Where the 109 marks went

NEET Physics Mock · Ray & Wave Optics and Dual Nature of Radiation & Matter · sat 10 November 2026

Aamirah Fathima · NEET 2027

SCORE211of 320

Correct58of 80

Wrong21−21 marks

Blank1Q73

Percentage66%

Twenty-one questions went wrong. Seven of them were a single missing or spare factor of 2 — nothing to do with understanding the physics. Another two were electron-volts reported as joules. Fix those two habits alone and this paper is 256 / 320, not 211. That is the difference between sitting below the Tamil Nadu government-college line and sitting comfortably above it.

Optics 115 / 160 · Dual Nature 96 / 160

As it stands   58 right, 21 wrong, 1 blank = 211 / 320 (65.9%)
Factor-of-2 errors corrected   65 right, 14 wrong = 246 / 320 (76.9%)   +35
eV / J errors corrected too   67 right, 12 wrong = 256 / 320 (80.0%)   +45
Every question you turn from wrong to right is worth 5 marks, not 4 — you gain the 4 and stop losing the 1.

The twenty-one, grouped by cause

7The factor of 235 marks

Seven of your twenty-one errors are a single missing or spare factor of 2. Not one of them is a misunderstanding of the physics — in every case you had the right method and the right numbers. This one habit is worth 35 marks.

Q26 6 mm → 12 mm

You found λD/a, the distance from the centre to the first minimum. The width of the central maximum is twice that.

Fix Central maximum: width = 2λD/a. Half-width = λD/a. The word 'width' always means the full span.

Q38 60° → 45°

Unpolarised light loses half its intensity at the first polaroid. You solved I₀cos²θ = I₀/4 and got 60°; the equation is (I₀/2)cos²θ = I₀/4.

Fix Unpolarised in → halve first, then Malus.

Q40 I₀/8 → I₀/4

Here the light was already plane-polarised, so there is no half to take — and you took one anyway. This is the exact mirror of Q38.

Fix Already polarised → Malus alone, no ½.

Q54 6.6 × 10⁵ → 9.4 × 10⁵

K = 2.5 eV was right. You then used v = √(K/m) instead of √(2K/m) — your answer is exactly √2 times too small.

Fix K = ½mv², so v = √(2K/m). The 2 comes up from the ½.

Q50 0.5 eV → 1.0 eV

ΔK = 1243(1/248 − 1/310) = 1.00 eV. You wrote half of it.

Fix Compute both photon energies in full (5.01 and 4.01 eV) and subtract.

Q63 600 nm → 300 nm

λ₀ = 1243/4.14 = 300 nm. You doubled it.

Fix λ₀ = hc/φ, nothing more.

Q75 1200 V → 600 V

V = 604 V. You doubled it. Q73 — the same formula run backwards — you left blank.

Fix K(eV) = 150.4/λ(Å)². For 0.5 Å: 150.4/0.25 = 602.

2Electron-volts and joules10 marks

Both of these are the same slip: a number that was in joules got reported as electron-volts, or the ×1.6 × 10⁻¹⁹ was never applied.

Q52 1.5 × 10⁻¹⁹ J → 2.4 × 10⁻¹⁹ J

K = 1.5 eV. You attached × 10⁻¹⁹ to the number 1.5 without multiplying by 1.6.

Fix eV → J means × 1.6 × 10⁻¹⁹, every time.

Q60 3.98 eV → 2.49 eV

φ = hν₀ = 3.978 × 10⁻¹⁹ J. You read the mantissa 3.98 straight off as the answer in eV. The paper offered 3.98 as an option precisely to catch that.

Fix Finish the conversion: 3.978 × 10⁻¹⁹ / 1.6 × 10⁻¹⁹ = 2.49 eV.

2Which de Broglie relation applies10 marks

There are three different constraints and three different laws. You used the wrong one each time. Learn them as a set, not one by one.

Q71 1 : 2 → 4 : 1

Same speed. Then λ = h/mv, so λ ∝ 1/m and λ_p:λ_α = 4:1. You used the equal-energy law λ ∝ 1/√m, and inverted it as well.

Fix Same speed → λ ∝ 1/m.

Q74 √2 : 1 → 2 : 1

Same potential difference. Then λ = h/√(2mqV). You used the masses (√(4/2) = √2) and forgot that the α-particle also carries twice the charge.

Fix Same V → λ ∝ 1/√(mq). Charge counts too.

2One change followed, the other missed10 marks

In both of these, two things changed at once. You tracked one of them and stopped.

Q28 doubles → unchanged

Width = 2λD/a. Doubling D doubles it; doubling a halves it. The two cancel exactly.

Fix Write the formula, change every symbol the question changes, then look.

Q56 2.62 V → 4.42 V

You doubled the stopping potential at 400 nm (1.31 V) to get 2.62 V. Halving λ doubles the photon energy, not the stopping potential — because φ is subtracted only once.

Fix Recompute from scratch: 1243/200 − 1.8 = 4.42 V. eV₀ is not proportional to 1/λ.

2Counting to the boundary10 marks

Both are coaxial-source questions, and both turn on what happens at the edge of the range.

Q33 11 → 10

Maxima need d cosθ = nλ, so cosθ = n/10. The n = 0 fringe would sit at θ = 90°, parallel to the screen, and never lands on it. n runs 1 to 10.

Fix Check whether the end member of a count is physically reachable.

Q34 cos⁻¹(1/9) → cos⁻¹(8/9)

You used n = 0.5, as though the path difference were measured from the centre. It is absolute: d cosθ = nλ with d = 4.5λ, so the first ring is the largest whole n below 4.5, which is 4.

Fix cosθ = nλ/d, n whole. First ring = largest n that fits.

1Direction and side5 marks

Still the family that has not shrunk. The arithmetic was perfect; the answer was placed on the wrong side of the surface.

Q15 90 cm on the object side (virtual) → 90 cm inside the glass (real)

You solved 1.5/v + 1/30 = 0.05 correctly and got v = +90 cm. A positive v is on the far side, where the light is going: real, inside the glass.

Fix After every sign-convention problem, say out loud which side the sign puts the image on before you look at the options.

5One-off slips25 marks

Five errors with nothing in common. These are the cheapest to fix.

Q23 A → B

Magnifying power rises with a longer objective focal length (A is true). Resolving power depends on aperture (R is true). But resolving power has nothing to do with magnifying power, so R is not the explanation.

Fix In assertion–reason, check the link separately from the two statements. A true reason is not automatically the reason.

Q36 200 nm → 100 nm

You used 2μt = mλ, which is the condition for the film to look dark. Reflection off the denser medium adds an extra half-wave, so bright needs 2μt = (m + ½)λ.

Fix Reflected bright: 2μt = (m + ½)λ. Reflected dark: 2μt = mλ. Q37, which uses the dark one, you got right.

Q55 0.34 cm → 3.4 cm

r = mv/qB = 0.0337 m = 3.37 cm. A clean decade slip.

Fix Carry the power of ten through the working rather than at the end.

Q66 1.3 × 10¹⁸ → 1.6 × 10¹⁹

1.99 × 10²⁰ photons leave the lamp each second. You divided by 4πr² twice.

Fix One division by the sphere area, at the end.

Q57 6λ → 4λ

3eV₀ = hc/λ − φ and eV₀ = hc/2λ − φ. Eliminating V₀ gives 2φ = hc/2λ, so φ = hc/4λ and λ₀ = 4λ.

Fix Multiply the second equation by 3 before subtracting, and write every line out.

What is already solid

Thirty-six of your fifty-eight correct answers fall in these blocks. None of this needs re-studying — the concepts are in place, and the marks are being lost in execution.

Topic	Questions	Score	What it covered
Mirrors, slabs and refraction at a surface	Q1–Q14	14 / 14	Two-mirror deviation, convex and concave mirror magnification, lateral shift, apparent depth through two media, critical angle. Not one error.
Lenses	Q18–Q21	4 / 4	Lensmaker's formula in a liquid, when a convex lens turns diverging, μ from f, and a two-lens combination.
Photoelectric effect — the concepts	Q41–Q49	9 / 9	Emission types, what intensity controls and what frequency controls, below-threshold behaviour, saturation current.
Reading graphs	Q59, Q61, Q64, Q65, Q67	5 / 5	Including the h/e-versus-h slope distinction, which catches most people.
de Broglie basics	Q68, Q69, Q70, Q72	4 / 4	Equal-energy and equal-wavelength comparisons both handled correctly.

What to do before the next paper

The factor-of-2 pass. After finishing any numerical answer, ask one question: is there a 2 or a ½ that belongs here? Check ½mv², the full-versus-half width of a maximum, the ½ that unpolarised light loses, and the 2 in 2μt.
Never write 10⁻¹⁹ without the 1.6. If an answer is wanted in joules and you have electron-volts, write the multiplication down as a separate line. Q52 and Q60 were both lost inside that one step.
The three de Broglie cases, on one card. Same speed → λ ∝ 1/m. Same kinetic energy → λ ∝ 1/√m. Same accelerating potential → λ ∝ 1/√(mq). Then redo Q69, Q70, Q71 and Q74 together; you got the first two right and the second two wrong, and the only difference is which line applies.
Q73 was left blank and Q75 was doubled — the same relation, K(eV) = 150.4/λ(Å)², both times. Learn that one line and both marks come back.
Say the side out loud. On Q15 your arithmetic was exact and the image went on the wrong side of the glass. Before reading the options, finish the sentence: “v is positive, so the image is real, inside the glass.”