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Modern Physics and Optics for NEET: Key Ideas, Formulas and Worked Examples
Modern physics for NEET covers three units of the official syllabus: dual nature of matter and radiation, atoms and nuclei, and electronic devices (semiconductors). Optics is a fourth unit, split into ray optics and wave optics. All four are Class 12 topics built on a small set of formulas, and their questions are usually short and direct, which makes them some of the most reliable scoring areas in NEET physics. One warning: the NEET syllabus includes a few topics the current NCERT textbook no longer covers, such as Zener diodes, logic gates and Brewster's law.
This guide is part of our NEET Physics home tutor in Gurgaon series. For each unit, it gives the key ideas, a formula table, the traps our tutors see most often, and worked examples we have checked step by step. It starts with what the NEET (UG) 2026 syllabus includes and leaves out, because that is where many students either waste time or lose easy marks.
Why modern physics and optics matter in NEET
Modern physics and optics reward preparation because the ideas are few and most questions take one or two steps. According to the NEET (UG) 2026 Information Bulletin, published on the official NEET (UG) website of the National Testing Agency, the paper has 45 physics questions worth 180 marks, all multiple-choice with four options. A correct answer earns +4 and a wrong answer costs −1. The whole paper has 180 questions in 180 minutes, and in 2026 it was a pen-and-paper exam. Check the bulletin for your own exam year, as NTA can change the pattern.
We do not give chapter-wise question counts here, because we only use counts taken from official papers. What matters for planning is that a question on the photoelectric equation, the Bohr model, the lens formula or fringe width can usually be done in well under a minute by a student who knows the formula and when it applies. The time saved goes to harder mechanics and electricity questions. The Bohr model, de Broglie wavelength and photoelectric effect also appear in NEET chemistry's atomic structure unit, so this work pays off twice.
What the NEET 2026 syllabus includes and leaves out
The NEET (UG) syllabus is set by the National Medical Commission (NMC). Its Undergraduate Medical Education Board notified the NEET (UG) 2026 syllabus on 22 December 2025, and NTA hosts it as the Syllabus for NEET (UG)-2026. As of 29 September 2026 we have not found a separate 2027 syllabus on the NTA NEET site, so this guide uses the 2026 syllabus. Check for a new notice before you finalise a 2027 plan.
The four units here are Units 16 to 19 of the NEET physics syllabus. We compared them with the current NCERT Class 12 Physics Part II textbook (Chapters 9 to 14), chapter by chapter.
| NEET 2026 unit | In the NEET syllabus but not in the current NCERT chapter | In NCERT but not named in the NEET syllabus |
|---|---|---|
| Unit 16: Optics (ray optics, microscope and telescope, wave optics, polarisation) | Brewster's law (NCERT covers Polaroids and Malus' law only) | Malus' law is not named, but it follows directly from Polaroids, so learn it |
| Unit 17: Dual nature of matter and radiation | None found | The experimental study of the photoelectric effect is not named separately, but its graphs follow from Einstein's equation |
| Unit 18: Atoms and nuclei | None found | Nuclear force; radioactivity (a short descriptive section in NCERT); de Broglie's explanation of Bohr's second postulate |
| Unit 19: Electronic devices | LED, photodiode, solar cell, Zener diode as a voltage regulator, and logic gates (OR, AND, NOT, NAND, NOR). The NCERT chapter stops at the diode as a rectifier | None found |
Two practical points follow. First, NEET students must learn Zener diodes, optoelectronic devices, logic gates and Brewster's law from a source other than the current NCERT textbook. The NCERT Exemplar Problems book for Class 12 has questions on Zener diodes and other devices, but it also covers transistors and communication systems, which the NEET 2026 syllabus does not list. Second, the CBSE 2026–27 Class 12 physics curriculum leaves out polarisation and stops at the diode as a rectifier, so a topic skipped for the board exam may still be tested in NEET. Our guide to the NEET physics syllabus and exam pattern covers every unit.
Dual nature of matter and radiation
This unit shows that light behaves as particles (photons) in the photoelectric effect, and that particles such as electrons behave as waves. It is short, formula-driven and closely tied to graphs.
Key ideas
- Einstein's photoelectric equation: the maximum kinetic energy of emitted electrons equals the photon energy minus the work function. There is no emission below the threshold frequency, however bright the light.
- Stopping potential: a graph of V0 against frequency is a straight line with slope h/e for every metal; its intercept on the frequency axis is the threshold frequency.
- Intensity vs frequency: more intensity means more photoelectrons (a larger saturation current), not faster ones. Higher frequency means a larger Kmax and stopping potential.
- Matter waves: every moving particle has a de Broglie wavelength λ = h/p. For an electron accelerated from rest through V volts, λ ≈ 1.227/√V nm.
| Quantity | Formula | Note |
|---|---|---|
| Photon energy | E = hν = hc/λ | E (eV) ≈ 1240/λ (nm) |
| Photon momentum | p = h/λ = E/c | A photon has no rest mass |
| Photoelectric equation | Kmax = hν − φ0 = eV0 | Valid only when ν > ν0 |
| Threshold | ν0 = φ0/h; λ0 = hc/φ0 | Wavelengths longer than λ0 emit nothing |
| de Broglie wavelength | λ = h/p = h/√(2mK) | K = qV for a charge accelerated through V |
Traps
- Mixing units: using λ in metres with the 1240 shortcut, or joules with eV. Choose eV and nm at the start.
- Forgetting that a proton and an electron with the same kinetic energy have different de Broglie wavelengths, because λ depends on √m.
Worked example 1: Light of wavelength 300 nm falls on a metal with work function 2.3 eV. Find the maximum kinetic energy of the photoelectrons, the stopping potential and the threshold wavelength (take hc = 1240 eV nm).
Answer: Kmax ≈ 1.83 eV; V0 ≈ 1.83 V; λ0 ≈ 539 nm. Photon energy = 1240/300 ≈ 4.13 eV, so Kmax = 4.13 − 2.3 = 1.83 eV. Since eV0 = Kmax, the stopping potential is 1.83 V. Threshold wavelength = 1240/2.3 ≈ 539 nm; longer wavelengths release no electrons from this metal.
Worked example 2: An electron starting from rest is accelerated through 100 V. Find its de Broglie wavelength.
Answer: about 0.123 nm. Using the electron shortcut, λ = 1.227/√100 = 0.1227 nm. From first principles, λ = h/√(2meV) = 6.63 × 10−34 / √(2 × 9.11 × 10−31 × 1.6 × 10−19 × 100) ≈ 1.23 × 10−10 m, similar to the spacing of atoms in a crystal. Trap: using the shortcut for a proton or alpha particle; it holds only for electrons.
Atoms: Rutherford, Bohr and the hydrogen spectrum
The atoms part of Unit 18 moves from Rutherford's nuclear model to the Bohr model of hydrogen and its line spectrum. Almost every NEET question here uses one of three dependences on n and Z.
Key ideas
- Alpha-particle scattering: most alpha particles pass straight through gold foil and very few bounce back, so the positive charge and most of the mass sit in a tiny nucleus.
- Bohr's postulates: electrons move in fixed orbits without radiating; angular momentum is quantised, mvr = nh/2π; a photon is emitted or absorbed only in a jump between levels, with hν = Ei − Ef.
- Spectral series of hydrogen: Lyman (ending at n = 1, ultraviolet), Balmer (n = 2, visible), then Paschen, Brackett and Pfund (n = 3, 4 and 5, infrared).
| Quantity (hydrogen-like atom) | Formula | Hydrogen ground state |
|---|---|---|
| Orbit radius | rn = 0.529 n2/Z Å | 0.529 Å |
| Electron speed | vn ≈ 2.18 × 106 Z/n m/s | 2.18 × 106 m/s |
| Total energy | En = −13.6 Z2/n2 eV | −13.6 eV |
| Kinetic and potential energy | K = −E; U = 2E | K = 13.6 eV; U = −27.2 eV |
| Emitted wavelength | 1/λ = RZ2(1/nf2 − 1/ni2) | R ≈ 1.097 × 107 m−1 |
Traps
- Forgetting the Z2 for He+ or Li2+. The ground-state energy of He+ is −54.4 eV, not −13.6 eV.
- Thinking the longest wavelength in a series comes from the largest jump. It comes from the smallest energy difference.
Worked example 3: Find the ratio of the longest to the shortest wavelength in the Balmer series of hydrogen.
Answer: 9 : 5 (1.8). Balmer lines end at n = 2. The longest wavelength is the smallest jump, 3 → 2: 1/λmax = R(1/4 − 1/9) = 5R/36. The shortest is the largest jump, ∞ → 2: 1/λmin = R/4. So λmax/λmin = (36/5R)/(4/R) = 9/5. With R = 1.097 × 107 m−1, these are about 656 nm and 365 nm, but R cancels in the ratio, so there is no need to substitute it.
Nuclei: size, mass defect, binding energy, fission and fusion
The nuclei part of Unit 18 explains why nuclei hold together and where nuclear energy comes from. The central idea is mass defect: a nucleus weighs less than its separate protons and neutrons, and the missing mass is the binding energy.
Key ideas
- Composition: a nucleus with mass number A and atomic number Z has Z protons and A − Z neutrons. Isotopes share Z; isobars share A; isotones share the neutron number.
- Size: radius grows as A1/3, so volume grows as A and nuclear density is roughly the same for all nuclei.
- Binding energy per nucleon is highest (about 8.8 MeV) near iron and lower for very light and very heavy nuclei. Fission of a heavy nucleus and fusion of light nuclei both move towards that peak, so both release energy.
| Quantity | Formula | Note |
|---|---|---|
| Nuclear radius | R = R0A1/3 | R0 ≈ 1.2 × 10−15 m |
| Mass defect | Δm = Zmp + (A − Z)mn − M | With atomic masses, use m(1H) in place of mp so electron masses cancel |
| Binding energy | BE = Δm c2 = Δm (in u) × 931.5 MeV | BE per nucleon = BE/A |
| Energy released (Q value) | Q = (mass before − mass after) c2 | Also Q = total BE after − total BE before |
Traps
- Judging stability by total binding energy. Use binding energy per nucleon.
- Rounding the mass defect early. It is a small difference of large numbers, so keep every decimal place until the end.
Worked example 4: Find the binding energy and the binding energy per nucleon of helium-4. Take m(1H) = 1.007825 u, mn = 1.008665 u, m(4He) = 4.002603 u and 1 u = 931.5 MeV.
Answer: about 28.3 MeV; about 7.07 MeV per nucleon. Mass of the separate parts = 2 × 1.007825 + 2 × 1.008665 = 4.032980 u. Mass defect = 4.032980 − 4.002603 = 0.030377 u. Binding energy = 0.030377 × 931.5 ≈ 28.30 MeV, or 28.30/4 ≈ 7.07 MeV per nucleon.
From our tutors: in atoms and nuclei, we ask students to write a one-line "scaling card" before solving anything: r ∝ n2/Z, v ∝ Z/n, E ∝ Z2/n2, R ∝ A1/3. Most NEET questions in these chapters are ratio questions, and a student who writes the scaling first rarely needs to substitute constants at all. The students who struggle are usually trying to recall a separate formula for every ratio.
Semiconductors and electronic devices
Unit 19 of the NEET 2026 syllabus is wider than the current NCERT chapter. NCERT covers semiconductors, the p-n junction, the diode and the rectifier; NEET adds the LED, photodiode, solar cell, Zener diode and five logic gates.
Key ideas
- Energy bands: conductors have overlapping bands, insulators a large band gap, and semiconductors a small gap (about 1.1 eV for silicon).
- Doping: a pentavalent impurity gives an n-type semiconductor (electrons are the majority carriers); a trivalent impurity gives p-type (holes are the majority). Both stay electrically neutral.
- p-n junction diode: in forward bias the barrier falls and current rises sharply after the knee voltage; in reverse bias only a tiny current flows until breakdown.
- Rectifiers: a half-wave rectifier's output ripple has the input frequency; a full-wave rectifier's has twice the input frequency.
- Logic gates: OR (Y = A + B), AND (Y = A·B), NOT, NAND (NOT of AND) and NOR (NOT of OR). NAND and NOR are universal gates: any other gate can be built from them.
| Device | Bias in use | What to remember for NEET |
|---|---|---|
| p-n junction diode | Forward (conducts) or reverse (blocks) | I–V curve; knee voltage; use as a rectifier |
| Zener diode | Reverse, in breakdown | Heavily doped; voltage across it stays nearly constant, so it regulates. Series-resistor current = IZ + IL |
| LED | Forward | Emits light as electrons and holes recombine; photon energy ≈ band gap |
| Photodiode | Reverse | Current changes with light intensity; used as a light detector |
| Solar cell | No external bias | Generates an emf from light; I–V curve drawn in the fourth quadrant |
Traps
- Saying a p-type semiconductor is positively charged. Doping adds carriers, not net charge.
- In Zener questions, forgetting that the series resistor carries the sum of the Zener and load currents.
Worked example 5: A Zener diode with breakdown voltage 6 V is used with a 10 V supply, a 200 Ω series resistor and a 1 kΩ load across the Zener. Find the current through the Zener diode.
Answer: 14 mA. The Zener holds the load voltage at 6 V, so the series resistor has 10 − 6 = 4 V across it and carries 4/200 = 20 mA. The load current is 6/1000 = 6 mA, so the Zener carries 20 − 6 = 14 mA. If the load resistance fell, the load would draw more current and the Zener less, while the load voltage stayed at 6 V. That is how it regulates.
Worked example 6: The output of a NAND gate is fed into a NOT gate. Which single gate does the combination act as? And what does a NOR gate do if both its inputs are joined together?
Answer: an AND gate; a NOT gate. For inputs 00, 01, 10 and 11, NAND gives 1, 1, 1, 0; the NOT gate turns this into 0, 0, 0, 1, which is the AND truth table. If both NOR inputs are the same (A = B), Y = NOT(A + A) = NOT A, so it acts as a NOT gate. Build every combined-gate answer from a truth table; it takes seconds and removes guesswork.
Is your child losing marks in modern physics or optics because of topics their school textbook skipped? Book a free NEET Physics demo class in Gurgaon. Our tutor will check the NEET-only topics, from Zener diodes to Brewster's law, and show you where the gaps are.
Book a Free NEET Physics Demo +91 92204 75088Ray optics and optical instruments
Ray optics treats light as rays that reflect and refract. Most NEET questions here use one of four tools: the mirror formula, the lens formula, the prism formula or the critical angle. Most mistakes come down to signs.
Key ideas
- Sign convention: NCERT uses the Cartesian sign convention. Measure distances from the pole or optical centre: along the incident light is positive, against it negative. With light travelling left to right, a real object has a negative u.
- Refraction: Snell's law, n1 sin i = n2 sin r. Seen from directly above, apparent depth = real depth/n.
- Total internal reflection happens only when light goes from a denser to a rarer medium at more than the critical angle. Optical fibres depend on it.
- Instruments: a compound microscope magnifies a small nearby object; an astronomical telescope magnifies a distant one. The syllabus names both, with reflecting and refracting telescopes.
| Situation | Formula | Watch for |
|---|---|---|
| Spherical mirror | 1/v + 1/u = 1/f; f = R/2; m = −v/u | f is negative for a concave mirror in the Cartesian convention |
| Thin lens | 1/v − 1/u = 1/f; m = v/u | Note the minus sign, unlike the mirror formula |
| Lens maker's formula | 1/f = (nlens/nmedium − 1)(1/R1 − 1/R2) | A glass lens in water has a longer focal length |
| Power and combination | P = 1/f (f in metres); P = P1 + P2 | A 20 cm lens has power 5 D, not 0.05 D |
| Critical angle | sin C = nrarer/ndenser | Only from denser to rarer |
| Prism | A + δ = i + e; n = sin[(A + δm)/2]/sin(A/2) | Thin prism: δ ≈ (n − 1)A |
| Microscope (image at near point) | m ≈ (L/fo)(D/fe) | D = 25 cm; L = tube length, as NCERT defines it |
| Telescope (normal adjustment) | m = fo/fe; length = fo + fe | Long-focus objective, short-focus eyepiece |
Traps
- Using the mirror formula's plus sign in the lens formula, or the other way round.
- Assuming a convex lens always converges. In a liquid of higher refractive index than the lens, it diverges.
Worked example 7: An object is placed 30 cm in front of a convex lens of focal length 20 cm. Find the position and nature of the image. If a concave lens of focal length 40 cm is then placed in contact with it, what is the power of the combination?
Answer: 60 cm behind the lens, real, inverted and twice the size; combined power +2.5 D. With u = −30 cm and f = +20 cm: 1/v = 1/f + 1/u = 1/20 − 1/30 = 1/60, so v = +60 cm. Magnification m = v/u = 60/(−30) = −2. For the pair, P = 1/0.20 + 1/(−0.40) = 5 − 2.5 = +2.5 D, a converging combination of focal length 40 cm.
Worked example 8: A prism of angle 60° gives a minimum deviation of 30°. Find its refractive index and the critical angle for this glass.
Answer: n = √2 ≈ 1.41; critical angle 45°. n = sin[(60° + 30°)/2] / sin(60°/2) = sin 45° / sin 30° = (1/√2)/(1/2) = √2. Then sin C = 1/n = 1/√2, so C = 45°. For comparison, glass of n = 1.5 has a critical angle of about 41.8°. Trap: using A + δm without halving it.
Wave optics: interference, diffraction and polarisation
Wave optics explains what ray optics cannot: fringes, the spreading of light through a narrow slit, and polarisation. For NEET, the working tools are the fringe-width formula, the path-difference conditions, the width of the single-slit central maximum and Brewster's law.
Key ideas
- Huygens' principle: every point on a wavefront acts as a source of secondary wavelets, and the new wavefront is their common tangent. It gives the laws of reflection and refraction.
- Coherent sources keep a constant phase difference. Two independent bulbs are not coherent, so they give no steady fringes.
- Polarisation: only transverse waves can be polarised. A Polaroid passes the electric-field component along its pass axis, so unpolarised light loses half its intensity through the first Polaroid.
- Brewster's law: at the angle where tan iB = n, reflected light is completely plane-polarised and the reflected and refracted rays are at 90°.
| Situation | Formula | Watch for |
|---|---|---|
| YDSE fringe width | β = λD/d | In a medium of refractive index n, β divides by n |
| Bright and dark fringes | Path difference nλ (bright); (n + ½)λ (dark) | The central fringe is bright |
| Single-slit minima | a sinθ = nλ (n = 1, 2, …) | Same form as the YDSE bright condition, but these are dark |
| Central maximum width | Angular 2λ/a; linear 2λD/a | A narrower slit gives a wider central maximum |
| Malus' law | I = I0 cos2θ | Halve unpolarised light at the first Polaroid first |
| Brewster's law | tan iB = n; iB + r = 90° | Reflected ray is plane-polarised |
Traps
- Swapping d and D: d is the slit separation (a fraction of a millimetre); D is the distance to the screen (around a metre).
- Mixing up the single-slit minima condition with the double-slit bright-fringe condition.
Worked example 9: In a Young's double-slit experiment, the slits are 0.5 mm apart, the screen is 1 m away and the wavelength is 600 nm. Find the fringe width, and the fringe width if the apparatus is placed in water of refractive index 4/3.
Answer: 1.2 mm in air; 0.9 mm in water. β = λD/d = (600 × 10−9 × 1)/(0.5 × 10−3) = 1.2 × 10−3 m = 1.2 mm. In water the wavelength becomes λ/n, so β = 1.2/(4/3) = 0.9 mm and the fringes move closer together. Trap: leaving d in millimetres while λ is in metres.
Worked example 10: Light is reflected from glass of refractive index √3. At what angle of incidence is the reflected light completely polarised, and what is the angle of refraction? Separately, unpolarised light of intensity I0 passes through two Polaroids whose axes are at 60°. Find the final intensity.
Answer: 60° and 30°; I0/8. By Brewster's law, tan iB = √3, so iB = 60°. The reflected and refracted rays are then perpendicular, so r = 30° (check: sin 60° / sin 30° = √3). For the Polaroids, the first halves the intensity to I0/2 and the second multiplies it by cos260° = 1/4, giving I0/8.
From our tutors: in ray optics, we ask students to write the sign of every quantity before touching the formula: "u negative, f positive, v unknown". It feels slow for a week. After that, most sign errors disappear, and sign errors are the most common reason we see for a lost optics mark. In wave optics, the matching habit is converting every length to metres on the first line of working.
How to prepare these units for NEET
These units usually come at the end of Class 12, when board exams, school tests and NEET revision all compete for time. This order works for most students:
- Read the NCERT chapter first (Class 12 Part II, Chapters 9 to 14), including the in-text examples. NEET theory questions often test a sentence NCERT states clearly.
- Fill the NEET-only gaps: Brewster's law, the Zener regulator, LED, photodiode, solar cell and logic gates. Our guide to the best books for NEET physics explains where to find material for them.
- Keep one formula table per unit, like those above, beside you while practising. Our NEET physics formula sheet covers the whole syllabus.
- Practise numericals by type: photoelectric, Bohr ratios, binding energy, lenses and mirrors, prism, fringe width. Write units first, signs second and the formula third.
- Revise the experiments that NEET's Experimental Skills unit links to these chapters: focal length of mirrors and a convex lens by parallax, the prism's deviation against incidence curve, refractive index of a glass slab with a travelling microscope, and diode and Zener characteristic curves. Know what each graph looks like.
- Solve previous year questions chapter by chapter, checking answers against NTA's official answer keys where available, then move to mixed timed sets with negative marking in mind.
Students in Gurgaon and across Gurugram often reach these chapters late in Class 12, just before pre-boards. If time is short, start with dual nature, atoms and nuclei, which are compact and formula-driven, then ray optics, then wave optics and semiconductors.
Frequently asked questions
Which chapters count as modern physics in NEET?
Usually Unit 17 (dual nature of matter and radiation), Unit 18 (atoms and nuclei) and Unit 19 (electronic devices) of the NEET (UG) 2026 syllabus. In NCERT, these are Chapters 11 to 14 of Class 12 Physics Part II.
Is modern physics easy to score in NEET?
For most students, yes. Questions are usually short, and a small set of formulas covers most of them. Marks are lost mainly to unit errors, sign errors and forgetting the Z2 for hydrogen-like ions, all of which practice fixes.
Are Zener diodes and logic gates in the NEET 2026 syllabus?
Yes. Unit 19 lists the Zener diode as a voltage regulator, the LED, photodiode and solar cell, and the OR, AND, NOT, NAND and NOR gates. The current NCERT Class 12 chapter stops at the diode as a rectifier, so these need another source.
Is radioactivity in the NEET 2026 syllabus?
It is not named in Unit 18, which lists nuclear composition and size, atomic masses, the mass–energy relation, mass defect, binding energy per nucleon, fission and fusion. The current NCERT chapter has only a short descriptive section on it. Check the syllabus for your exam year before deciding how much time to give it.
Is Brewster's law in NCERT?
Not in the current NCERT Class 12 wave optics chapter, which covers Polaroids and Malus' law. Brewster's law is listed in Unit 16 of the NEET (UG) 2026 syllabus, so NEET students should learn it.
Can a home tutor help with modern physics and optics for NEET?
Yes. A one-to-one tutor can see whether a student's errors come from concepts, signs or units, and can teach the NEET-only topics the school textbook no longer covers. Our NEET Physics tutors in Gurgaon teach at home or online.
Want modern physics and optics taught to the NEET syllabus, not just the board syllabus? Book a free NEET Physics demo class with Ajay Vatsyayan Classes, Saraswati kunj II, Wazirabad, Sector 52, Gurugram, Haryana 122003. Male and female tutors are available, at home or online.
Book a Free NEET Physics Demo +91 92204 75088