{1}{f}

Magnification: m=v/um = v/u.

Sub-topic E: Lens Combinations and Power

For two thin lenses in contact:

1f=1f1+1f2\frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2}

Power: P=1/fP = 1/f (in diopters when ff in metres). For combinations:

P=P1+P2+P = P_1 + P_2 + \ldots

Sub-topic F: Prism

A ray entering and leaving a prism of angle AA at angles i1,i2i_1, i_2 (on each face) gives deviation δ=i1+i2A\delta = i_1 + i_2 - A.

Minimum deviation condition: i1=i2=ii_1 = i_2 = i. Then symmetric:

n=sin((A+δm)/2)sin(A/2)n = \frac{\sin((A + \delta_m)/2)}{\sin(A/2)}

Thin prism (AA small): deviation

δ=(n1)A\delta = (n - 1)A

Sub-topic G: Total Internal Reflection

Going from denser to lighter medium, beyond a critical angle there is no refraction — only reflection.

sinθc=n2/n1=1/n (light to denser)\sin\theta_c = n_2/n_1 = 1/n \text{ (light to denser)}

For water-air: θc=arcsin(1/1.33)48.6\theta_c = \arcsin(1/1.33)\approx 48.6^\circ.

Applications: optical fibres, sparkle of diamond (n=2.4n = 2.4, θc24.4\theta_c \approx 24.4^\circ).

Sub-topic H: Dispersion and Scattering

Dispersion: n(λ)n(\lambda), blue refracted more than red. Angular dispersion =(nvnr)A= (n_v - n_r)A for thin prism. Dispersive power ω=(nvnr)/(ny1)\omega = (n_v - n_r)/(n_y - 1).

Rayleigh scattering: intensity 1/λ4\propto 1/\lambda^4. Sky is blue (preferentially scatters blue out of sunlight); sunset is red (transmitted long-wavelength light).

Sub-topic I: Optical Instruments

Simple microscope (magnifying glass): M=1+D/fM = 1 + D/f where D=25D = 25 cm is least distance of distinct vision.

Compound microscope: objective + eyepiece. M=(L/fo)(1+D/fe)M = (L/f_o)(1 + D/f_e) where LL \approx tube length.

Refracting astronomical telescope (final image at infinity, normal adjustment): M=fo/feM = -f_o/f_e, tube length =fo+fe= f_o + f_e.

Reflecting telescope (Cassegrain/Newtonian): uses a concave mirror as objective — no chromatic aberration.

Resolving power of microscope: 1.22λ/(2nsinθ)1.22\lambda/(2n\sin\theta); telescope: 1.22λ/D1.22\lambda/D.

Eye defects:

  • Myopia (short sight): far image forms before retina. Correct with diverging (concave) lens, P<0P < 0.
  • Hypermetropia (long sight): near image forms behind retina. Correct with converging (convex) lens, P>0P > 0.
  • Astigmatism: cylindrical lens.
  • Presbyopia: bifocal.

Worked Problem 1

A convex lens of focal length 2020 cm is in contact with a concave lens of focal length 3030 cm. Find the equivalent focal length and power.

Solution. 1/f=1/201/30=(32)/60=1/601/f = 1/20 - 1/30 = (3 - 2)/60 = 1/60. So f=60f = 60 cm (converging). P=1/0.6=+1.67P = 1/0.6 = +1.67 D.

Worked Problem 2

A ray of light enters a prism of angle 6060^\circ at angle 4545^\circ and exits at 4545^\circ on the other side (symmetric, δm\delta_m). Find nn.

Solution. δm=45+4560=30\delta_m = 45 + 45 - 60 = 30^\circ. n=sin((60+30)/2)/sin(30)=sin45/sin30=(2/2)/(1/2)=2n = \sin((60+30)/2)/\sin(30) = \sin 45/\sin 30 = (\sqrt 2/2)/(1/2) = \sqrt 2.


Topic 2: Wave Optics

Sub-topic A: Huygens' Principle

Each point on a wavefront acts as a secondary source. The envelope of secondary wavelets at a later time is the new wavefront. From this:

Reflection: incident plane wavefront hits a flat mirror, secondary wavelets form a reflected plane wavefront; angle of reflection = angle of incidence.

Refraction: at the interface, secondary waves travel with different speeds in the two media. Construction shows Snell's law: n1sinθ1=n2sinθ2n_1\sin\theta_1 = n_2\sin\theta_2.

Sub-topic B: Young's Double-Slit Experiment

Two narrow slits separated by dd illuminated by monochromatic light of wavelength λ\lambda. Observation screen at distance DdD \gg d.

Path difference at a point at distance yy from the central axis:

Δx=dsinθdy/D\Delta x = d\sin\theta \approx d\,y/D

Constructive (bright fringe): Δx=mλ\Delta x = m\lambda. Destructive (dark fringe): Δx=(m+12)λ\Delta x = (m + \tfrac{1}{2})\lambda.

Fringe width:

β=λD/d\boxed{\beta = \lambda D/d}

Intensity distribution:

I(θ)=4I0cos2(πdsinθλ)I(\theta) = 4I_0\cos^2\left(\frac{\pi d\sin\theta}{\lambda}\right)

where I0I_0 is the intensity from one slit alone. Max intensity =4I0= 4I_0.

Sub-topic C: Missing Fringes (Slits with Finite Width)

For real slits of width aa, each slit produces a single-slit diffraction pattern (envelope). The double-slit pattern is modulated by this envelope. A fringe is missing when an interference maximum coincides with a diffraction minimum:

da=mintmdiff\frac{d}{a} = \frac{m_{\text{int}}}{m_{\text{diff}}}

Sub-topic D: Thin Film Interference

For a soap film of thickness tt, index nn, in air, viewed at angle θ\theta (in film):

Path difference between top-reflected and bottom-reflected ray: 2ntcosθ2nt\cos\theta. Plus phase shift of π\pi on reflection from the upper surface (going from lower to higher nn): equivalent to λ/2\lambda/2 extra.

Constructive: 2ntcosθ=(m+12)λ2nt\cos\theta = (m + \tfrac{1}{2})\lambda. Destructive: 2ntcosθ=mλ2nt\cos\theta = m\lambda.

(Why colourful soap bubbles: different colours satisfy condition at different points of varying tt.)

Sub-topic E: Single-Slit Diffraction

A slit of width aa illuminated by parallel light of wavelength λ\lambda. Diffraction minima at

asinθ=mλ,m=±1,±2,a\sin\theta = m\lambda,\quad m = \pm 1, \pm 2, \ldots

Width of central maximum (between m=+1m = +1 and m=1m = -1):

Δθ=2λ/a\Delta\theta = 2\lambda/a

or linear width on a screen at distance DD: 2λD/a2\lambda D/a.

Intensity distribution:

I=I0(sinββ)2,β=πasinθ/λI = I_0\left(\frac{\sin\beta}{\beta}\right)^2,\quad \beta = \pi a\sin\theta/\lambda

Sub-topic F: Resolving Power

Microscope (Abbe): dmin=1.22λ/(2nsinθ)d_{\min} = 1.22\lambda/(2n\sin\theta). Resolving power =1/dmin= 1/d_{\min}.

Telescope (Rayleigh): θmin=1.22λ/D\theta_{\min} = 1.22\lambda/D.

Sub-topic G: Polarization

Malus's Law: light of intensity I0I_0 polarized along an axis, then passed through a polarizer at angle θ\theta to that axis:

I=I0cos2θI = I_0\cos^2\theta

(For unpolarized light passing through the first polarizer, I0Iincident/2I_0 \to I_{\text{incident}}/2.)

Brewster's Angle: reflected light is completely polarized when reflected ray is perpendicular to refracted ray:

tanθB=n2/n1\tan\theta_B = n_2/n_1

At θB\theta_B, the reflected ray has E\vec E entirely perpendicular to plane of incidence (s-polarized).

Worked Problem 3

A YDSE has slits 0.50.5 mm apart, screen 11 m away. With λ=600\lambda = 600 nm find the fringe width.

Solution. β=λD/d=6×1071/(5×104)=1.2×103\beta = \lambda D/d = 6\times 10^{-7}\cdot 1/(5\times 10^{-4}) = 1.2\times 10^{-3} m =1.2= 1.2 mm.

Worked Problem 4

A 0.10.1 mm wide slit is illuminated by 600600 nm light. Find the width of the central diffraction maximum on a screen 11 m away.

Solution. Half-width angular = λ/a=6×103\lambda/a = 6\times 10^{-3} rad. Full width on screen =2λD/a=1.2×102= 2\lambda D/a = 1.2\times 10^{-2} m =1.2= 1.2 cm.


Topic 3: Photoelectric Effect

Sub-topic A: Phenomenology and Einstein's Explanation

When monochromatic light of frequency ν\nu strikes a metal:

  • Below a threshold frequency ν0\nu_0, no electrons are emitted, regardless of intensity.
  • Above ν0\nu_0, electrons are emitted instantaneously (no time lag).
  • Maximum KE of emitted electrons depends on ν\nu, not on intensity.
  • Number of photoelectrons per second is proportional to intensity (for ν>ν0\nu > \nu_0).

Classical wave theory cannot explain these. Einstein (1905): light is composed of photons, each of energy hνh\nu. A single photon either ejects an electron (if hν>ϕh\nu > \phi, the work function) or doesn't.

hν=ϕ+KEmax\boxed{h\nu = \phi + \text{KE}_{\max}}

Stopping potential VsV_s: applied retarding voltage that just stops the most energetic electron: KEmax=eVs\text{KE}_{\max} = eV_s, so

eVs=h(νν0)eV_s = h(\nu - \nu_0)

A plot of VsV_s vs ν\nu is a straight line of slope h/eh/e and xx-intercept ν0\nu_0.

Sub-topic B: Photon Properties

  • Energy E=hν=hc/λE = h\nu = hc/\lambda.
  • Momentum p=E/c=h/λp = E/c = h/\lambda (massless particle).
  • Mass equivalence m=E/c2=hν/c2m = E/c^2 = h\nu/c^2 (relativistic concept; rest mass is zero).

Sub-topic C: de Broglie Hypothesis

Particles also have wave nature:

λ=h/p\boxed{\lambda = h/p}

For an electron accelerated through potential VV: 12mv2=eVp=2meV\tfrac{1}{2}m v^2 = eV \Rightarrow p = \sqrt{2meV}. So

λ=h2meV=12.27VA˚ (electron, V in volts)\lambda = \frac{h}{\sqrt{2meV}} = \frac{12.27}{\sqrt{V}}\,\text{Å (electron, V in volts)}

Sub-topic D: Davisson–Germer Experiment

Electrons scattering off a Ni crystal showed Bragg-like diffraction peaks, confirming the wave nature of matter. The measured λ\lambda matched de Broglie's h/ph/p exactly.

Worked Problem 5

Light of wavelength 300300 nm falls on a metal of work function 22 eV. Find KE of the most energetic photoelectron and stopping potential.

Solution. Eγ=1240/300=4.13E_\gamma = 1240/300 = 4.13 eV. KE = 4.132=2.134.13 - 2 = 2.13 eV. Vs=V_s = KE /e=2.13/e = 2.13 V.

Worked Problem 6

An electron is accelerated through 100100 V. Find its de Broglie wavelength.

Solution. λ=12.27/100=1.227\lambda = 12.27/\sqrt{100} = 1.227 Å.


Topic 4: Bohr Model of the Atom

Sub-topic A: Postulates

  1. Electrons revolve in circular orbits around the nucleus under Coulomb attraction.
  2. Only orbits with angular momentum L=n=nh/2πL = n\hbar = nh/2\pi are allowed (quantization).
  3. While in an allowed orbit, the electron does not radiate. When it transitions, it emits/absorbs a photon of energy hν=EiEfh\nu = E_i - E_f.

Sub-topic B: Derivation of Orbital Parameters

Centripetal balance for an electron (charge e-e) orbiting a nucleus of charge +Ze+Ze:

mv2r=kZe2r2mv2=kZe2/r\frac{m v^2}{r} = \frac{kZe^2}{r^2}\Rightarrow m v^2 = kZe^2/r

Angular momentum quantization: mvr=nmvr = n\hbarv=n/(mr)v = n\hbar/(mr).

Substitute:

m(nmr)2=kZe2rr=n22kZe2m=n2a0Zm\left(\frac{n\hbar}{mr}\right)^2 = \frac{kZe^2}{r}\Rightarrow r = \frac{n^2\hbar^2}{kZe^2 m} = \frac{n^2 a_0}{Z}

where a0=2/(kme2)=0.529a_0 = \hbar^2/(kme^2) = 0.529 Å is the Bohr radius for hydrogen (Z=1,n=1Z = 1, n = 1).

Velocity:

v=kZe2n=Zαcn,α=ke2/(c)1/137v = \frac{kZe^2}{n\hbar} = \frac{Z\alpha c}{n},\quad \alpha = ke^2/(\hbar c) \approx 1/137

Energy (kinetic + potential):

En=12kZe2rn=mk2Z2e422n2=13.6Z2n2eVE_n = -\frac{1}{2}\cdot \frac{kZe^2}{r_n}=-\frac{m k^2 Z^2 e^4}{2\hbar^2 n^2}=-\frac{13.6\,Z^2}{n^2}\,\text{eV}

For hydrogen: E1=13.6E_1 = -13.6 eV (ground state), E2=3.4E_2 = -3.4 eV, E3=1.51E_3 = -1.51 eV, …, E=0E_\infty = 0.

Sub-topic C: Spectral Series — Rydberg Formula

A transition from nin_i to nfn_f (nf<nin_f < n_i) emits a photon with

hν=EniEnf=13.6Z2(1nf21ni2)eVh\nu = E_{n_i} - E_{n_f}=13.6 Z^2 \left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right)\,\text{eV} 1λ=RHZ2(1nf21ni2),RH=1.097×107m1\frac{1}{\lambda} = R_H Z^2\left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right),\quad R_H = 1.097\times 10^7\,\text{m}^{-1}

Spectral series of hydrogen:

  • Lyman (nf=1n_f = 1): UV. First line (Lyman-α): n=21n = 2 \to 1, λ=121.6\lambda = 121.6 nm.
  • Balmer (nf=2n_f = 2): visible. H-α (323\to 2): λ=656.3\lambda = 656.3 nm (red).
  • Paschen (nf=3n_f = 3): IR.
  • Brackett (nf=4n_f = 4), Pfund (nf=5n_f = 5): far IR.

Sub-topic D: X-rays and Moseley's Law

X-rays are emitted when high-energy electrons strike a metal target:

  • Continuous (Bremsstrahlung): slowing-down of electrons in the target produces a broad spectrum, with sharp short-wavelength cutoff at λmin=hc/(eV)\lambda_{\min} = hc/(eV).
  • Characteristic spectrum: discrete lines from electron transitions to inner shells. K-α: LKL\to K; K-β: MKM\to K.

Moseley's law (semi-empirical):

ν=a(Zb)\sqrt\nu = a(Z - b)

For K-α, b=1b = 1 (one inner electron shields the nuclear charge by 1e-1e). Established the central role of atomic number ZZ.

Worked Problem 7

The wavelength of the first line in the Lyman series of hydrogen is approximately:

Solution. 1/λ=R(11/4)=3R/41/\lambda = R(1 - 1/4) = 3R/4. λ=4/(3R)=4/(31.097×107)1.22×107\lambda = 4/(3R) = 4/(3\cdot 1.097\times 10^7)\approx 1.22\times 10^{-7} m =122= 122 nm.


Topic 5: Nuclei and Radioactivity

Sub-topic A: Nuclear Composition and Binding Energy

A nucleus ZAX^A_Z X has ZZ protons and N=AZN = A-Z neutrons. Mass of the bound nucleus MM is less than the sum of free masses: the mass defect Δm=(Zmp+(AZ)mn)M\Delta m = (Z m_p + (A-Z)m_n) - M shows up as binding energy:

B=Δmc2B = \Delta m\, c^2

In practice, masses are expressed in u (atomic mass units), where 11 u =931.5= 931.5 MeV/c².

Binding energy per nucleon (B/A) curve:

  • Rises sharply for light nuclei.
  • Peaks near A=56A = 56 (iron) at 8.8\approx 8.8 MeV/nucleon.
  • Falls slowly for heavier nuclei.

This shape explains:

  • Fission of heavy nuclei (e.g. ²³⁵U) into medium nuclei releases energy.
  • Fusion of light nuclei (e.g. H → He in stars) also releases energy.

Sub-topic B: Radioactive Decay Law

The rate of decay of a sample is proportional to the number of unstable nuclei:

dNdt=λNN(t)=N0eλt\frac{dN}{dt}=-\lambda N \Rightarrow N(t)=N_0 e^{-\lambda t}

where λ\lambda is the decay constant (probability per nucleus per unit time).

Half-life:

T1/2=ln2λT_{1/2} = \frac{\ln 2}{\lambda}

Mean lifetime:

τ=0tet/τ0τ0dt=1λ=T1/2/ln2\tau = \int_0^\infty t\cdot \frac{e^{-t/\tau_0}}{\tau_0}dt = \frac{1}{\lambda} = T_{1/2}/\ln 2

Activity: A=dN/dt=λN=λN0eλtA = -dN/dt = \lambda N = \lambda N_0 e^{-\lambda t}. Unit: becquerel (Bq, 11 disintegration/s); curie (Ci) =3.7×1010= 3.7\times 10^{10} Bq.

Sub-topic C: Types of Decay

  • α-decay: nucleus emits ⁴He. AA4A\to A-4, ZZ2Z\to Z-2. Range in air few cm; stopped by paper.
  • β⁻-decay: np+e+νˉen\to p + e^- + \bar\nu_e. AA same, ZZ+1Z\to Z+1. Range ~m in air.
  • β⁺-decay: pn+e++νep\to n + e^+ + \nu_e.
  • γ-emission: excited nucleus releases photon; A,ZA, Z unchanged. Highly penetrating; needs lead.

Sub-topic D: Fission and Fusion

Fission (²³⁵U + slow neutron): releases ≈ 200 MeV per fission; harnessed in reactors and bombs.

Fusion (in stars):

  • pp-chain: 4¹H → ⁴He + 2e⁺ + 2ν + 26.7 MeV.
  • CNO cycle: catalysed by C, N, O.
  • Earth-based fusion: deuterium-tritium D + T → ⁴He + n + 17.6 MeV.

Worked Problem 8

A radioactive sample has half-life of 5 years. What fraction remains after 15 years?

Solution. 1515 y =3= 3 half-lives. Fraction =(1/2)3=1/8=12.5%= (1/2)^3 = 1/8 = 12.5\%.

Worked Problem 9

The activity of a sample at t=0t = 0 is 8×1048\times 10^4 Bq and falls to 2×1042\times 10^4 Bq in 6 hours. Find half-life and decay constant.

Solution. A/A0=1/4=(1/2)t/T1/2t/T1/2=2T1/2=3A/A_0 = 1/4 = (1/2)^{t/T_{1/2}}\Rightarrow t/T_{1/2} = 2 \Rightarrow T_{1/2} = 3 h. λ=ln2/T1/2=0.231/\lambda = \ln 2/T_{1/2} = 0.231/h.


Topic 6: Semiconductors and Electronics

Sub-topic A: Band Theory

In solids, atomic energy levels broaden into bands (valence and conduction) separated by a forbidden gap EgE_g.

MaterialEgE_g (eV)Behaviour at room T
Conductor (metal)0 (overlap)High conductivity
Semiconductor~1 (Si: 1.1, Ge: 0.7)Moderate, rises with T
Insulator> 3Very low

Sub-topic B: Intrinsic Semiconductors

Pure Si or Ge. At T>0T > 0, some electrons thermally excited from valence band to conduction band, leaving behind holes.

n=p=nin = p = n_i (intrinsic carrier density). At room T, ni(Si)1016n_i(\text{Si})\approx 10^{16}/m³ and ni(Ge)2.4×1019n_i(\text{Ge})\approx 2.4\times 10^{19}/m³.

Conductivity rises rapidly with T (exponentially): niT3/2eEg/(2kBT)n_i\propto T^{3/2}e^{-E_g/(2k_BT)}.

Sub-topic C: Extrinsic Semiconductors (Doping)

Add traces of impurity to control carrier type and density.

n-type: dope Si with group V (P, As). Each donor adds a loosely-bound electron — promoted easily into conduction band. Majority carriers: electrons; minority: holes.

p-type: dope Si with group III (B, Al). Each acceptor needs an extra electron to bond — creates a hole. Majority: holes; minority: electrons.

For both: np=ni2np = n_i^2 (mass-action law).

Sub-topic D: pn-junction

Junction of p- and n-type semiconductors. Initially diffusion creates depletion region with built-in potential V00.7V_0 \approx 0.7 V for Si.

Forward bias (p more positive): depletion region shrinks, large current flows. I-V characteristic:

I=I0(eeV/kT1)I = I_0(e^{eV/kT} - 1)

Reverse bias (n more positive): depletion widens, only tiny saturation current I0I_0. At a critical reverse voltage (breakdown), the junction conducts (avalanche or Zener).

Sub-topic E: Diode Applications

Half-wave rectifier: a single diode in series with a load and AC source. Passes only positive half-cycles. DC component Vdc=Vm/πV_{dc} = V_m/\pi where VmV_m is peak.

Full-wave rectifier:

  • Centre-tap: two diodes and centre-tapped transformer.
  • Bridge: four diodes, no centre tap needed.

In both, Vdc=2Vm/πV_{dc} = 2V_m/\pi, ripple factor much lower than half-wave.

Zener diode: operates in reverse breakdown at a sharp voltage. Used as a voltage regulator — output stays at VzV_z regardless of input variation, provided the current is within rated range.

LED, photodiode, solar cell — all variants of the pn-junction with light coupling.

Sub-topic F: Logic Gates

A logic gate takes binary inputs and produces a binary output.

GateSymbolTruth Table (2-input)Boolean
NOTA → ¬A01,  100\to 1,\; 1\to 0Y=AˉY = \bar A
ANDA,BYA, B \to YY=1Y = 1 iff both 1Y=ABY = AB
ORA,BYA, B \to YY=1Y = 1 iff at least one 1Y=A+BY = A + B
NANDA,BYA, B \to YY=ABY = \overline{AB}NOT–AND
NORA,BYA, B \to YY=A+BY = \overline{A + B}NOT–OR
XORA,BYA, B \to YY=1Y = 1 iff A≠BY=ABˉ+AˉBY = A\bar B + \bar A B

NAND and NOR are universal (can build any logic from them).

Worked Problem 10

A common-emitter NPN transistor amplifier has β=100\beta = 100. If IB=50I_B = 50 μA, find ICI_C and IEI_E.

Solution. IC=βIB=5I_C = \beta I_B = 5 mA. IE=IC+IB=5.05I_E = I_C + I_B = 5.05 mA.

(Note: detailed transistor analysis is mostly out of the JEE-Advanced syllabus 2025 onwards, but β and the I_C/I_B ratio still appear in JEE-Main.)


Problem-Solving Heuristics

  1. Sign convention is everything in geometric optics — mark distances from the pole/lens, sign per direction.
  2. For a lens combination, add powers; for mirror–lens combinations, treat each surface in turn.
  3. At a flat surface, apparent depth scales by n2/n1n_2/n_1.
  4. Total internal reflection requires going from denser to lighter and θ>θc\theta > \theta_c.
  5. For minimum deviation in a prism, light goes symmetrically; i1=i2i_1 = i_2.
  6. In YDSE, fringe width is β=λD/d\beta = \lambda D/d — independent of which fringe.
  7. For single-slit diffraction, central max is twice the width of side maxima.
  8. Photoelectric: stopping potential depends only on ν\nu, not intensity. Plot VsV_s vs ν\nu for h/eh/e.
  9. de Broglie: λ=12.27/V\lambda = 12.27/\sqrt V Å for electrons in V volts.
  10. Bohr levels: En=13.6Z2/n2E_n = -13.6 Z^2/n^2 eV, rn=n2a0/Zr_n = n^2 a_0/Z.
  11. For X-rays, short-wavelength cutoff is λmin=hc/(eV)\lambda_{\min} = hc/(eV), independent of target.
  12. Half-life vs mean lifetime: T1/2=τln20.693τT_{1/2} = \tau\ln 2 \approx 0.693\tau.
  13. Universal gates: any circuit reducible to NAND-only or NOR-only.
  14. For a Zener regulator, the load voltage equals VzV_z as long as the Zener is in reverse breakdown and current is within rating.

Common Traps & Mistakes

  • Sign of focal length and image distance. Be consistent — and check which side the image is on physically.
  • Lens formula vs mirror formula: signs of u,vu, v differ between conventions; pick one (Cartesian recommended) and stick with it.
  • Critical angle is from denser to lighter, not the other way.
  • In YDSE, increasing slit separation dd shrinks the fringe width — common reversed answer.
  • Single-slit diffraction: minimum (not maximum) at asinθ=mλa\sin\theta = m\lambda.
  • Malus law uses cos2\cos^2, not cos\cos.
  • Brewster's angle gives 100% polarized reflected light, but the refracted light is still partially polarized.
  • Photoelectric: stopping potential is Vs=(hνϕ)/eV_s = (h\nu - \phi)/enot (hν)/e(h\nu)/e.
  • Bohr orbits don't apply to multi-electron atoms (only one-electron systems and hydrogen-like ions like He⁺, Li²⁺).
  • Half-life and mean life are different: T1/2=0.693τT_{1/2} = 0.693\tau, not equal.
  • Activity decays at the same rate as NN: A(t)=A0eλtA(t) = A_0 e^{-\lambda t}.
  • Doping doesn't change resistivity by a constant factor — depends on dopant concentration and temperature.
  • Zener diode in forward bias behaves like a normal diode; only in reverse breakdown does it regulate.

Quick Revision Card

Ray Optics:

  • Mirror: 1/v+1/u=1/f1/v + 1/u = 1/f, m=v/um = -v/u.
  • Lens: 1/v1/u=1/f1/v - 1/u = 1/f, m=v/um = v/u.
  • Lens-maker: 1/f=(n1)(1/R11/R2)1/f = (n-1)(1/R_1 - 1/R_2).
  • Refraction at sphere: n2/vn1/u=(n2n1)/Rn_2/v - n_1/u = (n_2 - n_1)/R.
  • Prism: thin δ=(n1)A\delta = (n-1)A; min n=sin((A+δm)/2)/sin(A/2)n = \sin((A+\delta_m)/2)/\sin(A/2).
  • TIR: sinθc=1/n\sin\theta_c = 1/n.

Wave Optics:

  • YDSE: β=λD/d\beta = \lambda D/d, intensity 4I0cos2(πdsinθ/λ)4I_0\cos^2(\pi d\sin\theta/\lambda).
  • Single-slit: minima asinθ=mλa\sin\theta = m\lambda; central width 2λD/a2\lambda D/a.
  • Malus: I=I0cos2θI = I_0\cos^2\theta. Brewster: tanθB=n\tan\theta_B = n.

Modern:

  • Photoelectric: hν=ϕ+KEh\nu = \phi + KE; eVs=h(νν0)eV_s = h(\nu - \nu_0).
  • de Broglie: λ=h/p\lambda = h/p; electron, VV volts: λ=12.27/V\lambda = 12.27/\sqrt V Å.
  • Bohr: En=13.6Z2/n2E_n = -13.6 Z^2/n^2 eV, rn=n2a0/Zr_n = n^2 a_0/Z, a0=0.529a_0 = 0.529 Å.
  • Lyman (nf=1n_f=1, UV), Balmer (nf=2n_f=2, visible), Paschen (nf=3n_f=3, IR).
  • Moseley: ν(Zb)\sqrt\nu \propto (Z - b).
  • Decay: N=N0eλtN = N_0 e^{-\lambda t}, T1/2=ln2/λT_{1/2} = \ln 2/\lambda, τ=1/λ\tau = 1/\lambda.

Electronics:

  • Semiconductors: EgE_g(Si) = 1.1 eV, EgE_g(Ge) = 0.7 eV.
  • np=ni2np = n_i^2 (mass-action).
  • Half-wave rectifier: Vdc=Vm/πV_{dc} = V_m/\pi. Full-wave: Vdc=2Vm/πV_{dc} = 2V_m/\pi.
  • Zener: voltage regulator.
  • NAND, NOR are universal gates.

Formula Sheet

ConceptFormula
Mirror formula1/v+1/u=1/f1/v + 1/u = 1/f, f=R/2f = R/2
Magnification (mirror)m=v/um = -v/u
Snell's lawn1sinθ1=n2sinθ2n_1\sin\theta_1 = n_2\sin\theta_2
Apparent depthda=drn2/n1d_a = d_r\cdot n_2/n_1
Refraction at spheren2/vn1/u=(n2n1)/Rn_2/v - n_1/u = (n_2-n_1)/R
Lens-maker1/f=(n1)(1/R11/R2)1/f = (n-1)(1/R_1 - 1/R_2)
Thin-lens formula1/v1/u=1/f1/v - 1/u = 1/f
Combination of lenses1/f=1/fi1/f = \sum 1/f_i, P=PiP = \sum P_i
Prism deviationδ=i1+i2A\delta = i_1 + i_2 - A
Thin prismδ=(n1)A\delta = (n-1)A
Min-deviationn=sin((A+δm)/2)/sin(A/2)n = \sin((A+\delta_m)/2)/\sin(A/2)
Critical anglesinθc=1/n\sin\theta_c = 1/n
Microscope (simple)M=1+D/fM = 1 + D/f
Telescope (normal)M=fo/feM = f_o/f_e, length L=fo+feL = f_o + f_e
YDSE fringe widthβ=λD/d\beta = \lambda D/d
YDSE intensityI=4I0cos2(πdsinθ/λ)I = 4I_0\cos^2(\pi d\sin\theta/\lambda)
Single-slit minimumasinθ=mλa\sin\theta = m\lambda
Central diff. widthW=2λD/aW = 2\lambda D/a
Resolving microscopedmin=1.22λ/(2nsinθ)d_{\min}=1.22\lambda/(2n\sin\theta)
Resolving telescopeθmin=1.22λ/D\theta_{\min}=1.22\lambda/D
MalusI=I0cos2θI = I_0\cos^2\theta
BrewstertanθB=n\tan\theta_B = n
Photoelectrichν=ϕ+KEmaxh\nu = \phi + KE_{\max}, eVs=h(νν0)eV_s = h(\nu - \nu_0)
de Broglieλ=h/p\lambda = h/p
Bohr energyEn=13.6Z2/n2E_n = -13.6Z^2/n^2 eV
Bohr radiusrn=n2a0/Zr_n = n^2 a_0/Z, a0=0.529a_0 = 0.529 Å
Rydberg1/λ=RZ2(1/nf21/ni2)1/\lambda = RZ^2(1/n_f^2 - 1/n_i^2), R=1.097×107R = 1.097\times 10^7 /m
Moseleyν=a(Zb)\sqrt\nu = a(Z - b)
Mass-energyE=mc2E = mc^2
Binding energyB=Δmc2B = \Delta m c^2
Decay lawN=N0eλtN = N_0 e^{-\lambda t}
Half-lifeT1/2=ln2/λT_{1/2} = \ln 2/\lambda
Mean lifeτ=1/λ=T1/2/ln2\tau = 1/\lambda = T_{1/2}/\ln 2
ActivityA=λNA = \lambda N
Half-wave rectifierVdc=Vm/πV_{dc} = V_m/\pi
Full-wave rectifierVdc=2Vm/πV_{dc} = 2V_m/\pi
Mass-action lawnp=ni2np = n_i^2
Diode I-VI=I0(eeV/kBT1)I = I_0(e^{eV/k_BT} - 1)

Sub-topics

6 pages

Practice quiz

Quiz
Unit 12: Optics, Modern Physics & Electronics — JEE Quiz
15 questions · pick the best answer
Q1

An object is placed 30 cm in front of a convex lens of focal length 20 cm. The image is:

Q2

A prism of angle 60° produces minimum deviation of 30°. The refractive index is:

Q3

Critical angle for a glass-air interface is 42°. Refractive index of glass:

Q4

In YDSE, slits 0.5 mm apart, screen 1 m away. λ = 600 nm. Fringe width:

Q5

Single-slit diffraction with slit width 0.1 mm and λ = 600 nm. Angular width of central maximum:

Q6

Light of wavelength 300 nm strikes a metal of work function 2eV.KEmax2 eV. KE_max of photoelectrons (use hc = 1240 eV·nm):

Q7

de Broglie wavelength of an electron accelerated through 100 V is:

Q8

Wavelength of first Lyman series (Hydrogen) is:

Q9

Energy of an electron in n = 3 state of Li²⁺ ion (Z = 3):

Q10

Radioactive sample has half-life 10 years. Fraction left after 40 years:

Q11

Activity of a sample falls from 1000 Bq to 250 Bq in 30 minutes. Half-life:

Q12

Width of central maximum in single-slit diffraction is doubled by:

Q13

In a half-wave rectifier with peak input 311 V, the DC output voltage is approximately:

Q14

The boolean expression for the circuit consisting of two NAND gates: input to first NAND (A, A), then output fed with B into second NAND:

Q15

The number of photons per second emitted by a 100 W lamp emitting light of wavelength 600 nm: