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Ray & Wave Optics: Error Notes

Aamirah Fathima · NCERT Class 12 Physics · 100 questions · +4 / −1 / 0

Score275 / 400
Correct74
Wrong21
Skipped5
Accuracy78%
Target 300: missed by 25 marks. Above the danger line of 240 by 35.
Allowed wrong 20: 21 wrong, one over the limit. The 21 wrong answers cost 105 marks (21 × 4 lost + 21 negative).
Quickest gain: the five “coherent sources on the axis” questions (Q85–89) all use one idea, path difference = d cosθ. Getting the five right adds up to 21 marks.
Strong areas: Prisms & dispersion (8/8) and Velocity of the image (4/4) were perfect, and YDSE was 13/14.

Section-wise marks

SectionQs✔✘–Marks
Plane mirrors & reflectionQ1–541015/20
Spherical mirrorsQ6–1463021/36
Refraction at plane surfacesQ15–2362122/36
Total internal reflectionQ24–2841015/20
Refraction at a spherical surfaceQ29–312107/12
Thin lensesQ32–3641015/20
Lens combinations, power & silvered lensQ37–4251019/24
Prisms & dispersionQ43–5080032/32
Optical instrumentsQ51–5651019/24
Velocity of the imageQ57–6040016/16
Young's double slitQ61–74131051/56
Diffraction & resolutionQ75–8455015/40
Coherent sources on the axisQ85–89014-1/20
Huygens principle & superpositionQ90–9661023/28
PolarisationQ97–1002206/16

Answer grid

14223442→4516272→383→492104→311112413114315216117118–→2191→3202→4214221232243254263271283→2291304→2313324332→4341352363372382→3394401413421434442454463471482493503511522534541554561→2572583593602614621633641652664671683694702713723→4732741752→3762774781792801→4814→1821→3832842→385–→1863→487–→188–→389–→2903914924931942953964→1974983→2991→31001
Correct (her answer)Wrong (hers → key)Skipped (– → key)Click a red or grey box to jump to its note.

Error patterns

PatternQuestions
Stopping one step early / answering a different quantityQ10 (gave v not f), Q20 (apparent depth not shift), Q75 (half-width), Q99 (stopped after P₁)
Sign-convention slipsQ7, Q8, Q30
Formula half-rememberedQ4 (2θ vs 360° − 2θ), Q38 (extra factor 2), Q72 (φ instead of φ/2), Q96 (missed 2 and π-shift), Q98
EXCEPT / Assertion–Reason readingQ28, Q81 (picked a true statement), Q56 (definition ≠ explanation)
Concept gapsQ19 (λ in medium), Q33 (lens in denser liquid), Q80, Q82, Q84 (resolution vs magnification), Q86
Skipped: whole new topic blockQ85, Q87, Q88, Q89 (coherent sources on the axis) + Q18 (lateral shift)

Error notes (21 wrong + 5 skipped)

Q4Plane mirrorsWrong · −1

A ray of light is reflected successively from two plane mirrors inclined to each other at 50°. The total deviation suffered by the ray, independent of its angle of incidence, is

  1. (1)130°
  2. (2)100°your answer
  3. (3)80°
  4. (4)260°correct
What went wrong: 100° = 2 × 50°. You used 2θ, which is the angle between the incident and emergent lines, not the deviation.

Solution

  1. One reflection at angle of incidence i deviates the ray by 180° − 2i.
  2. For two mirrors at angle θ, adding the two deviations gives δ = 360° − 2θ (independent of i).
  3. δ = 360° − 2(50°) = 260°.
θ = 50° incidentemergent Deviation after two reflections: δ = 360° − 2θ
Remember: Two mirrors at θ → deviation 360° − 2θ. If θ = 90°, δ = 180° (the ray comes back parallel).
Q7Spherical mirrorsWrong · −1

An object is placed 60 cm from a convex mirror of focal length 20 cm. The image is

  1. (1)virtual, erect and of the same size, 60 cm behind the mirror
  2. (2)virtual, erect and 50% of the size of the object, 30 cm behind the mirroryour answer
  3. (3)virtual, erect and 25% of the size of the object, 15 cm behind the mirrorcorrect
  4. (4)real, inverted and 25% of the size of the object, 15 cm in front of the mirror
What went wrong: Sign error: you computed 1/v = 1/20 − 1/60 = 1/30. With u = −60 the formula gives 1/20 + 1/60.

Solution

  1. Convex mirror: f = +20 cm, u = −60 cm.
  2. 1/v = 1/f − 1/u = 1/20 − 1/(−60) = 1/20 + 1/60 = 4/60 = 1/15 → v = +15 cm (behind the mirror).
  3. m = −v/u = −15/(−60) = +¼ → erect, 25% size, virtual.
O (u = −60) I (+15) F (+20) Image: behind the mirror, erect, m = +¼ (scale 2 px per cm)
Remember: Always put the sign on u before substituting. A convex mirror always gives |v| < f.
Q8Spherical mirrorsWrong · −1

A mirror forms an erect image one-third the size of an object placed 40 cm in front of it. The radius of curvature of the mirror is

  1. (1)40 cm, and the mirror is concave
  2. (2)20 cm, and the mirror is convex
  3. (3)80 cm, and the mirror is convexyour answer
  4. (4)40 cm, and the mirror is convexcorrect
What went wrong: You identified the convex mirror correctly but doubled twice (took f = 40 or R = 4f).

Solution

  1. Erect + diminished → only a convex mirror.
  2. m = −v/u = +1/3 with u = −40 → v = +40/3 cm.
  3. 1/f = 1/v + 1/u = 3/40 − 1/40 = 2/40 → f = +20 cm.
  4. R = 2f = 40 cm, convex.
Remember: Find f first, then R = 2f. Write f and R separately so you don't mix them up.
Q10Spherical mirrorsWrong · −1

The distance between an object and its four-times magnified virtual image produced by a curved mirror is 45 cm. The focal length of the mirror is

  1. (1)+12 cm
  2. (2)−9 cm
  3. (3)−12 cmcorrect
  4. (4)−36 cmyour answer
What went wrong: −36 cm is the image distance, not f. You stopped one step early.

Solution

  1. Virtual and magnified → concave mirror, m = +4.
  2. Let u = −x; then v = +4x (behind the mirror). Object and image are on opposite sides, so separation = x + 4x = 5x = 45 → x = 9.
  3. u = −9 cm, v = +36 cm.
  4. 1/f = 1/v + 1/u = 1/36 − 1/9 = −3/36 → f = −12 cm.
F (12) O: 9 cm I: 36 cm behind Object 9 cm in front + image 36 cm behind = 45 cm apart
Remember: Before marking, check that you've answered what the question asks for (f, not v).
Q18Refraction at plane surfacesSkipped · 0

A ray strikes a parallel-sided glass slab of thickness 6 cm and refractive index √3 at an angle of incidence i = 60°. The perpendicular distance between the incident ray produced and the emergent ray is

  1. (1)√3 cm
  2. (2)2√3 cmcorrect
  3. (3)3√3 cm
  4. (4)6 cm
What went wrong: Not attempted. The question looks long, but it only needs the lateral-shift formula plus Snell's law.

Solution

  1. Snell: sin r = sin 60°/√3 = (√3/2)/√3 = ½ → r = 30°.
  2. Lateral shift d = t·sin(i − r)/cos r = 6 × sin 30°/cos 30° = 6 × (½)/(√3/2) = 6/√3.
  3. d = 2√3 cm.
μ = √3t = 6 cm i = 60°r = 30° d d = t·sin(i − r)/cos r ; dashed = incident ray produced
Remember: Lateral shift: d = t sin(i − r)/cos r. Memorise it together with the normal shift t(1 − 1/μ).
Q19Refraction at plane surfacesWrong · −1

A monochromatic wave of wavelength λ₁ and frequency ν₁ in air enters a medium. The angles of incidence and refraction are 45° and 30°. The wavelength λ₂ and frequency ν₂ in the medium are

  1. (1)λ₂ = √2 λ₁, ν₂ = ν₁your answer
  2. (2)λ₂ = λ₁, ν₂ = √2 ν₁
  3. (3)λ₂ = λ₁/√2, ν₂ = ν₁correct
  4. (4)λ₂ = λ₁, ν₂ = ν₁/√2
What went wrong: Frequency was right, but the wavelength gets shorter in a denser medium. You multiplied by μ instead of dividing.

Solution

  1. μ = sin 45°/sin 30° = (1/√2)/(½) = √2.
  2. Frequency is set by the source, so ν₂ = ν₁.
  3. v = c/μ and λ = v/ν → λ₂ = λ₁/μ = λ₁/√2.
Remember: Entering a denser medium: speed ↓, wavelength ↓, frequency unchanged.
Q20Refraction at plane surfacesWrong · −1

A glass slab of thickness 12 cm and refractive index 1.5 is placed over an ink mark on a table. Viewed from directly above, the mark appears to be raised by

  1. (1)18 cm
  2. (2)8 cmyour answer
  3. (3)12 cm
  4. (4)4 cmcorrect
What went wrong: 8 cm is the apparent depth. The question asks how much the mark is raised.

Solution

  1. Apparent depth = t/μ = 12/1.5 = 8 cm.
  2. Shift = real depth − apparent depth = 12 − 8 = 4 cm (or t(1 − 1/μ)).
real mark appears here shift 4 cm t = 12 Apparent depth = t/μ = 8 cm → raised by t − t/μ = 4 cm
Remember: Read the last word of the question: “apparent depth” vs “raised by” / “shift”.
Q28Total internal reflectionWrong · −1

Each of the following is true of total internal reflection, EXCEPT one. Identify the exception.

  1. (1)The critical angle is smaller for a medium of larger refractive index
  2. (2)It can occur when light passes from a rarer medium into a denser mediumcorrect
  3. (3)It requires the angle of incidence to exceed the critical angleyour answer
  4. (4)It is the principle on which an optical fibre guides light along its core
What went wrong: Option 3 is a TRUE condition for TIR. In an EXCEPT question you have to pick the false statement.

Solution

  1. TIR needs two conditions: (i) light goes from denser → rarer, (ii) i > critical angle.
  2. sin C = 1/μ → larger μ gives smaller C (true). Optical fibres work by TIR (true).
  3. Rarer → denser can never give TIR → statement (2) is the exception.
Remember: For EXCEPT/NOT questions, write T or F next to each option before choosing.
Q30Refraction at a spherical surfaceWrong · −1

An object O stands 10 cm from the pole P of a spherical surface of radius of curvature 20 cm separating a medium of index n₁ = 1 from one of index n₂ = 2. After a single refraction the image lies

  1. (1)40 cm from P, on the far side of the surface
  2. (2)40 cm from P, on the same side as the objectcorrect
  3. (3)80 cm from P, on the far side of the surface
  4. (4)20 cm from P, on the same side as the objectyour answer
What went wrong: A sign slip with u or R. Here the centre of curvature is in the denser medium, so R = +20.

Solution

  1. n₂/v − n₁/u = (n₂ − n₁)/R with u = −10, R = +20.
  2. 2/v + 1/10 = 1/20 → 2/v = −1/20 → v = −40 cm.
  3. Negative v → image on the same side as the object, i.e. virtual, 40 cm from P.
n₁ = 1n₂ = 2 P C (+20) O (−10) I (−40) n₂/v − n₁/u = (n₂ − n₁)/R → v = −40 cm: virtual, same side as O
Remember: Object inside the focal length of a refracting surface → virtual image (like a magnifying glass).
Q33Thin lensesWrong · −1

A thin convex lens made of crown glass is immersed in a liquid whose refractive index is greater than that of the glass. The lens then behaves as

  1. (1)a converging lens of longer focal length
  2. (2)a plane sheet of glass, producing no convergence or divergenceyour answer
  3. (3)a converging lens of shorter focal length
  4. (4)a diverging lenscorrect
What went wrong: “Plane sheet” is right only when n(liquid) = n(glass). Here n(liquid) is greater.

Solution

  1. Lens-maker's formula in a medium: 1/f = (n_lens/n_medium − 1)(1/R₁ − 1/R₂).
  2. n_lens/n_medium < 1 → the bracket is negative → f < 0.
  3. A convex lens becomes a diverging lens.
liquid: n > n(glass) (n_lens/n_liquid − 1) < 0 → focal length negative → diverging
Remember: Compare the medium with the glass: equal → no effect; less dense → behaviour kept (f longer); denser → behaviour reversed.
Q38Lens combinationsWrong · −1

A plano-convex lens fits exactly into a plano-concave lens, their plane faces parallel. The curved faces share a radius of curvature R, and the lenses have refractive indices μ₁ and μ₂. The focal length of the combination is

  1. (1)2R/(μ₁ − μ₂)
  2. (2)R/[2(μ₁ − μ₂)]your answer
  3. (3)R/(μ₁ − μ₂)correct
  4. (4)R/(μ₁ + μ₂)
What went wrong: You counted two curved surfaces for each lens. Each lens has only one curved face (the other face is plane, R = ∞).

Solution

  1. Plano-convex: 1/f₁ = (μ₁ − 1)(1/R − 1/∞) = (μ₁ − 1)/R.
  2. Plano-concave: 1/f₂ = −(μ₂ − 1)/R.
  3. In contact: 1/F = 1/f₁ + 1/f₂ = (μ₁ − μ₂)/R → F = R/(μ₁ − μ₂).
μ₁μ₂ common radius R 1/f = (μ₁−1)/R− (μ₂−1)/R = (μ₁−μ₂)/R
Remember: A plane face contributes 1/∞ = 0. Only the factor 2 for a symmetric biconvex lens gives 2/R.
Q56Optical instrumentsWrong · −1

Assertion (A): Increasing the diameter of the objective increases the range of an astronomical telescope.
Reason (R): The range of a telescope is a statement of how far away a star of given brightness can still be detected.

  1. (1)Both A and R are true and R is the correct explanation of Ayour answer
  2. (2)Both A and R are true but R is NOT the correct explanation of Acorrect
  3. (3)A is true but R is false
  4. (4)A is false but R is true
What went wrong: R only defines range. A definition does not explain why a larger objective helps.

Solution

  1. A is true: a bigger objective gathers more light, so fainter (more distant) stars become visible.
  2. R is true: it is simply the meaning of ‘range’.
  3. The real reason is that light-gathering power ∝ D². R doesn't say this → both true, R is not the explanation.
Remember: In A–R questions, ask: “Does R answer why A happens?” A definition almost never does.
Q72Young's double slitWrong · −1

Two coherent beams, each of intensity I₀, meet at a point with a phase difference of π/3. The resultant intensity there is

  1. (1)4I₀
  2. (2)2I₀
  3. (3)I₀your answer
  4. (4)3I₀correct
What went wrong: You used 4I₀cos²(φ) = 4I₀ × ¼. The formula uses φ/2.

Solution

  1. I = I₁ + I₂ + 2√(I₁I₂) cos φ = 2I₀ + 2I₀ cos 60° = 2I₀ + I₀ = 3I₀.
  2. Or: I = 4I₀ cos²(φ/2) = 4I₀ cos²30° = 4I₀ × ¾ = 3I₀.
aaφ = 60° R = 2a cos(φ/2) = √3 a I ∝ R² = 3a² → 3I₀
Remember: Check with known cases: φ = 0 gives 4I₀ and φ = π gives 0. φ/2 inside the cos² makes both work.
Q75DiffractionWrong · −1

Light of wavelength 6000 Å falls normally on a single slit of width 0.2 mm. The angular width of the central maximum is

  1. (1)1.2 × 10⁻² rad
  2. (2)3 × 10⁻³ radyour answer
  3. (3)6 × 10⁻³ radcorrect
  4. (4)2 × 10⁻³ rad
What went wrong: 3 × 10⁻³ rad = λ/a is only the angle to the first minimum (half-width).

Solution

  1. First minimum at sinθ ≈ θ = λ/a = 6 × 10⁻⁷/2 × 10⁻⁴ = 3 × 10⁻³ rad.
  2. The central maximum spans −θ to +θ → angular width = 2λ/a = 6 × 10⁻³ rad.
0λ3λ/22λ −λ central max: 2λD/a path difference between the two edges (a sinθ) at each point
Remember: ‘Width of the central maximum’ means the full width, 2λ/a (or 2λD/a on the screen).
Q80DiffractionWrong · −1

In a single-slit diffraction experiment the width of the slit is doubled while the wavelength is kept the same. The central maximum becomes

  1. (1)half as wide and fainteryour answer
  2. (2)unchanged in width but brighter
  3. (3)twice as wide and fainter
  4. (4)half as wide and brightercorrect
What went wrong: You got the width right but forgot that a wider slit lets more light through, and the light is squeezed into a narrower band.

Solution

  1. Width of central maximum = 2λD/a → doubling a halves it.
  2. Wider slit → more light energy enters, spread over a smaller region → brighter.
0λ3λ/22λ −λ central max: 2λD/a path difference between the two edges (a sinθ) at each point
Remember: Slit wider → pattern narrower and brighter. Slit narrower → pattern wider and dimmer.
Q81DiffractionWrong · −1

Each of the following is true of the single-slit diffraction pattern, EXCEPT one. Identify the exception.

  1. (1)The secondary maxima are of the same intensity as the central maximumcorrect
  2. (2)The minima occur where a sinθ is a whole number of wavelengths
  3. (3)Making the slit narrower spreads the pattern out
  4. (4)The central maximum is twice as wide as each secondary maximumyour answer
What went wrong: Option 4 is TRUE: the central maximum (−λ/a to +λ/a) is twice as wide as each secondary maximum.

Solution

  1. Secondary maxima fall off rapidly: the first is only about 4% of the central maximum → statement (1) is false.
  2. Minima: a sinθ = nλ (true). Narrower slit → wider pattern (true). Central = 2 × secondary width (true).
0λ3λ/22λ −λ central max: 2λD/a path difference between the two edges (a sinθ) at each point
Remember: Single slit: bright central maximum, with secondary maxima that get dimmer and dimmer. Double slit: fringes of equal brightness.
Q82DiffractionWrong · −1

In single-slit diffraction, match each point with the path difference between the waves from the two edges of the slit.
A. Centre · B. First minimum · C. First secondary maximum · D. Second minimum
I. λ · II. 3λ/2 · III. Zero · IV. 2λ

  1. (1)A-III, B-I, C-IV, D-IIyour answer
  2. (2)A-I, B-III, C-II, D-IV
  3. (3)A-III, B-I, C-II, D-IVcorrect
  4. (4)A-III, B-II, C-I, D-IV
What went wrong: You swapped C and D. Secondary maxima are at half-integral multiples (3λ/2), and minima at whole multiples (2λ).

Solution

  1. Centre: path difference 0 (III).
  2. Minima: a sinθ = nλ → first λ (I), second 2λ (IV).
  3. Secondary maxima: a sinθ ≈ (n + ½)λ → first 3λ/2 (II).
  4. → A-III, B-I, C-II, D-IV.
0λ3λ/22λ −λ central max: 2λD/a path difference between the two edges (a sinθ) at each point
Remember: Single slit is the reverse of YDSE: minima at nλ, maxima at (n + ½)λ.
Q84ResolutionWrong · −1

The resolving power of an astronomical telescope can be increased by

  1. (1)using an eyepiece of shorter focal length
  2. (2)lengthening the tube between the two lensesyour answer
  3. (3)using an objective of larger diametercorrect
  4. (4)using light of longer wavelength
What went wrong: Lengthening the tube or changing the eyepiece changes magnification, not resolution.

Solution

  1. Resolving power = D/(1.22λ).
  2. Only a larger D (objective diameter) or a smaller λ increases it. Longer λ decreases it.
  3. → larger objective diameter.
Remember: Magnifying power depends on fₒ/fₑ; resolving power depends on D/λ. Keep the two apart.
Q85Coherent sources on the axisSkipped · 0

Two coherent point sources of wavelength λ are placed a distance d apart on the line normal to a distant screen, one behind the other, in phase. O is where that line meets the screen, D ≫ d. Here d = 0.5 mm and λ = 5000 Å. The order of the bright fringe formed at O is

  1. (1)1000correct
  2. (2)100
  3. (3)500
  4. (4)10 000
What went wrong: Not attempted. The whole Q85–89 block uses one idea: path difference = d cosθ.

Solution

  1. At O (θ = 0) path difference = d.
  2. Order n = d/λ = 5 × 10⁻⁴/5 × 10⁻⁷ = 1000.
O S₁S₂ d θ Path difference = d cosθ θ = 0 (at O): Δ = d (maximum) Symmetric about S₁S₂ → equal-Δ points form circles on the screen
Remember: Sources one behind the other: Δ = d cosθ, largest at the centre O.
Q86Coherent sources on the axisWrong · −1

(Same set-up.) Here d = 3λ. The number of bright fringes on the screen and their shape are

  1. (1)three, and hyperbolic
  2. (2)three, and straight
  3. (3)an infinite number, and circularyour answer
  4. (4)three, and circularcorrect
What went wrong: The shape (circular) was right, but the number of fringes is limited because the path difference can never exceed d.

Solution

  1. Δ = d cosθ ranges from 3λ (at O) down to 0 (θ = 90°, which never reaches the screen).
  2. Bright fringes where Δ = 3λ, 2λ, λ → three.
  3. The set-up is symmetric about the axis → fringes are circles centred on O.
O S₁S₂ d θ Path difference = d cosθ θ = 0 (at O): Δ = d (maximum) Symmetric about S₁S₂ → equal-Δ points form circles on the screen
Remember: Maximum path difference = d → number of bright fringes ≤ d/λ.
Q87Coherent sources on the axisSkipped · 0

(Same set-up.) Here d = 7λ/2. The point O is

  1. (1)a point of minimum intensitycorrect
  2. (2)the centre of the seventh bright fringe
  3. (3)a point of intensity exactly half the maximum
  4. (4)a point of maximum intensity
What went wrong: Not attempted.

Solution

  1. Path difference at O = d = 7λ/2 = (2n − 1)λ/2 with n = 4 → odd multiple of λ/2.
  2. Destructive interference → minimum intensity (dark centre).
O S₁S₂ d θ Path difference = d cosθ θ = 0 (at O): Δ = d (maximum) Symmetric about S₁S₂ → equal-Δ points form circles on the screen
Remember: Whole λ → bright; half-odd λ → dark.
Q88Coherent sources on the axisSkipped · 0

(Same set-up.) How many of the following are correct?
A. Path difference at O equals d.
B. Path difference decreases as one moves outwards from O.
C. Fringes are circles centred on O.
D. Fringes are evenly spaced, exactly as in the usual YDSE.

  1. (1)Only one
  2. (2)Only two
  3. (3)Only threecorrect
  4. (4)All four
What went wrong: Not attempted.

Solution

  1. A: Δ = d cos0 = d ✔. B: cosθ falls as θ grows ✔. C: axial symmetry → circles ✔.
  2. D ✘: Δ = d cosθ is not linear in screen position, so spacing is not uniform.
  3. → only three.
O S₁S₂ d θ Path difference = d cosθ θ = 0 (at O): Δ = d (maximum) Symmetric about S₁S₂ → equal-Δ points form circles on the screen
Remember: Uniform fringe width β = λD/d holds only for the usual side-by-side slits.
Q89Coherent sources on the axisSkipped · 0

(Same set-up.) The fringes come out circular rather than straight because

  1. (1)diffraction at the sources dominates over interference
  2. (2)the whole arrangement is symmetric about the line joining the sources, so the path difference depends only on the angle from that linecorrect
  3. (3)the two sources have different intensities
  4. (4)the screen is curved rather than flat
What went wrong: Not attempted.

Solution

  1. Rotate the whole set-up about the S₁S₂ line and nothing changes.
  2. So Δ depends only on θ, and points of constant θ on a flat screen form circles → option 2.
O S₁S₂ d θ Path difference = d cosθ θ = 0 (at O): Δ = d (maximum) Symmetric about S₁S₂ → equal-Δ points form circles on the screen
Remember: Symmetry gives the fringe shape: side-by-side slits → straight lines; sources on the axis → circles.
Q96Thin filmsWrong · −1

A thin film of refractive index 1.5 is viewed in reflected light of wavelength 6000 Å at normal incidence. The smallest thickness for which the film appears bright is

  1. (1)1000 Åcorrect
  2. (2)2000 Å
  3. (3)1500 Å
  4. (4)4000 Åyour answer
What went wrong: You used μt = λ (4000 Å). This misses the factor 2 for the round trip and the extra π phase change on reflection.

Solution

  1. Reflection from the top (rarer → denser) adds a phase of π (λ/2).
  2. Bright in reflection: 2μt = (2n − 1)λ/2. Smallest: n = 1 → 2μt = λ/2.
  3. t = λ/(4μ) = 6000/(4 × 1.5) = 1000 Å.
μ = 1.5, t ray 1: phase π (reflects off denser) ray 2: extra path 2μt Bright in reflection: 2μt = (2n − 1)λ/2 → t_min = λ/4μ
Remember: Thin film in reflection: bright when 2μt = (n − ½)λ, dark when 2μt = nλ (the reverse of what you'd expect).
Q98PolarisationWrong · −1

Unpolarised light of intensity I₀ passes through two polaroids whose transmission axes make an angle of 30° with each other. The emergent intensity is

  1. (1)3I₀/4
  2. (2)3I₀/8correct
  3. (3)I₀/4your answer
  4. (4)I₀/8
What went wrong: You used cos²60° (= ¼) or cos 60°. The angle between the axes is 30°, so use cos²30° = ¾.

Solution

  1. First polaroid: I₀ → I₀/2 (always half for unpolarised light).
  2. Malus: I = (I₀/2) cos²30° = (I₀/2)(¾) = 3I₀/8.
P₁P₂ (45°)P₃ (90°) I₀I₀/2I₀/4I₀/8 Unpolarised → ½, then × cos²(angle to previous axis)
Remember: Use cos² of the angle between the two axes, and halve first.
Q99PolarisationWrong · −1

Unpolarised light of intensity I₀ passes through three polaroids. The axis of P₂ makes 45° with P₁, and P₃ is perpendicular to P₁. The intensity emerging from P₃ is

  1. (1)I₀/2your answer
  2. (2)I₀/4
  3. (3)I₀/8correct
  4. (4)zero
What went wrong: I₀/2 is only what comes out of P₁. You have to apply Malus's law at P₂ and at P₃.

Solution

  1. After P₁: I₀/2.
  2. After P₂ (45° to P₁): (I₀/2) cos²45° = I₀/4.
  3. After P₃ (45° to P₂, since P₃ ⟂ P₁): (I₀/4) cos²45° = I₀/8.
  4. (Without P₂, P₁ and P₃ are crossed → zero. The middle polaroid ‘rescues’ some light.)
P₁P₂ (45°)P₃ (90°) I₀I₀/2I₀/4I₀/8 Unpolarised → ½, then × cos²(angle to previous axis)
Remember: For each polaroid, use the angle relative to the previous polaroid, not the first one.