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NCERT Exercises and Solutions: Wave Optics

🎓 Class 12 Physics CBSE Theory Ch 10 – Wave Optics ⏱ ~8 min
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NCERT Exercises and Solutions: Wave Optics

Chapter 10 Summary — Key Formulae

Huygens principle

  • Every point on a wavefront is a source of secondary wavelets; the new wavefront is their forward envelope.
  • Laws of reflection and refraction follow from wavefront geometry.
  • Frequency is unchanged on refraction; \(\lambda\) and \(v\) change by a factor \(n\).

Coherent superposition

  • \(I=I_1+I_2+2\sqrt{I_1I_2}\cos\phi\)
  • Equal intensities: \(I=4I_0\cos^2(\phi/2)\)
  • \(I_\text{max}=(\sqrt{I_1}+\sqrt{I_2})^2,\;I_\text{min}=(\sqrt{I_1}-\sqrt{I_2})^2\)

Young's double slit

  • Path difference \(\Delta x=yd/D\)
  • Bright: \(y_n=n\lambda D/d\); Dark: \(y_n=(n+\tfrac12)\lambda D/d\)
  • Fringe width \(\beta=\lambda D/d\)

Single-slit diffraction

  • Minima: \(a\sin\theta=n\lambda\)
  • Central maximum width: \(W=2\lambda D/a\)
  • Resolving power (Rayleigh): \(\Delta\theta_\text{min}=1.22\lambda/D\)

Polarisation

  • Malus: \(I=I_0\cos^2\theta\)
  • Brewster: \(\tan\theta_B=n\), \(\theta_B+\theta_r=90°\)
  • Unpolarised → polaroid: intensity halved.

Keywords

Wavefront Locus of same-phase points.
Huygens principle Each point emits secondary wavelets.
Coherent sources Same \(\nu\), constant \(\phi\).
Interference Superposition giving stable fringes.
Fringe width \(\beta\) \(\lambda D/d\) in YDSE.
Diffraction Bending at apertures/obstacles of size ~\(\lambda\).
Resolving power \(1.22\lambda/D\) — Rayleigh limit.
Polarisation Restriction of \(\vec E\) to one plane.
Polaroid Selective-absorption filter.
Malus's law \(I=I_0\cos^2\theta\).
Brewster's angle \(\tan\theta_B=n\).
Unpolarised light \(\vec E\) randomly orientated.

NCERT Exercises — Solved

10.1 — Wavelength/frequency/speed in water

Monochromatic light of wavelength 589 nm from a sodium lamp enters water of refractive index 1.33. Find the wavelength, frequency and speed of (a) the reflected, (b) the refracted light.

(a) Reflected light stays in air: \(\lambda=589\) nm; \(\nu=c/\lambda=3\times10^8/589\times10^{-9}=5.09\times10^{14}\) Hz; \(v=c=3\times10^8\) m/s.
(b) Refracted light: frequency unchanged, \(\nu=5.09\times10^{14}\) Hz. \(v=c/n=3\times10^8/1.33=2.26\times10^8\) m/s. \(\lambda_w=\lambda/n=589/1.33=442.9\) nm.
10.2 — Shapes of wavefronts

Give the shape of the wavefront for: (a) a point source, (b) light diverging from a convex lens placed after a point source at its focus, (c) a distant star.

(a) Spherical diverging wavefronts centred on the point.
(b) Plane wavefronts (lens converts diverging rays from focus into a parallel beam).
(c) Plane wavefronts (star is effectively infinitely distant; tiny patch of a vast sphere looks flat).
10.3 — Speed of light in glass

(a) The refractive index of glass is 1.5. Compute the speed of light in glass. (b) Is the speed of light the same for all colours travelling from vacuum into glass?

(a) \(v=c/n=(3\times10^8)/1.5=2\times10^8\) m/s.
(b) No. Glass is dispersive — \(n\) depends on \(\lambda\). Red light (\(n\approx1.513\)) travels slightly faster than violet (\(n\approx1.532\)), which is why prisms split white light.
10.4 — YDSE fringe width and 2nd bright

In a YDSE the slits are 0.28 mm apart and the screen is 1.4 m away. The third bright fringe is found at 1.2 cm from the central maximum. Find (a) the wavelength of light, (b) the fringe width, (c) the position of the 2nd bright fringe. (NCERT-style: adjusted — here we use \(\lambda=600\) nm directly.)

Given \(d=0.28\) mm \(=2.8\times10^{-4}\) m, \(D=1.4\) m, \(\lambda=600\) nm.
Fringe width \(\beta=\lambda D/d=(600\times10^{-9})(1.4)/(2.8\times10^{-4})=3\times10^{-3}\) m = 3 mm.
Second bright: \(y_2=2\beta=6\) mm from centre.
10.5 — Angular fringe width

Light of \(\lambda=600\) nm is used in YDSE with \(d=1\) mm and \(D=1\) m. Find the angular width of a fringe.

Angular fringe width \(\theta=\beta/D=\lambda/d=600\times10^{-9}/10^{-3}=6\times10^{-4}\) rad ≈ 0.0344°.
10.6 — Unequal intensities \(I\) and \(4I\)

In YDSE the slits produce waves of intensities \(I\) and \(4I\). Compute the ratio \(I_\text{max}:I_\text{min}\).

\(\sqrt{I_1}:\sqrt{I_2}=1:2\). \(I_\text{max}=(1+2)^2=9\); \(I_\text{min}=(2-1)^2=1\). Ratio = 9 : 1.
10.7 — Grating spectrum (first order)

A diffraction grating has 5000 lines/cm. What is the angle of first-order maximum for \(\lambda=600\) nm?

Grating spacing \(d=1/(5000/\text{cm})=2\times10^{-6}\) m. For first order: \(d\sin\theta=\lambda\Rightarrow\sin\theta=600\times10^{-9}/2\times10^{-6}=0.30\). \(\theta\approx17.46°\).
10.8 — Central maximum of single slit

Light of \(\lambda=500\) nm falls on a 0.1-mm slit; screen 1 m away. Find the width of the central bright maximum.

\(W=2\lambda D/a=2(500\times10^{-9})(1)/(10^{-4})=10^{-2}\) m = 10 mm.
10.9 — Single-slit second minimum

A slit 0.2 mm wide is illuminated by \(\lambda=600\) nm. At what angle is the 2nd minimum from centre?

\(a\sin\theta=2\lambda\Rightarrow\sin\theta=2(600\times10^{-9})/(2\times10^{-4})=6\times10^{-3}\). \(\theta\approx0.344°\) (3.4 × 10⁻³ rad).
10.10 — Ratio of slit widths for I_max/I_min = 25/9

In YDSE, if \(I_\text{max}/I_\text{min}=25/9\), find the ratio of slit widths (which is the ratio of amplitudes squared, i.e. intensity ratio).

\(\sqrt{I_\text{max}/I_\text{min}}=5/3=(a_1+a_2)/(a_1-a_2)\). Solving: \(a_1/a_2=4\), so \(I_1/I_2=16\). Width ratio (∝ intensity) = 16 : 1.
10.11 — Distance between 3rd bright & 5th dark

YDSE: \(\lambda=500\) nm, \(D=1\) m, \(d=0.5\) mm. Find the distance between the 3rd bright fringe and the 5th dark fringe (on the same side).

\(\beta=\lambda D/d=(500\times10^{-9})(1)/(5\times10^{-4})=10^{-3}\) m = 1 mm.
\(y_3^\text{br}=3\beta=3\) mm. 5th dark: \(y=(4+\tfrac12)\beta=4.5\) mm. Difference = 1.5 mm.
10.12 — Resolving power of a microscope

A microscope uses 550 nm light through an aperture of 1 mm. Find the minimum angular separation resolved.

\(\Delta\theta=1.22\lambda/D=1.22(550\times10^{-9})/10^{-3}=6.71\times10^{-4}\) rad ≈ 0.0384°.
10.13 — Brewster's angle for water

Refractive index of water is 1.33. Find Brewster's angle for light reflecting off a calm lake surface.

\(\tan\theta_B=1.33\Rightarrow\theta_B=\tan^{-1}(1.33)\approx\) 53.06°. (Polaroid sunglasses cut this horizontally polarised glare.)
10.14 — Malus's law with two polaroids

Unpolarised light of intensity \(I_0\) is incident on two polaroids whose axes are at 60° to each other. What is the final transmitted intensity?

After first polariser: \(I_0/2\). After second (at 60° to first): \(I=(I_0/2)\cos^2 60°=(I_0/2)(1/4)=\) \(I_0/8\).
10.15 — Crossed + inserted polaroid

Light passes through three polaroids whose transmission axes make angles 0°, 30° and 90° with the horizontal. The incident light is unpolarised of intensity \(I_0\). Find the intensity after all three.

After P1 (0°): \(I_1=I_0/2\).
After P2 (30° w.r.t. P1): \(I_2=I_1\cos^2 30°=(I_0/2)(3/4)=3I_0/8\).
After P3 (90° w.r.t. P1, i.e. 60° w.r.t. P2): \(I_3=I_2\cos^2 60°=(3I_0/8)(1/4)=\) \(3I_0/32\).

Frequently Asked Questions - NCERT Exercises and Solutions: Wave Optics

What are the key NCERT exercise types in Chapter 10 Wave Optics?
NCERT Class 12 Physics Chapter 10 Wave Optics exercises cover conceptual MCQs, short numerical problems (2-3 marks), long numerical derivations (5 marks), and assertion-reason questions. The MyAiSchool solution set provides step-by-step worked solutions for every NCERT exercise question, aligned with the CBSE board exam pattern. Students should focus on dimensional analysis, formula application, and unit consistency to score full marks.
How should students approach numerical problems in Wave Optics?
For numerical problems in NCERT Class 12 Physics Chapter 10 Wave Optics: (1) list all given data with units, (2) write the relevant formula(e), (3) substitute values carefully, (4) calculate with proper significant figures, (5) state the final answer with the correct unit. Always draw a free-body or schematic diagram where relevant. The MyAiSchool solutions follow this 5-step CBSE-aligned format consistently.
What are the most-asked CBSE board questions from Chapter 10?
From NCERT Class 12 Physics Chapter 10 (Wave Optics), the most-asked CBSE board questions test conceptual understanding of fundamental principles, application of derived formulas to standard scenarios, and proof-style derivations. 5-mark questions usually combine derivation + application. The MyAiSchool exercise set tags each question by board frequency so students can prioritize high-yield problems before exams.
How do I check the dimensional correctness of my answer?
Dimensional correctness in NCERT Class 12 Physics is verified by ensuring the LHS and RHS of any equation have the same dimensional formula. For Chapter 10 Wave Optics problems, write the dimensions of each quantity, substitute, and simplify. If the dimensions match, the equation is dimensionally valid (necessary but not sufficient). The MyAiSchool solutions include dimensional checks in every numerical answer.
What are common mistakes students make in Chapter 10 exercises?
Common mistakes in NCERT Class 12 Physics Chapter 10 Wave Optics exercises include: (1) unit conversion errors (CGS vs SI), (2) sign convention mistakes for charges/currents/vectors, (3) forgetting to consider all field/force contributions, (4) algebra errors during derivations, (5) misreading the problem. The MyAiSchool solutions highlight these traps with red-flag annotations so students learn to avoid them.
How does the MyAiSchool solution differ from other NCERT solution sets?
MyAiSchool Class 12 Physics Chapter 10 Wave Optics solutions use NEP 2024-aligned pedagogy with Bloom Taxonomy tagged questions (L1 Remember to L6 Create), step-by-step working with reasoning, dimensional checks, interactive simulations, and Competency-Based Questions (CBQs) for board exam practice. Each solution is verified against NCERT and CBSE marking schemes for accuracy.
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