Chapter 7 — Wave Optics Exercise Solutions
Wave optics is concept plus formula. Learn the difference between interference and diffraction clearly.
How to Use This Page
- Read the Key Concepts table first to refresh definitions before attempting questions.
- Try each MCQ on your own before checking the bold correct answer and explanation.
- For numerical problems, follow the YDSE Numerical Pattern steps — always write the formula before substituting values.
- Use the Common Mistakes to Avoid section as a final check before submitting your exam paper.
- Revisit the Exam Checklist the night before your exam to make sure you have covered every essential point.
Key Concepts (Quick Recap)
| Term | Definition |
|---|
| Interference | Superposition of two coherent waves producing alternate bright and dark fringes; requires coherent sources with constant phase difference. |
| Coherent sources | Two sources that emit light with the same frequency and a constant phase difference. |
| Path difference | The difference in distance travelled by two waves from their sources to a point; determines whether the point is bright or dark. |
| Bright fringe | Constructive interference occurs when path difference = n lambda (n = 0, 1, 2, ...); intensity is maximum. |
| Dark fringe | Destructive interference occurs when path difference = (2n + 1) lambda / 2; intensity is minimum. |
| Fringe width | Distance between two consecutive bright (or dark) fringes; beta = lambda D / d. |
| Diffraction | Bending of light around the edges of an obstacle or aperture; produces a central maximum and alternating minima and maxima. |
| Single-slit diffraction | Diffraction pattern from a narrow slit; position of minima given by a sin theta = n lambda. |
| Resolving power | Ability of an optical instrument to distinguish between two closely spaced objects; improves with larger aperture. |
| Polarisation | Restriction of the vibrations of light waves to a single plane; proves the transverse nature of light. |
| Brewster's law | At the Brewster angle of incidence, reflected light is completely polarised; tan ip = n, where n is the refractive index. |
| Malus's law | Intensity of polarised light after passing through an analyser: I = I0 cos² theta. |
| Wavefront | Locus of all points in a medium that are in the same phase of vibration; can be spherical, cylindrical, or plane. |
| Huygens' principle | Every point on a wavefront acts as a source of secondary wavelets; the new wavefront is the envelope of these wavelets. |
Formula Recap
| Concept | Formula |
|---|
| Fringe width | beta = lambda D / d |
| Bright fringe | path difference = n lambda |
| Dark fringe | path difference = (2n + 1) lambda / 2 |
| Single slit minima | a sin theta = n lambda |
| Brewster law | tan ip = n |
| Malus's law | I = I0 cos² theta |
| Angular fringe width | theta = lambda / d |
YDSE Numerical Pattern
1. Identify wavelength lambda, distance D, slit separation d. 2. Convert wavelength to metre. 3. Use beta = lambda D / d. 4. Convert final answer to mm if needed. 5. Mention bright/dark condition if asked.
Short Answer Points
- Interference occurs due to superposition of coherent waves.
- Diffraction is bending of waves around obstacles.
- Polarisation proves that light is transverse.
- Coherent sources have constant phase difference.
I. Choose the Best Answer (MCQs)
1. In Young's double-slit experiment, the fringe width depends on:
Answer
(a) Only wavelength   (b) Only slit separation   (c) Wavelength, slit separation, and screen distance   (d) Only screen distance
Answer(c) Wavelength, slit separation, and screen distance — Fringe width beta = lambda D / d depends on all three quantities: wavelength (lambda), distance to screen (D), and slit separation (d).
2. Two coherent sources must have:
Answer
(a) Same amplitude only   (b) Same frequency and constant phase difference   (c) Same intensity only   (d) Different wavelengths
Answer(b) Same frequency and constant phase difference — Coherent sources must emit waves of the same frequency with a phase difference that remains constant over time.
3. If the slit separation in YDSE is doubled, the fringe width:
Answer
(a) Doubles   (b) Becomes half   (c) Remains the same   (d) Becomes four times
Answer(b) Becomes half — Since beta = lambda D / d, doubling d makes the fringe width half.
4. Diffraction of light is best observed when:
Answer
(a) The slit width is much larger than wavelength   (b) The slit width is comparable to the wavelength of light   (c) The slit width is infinite   (d) There is no slit
Answer(b) The slit width is comparable to the wavelength of light — Diffraction is most pronounced when the size of the obstacle or aperture is of the same order as the wavelength of light.
5. Polarisation of light proves that light is:
Answer
(a) Longitudinal   (b) Transverse   (c) Both longitudinal and transverse   (d) Neither longitudinal nor transverse
Answer(b) Transverse — Only transverse waves can be polarised; longitudinal waves cannot. The polarisation of light confirms its transverse nature.
6. The Brewster angle for a medium of refractive index 1.5 is approximately:
Answer
(a) 30°   (b) 45°   (c) 56.3°   (d) 60°
Answer(c) 56.3° — tan ip = n = 1.5, so ip = tan⁻¹(1.5) ≈ 56.3°.
7. In single-slit diffraction, the width of the central maximum:
Answer
(a) Is same as that of secondary maxima   (b) Is twice the width of secondary maxima   (c) Is half the width of secondary maxima   (d) Is independent of wavelength
Answer(b) Is twice the width of secondary maxima — The central maximum extends from n = -1 to n = +1, making it twice as wide as each secondary maximum.
8. According to Huygens' principle, each point on a wavefront acts as a source of:
Answer
(a) Plane waves only   (b) Secondary wavelets   (c) Longitudinal waves   (d) Standing waves
Answer(b) Secondary wavelets — Huygens' principle states that every point on a wavefront is a source of secondary wavelets, and the new wavefront is the forward envelope of these wavelets.
9. If white light is used in YDSE, the central fringe is:
Answer
(a) White   (b) Red   (c) Violet   (d) Dark
Answer(a) White — At the centre, the path difference is zero for all wavelengths, so all colours constructively interfere to produce a white fringe.
10. Malus's law gives the intensity of polarised light after passing through an analyser as:
Answer
(a) I = I0 sin² theta   (b) I = I0 cos² theta   (c) I = I0 tan² theta   (d) I = I0 / cos² theta
Answer(b) I = I0 cos² theta — Malus's law states that the transmitted intensity is proportional to the square of the cosine of the angle between the polariser and analyser transmission axes.
II. Fill in the Blanks
1. The distance between two consecutive bright fringes is called ______.
Answerfringe width
2. For constructive interference, the path difference must be an integral multiple of ______.
Answerwavelength (lambda)
3. Diffraction is more pronounced when the slit width is ______ to the wavelength of light.
Answercomparable
4. At the Brewster angle, the reflected and refracted rays are ______ to each other.
Answerperpendicular
5. According to Huygens' principle, the new wavefront is the ______ of all secondary wavelets.
Answerenvelope
6. The angular fringe width in YDSE is given by theta = ______.
Answerlambda / d
7. Sound waves cannot be polarised because they are ______ waves.
Answerlongitudinal
8. In single-slit diffraction, the condition for the first minimum is a sin theta = ______.
Answerlambda
III. True or False (with Corrections)
1. Interference and diffraction both require coherent sources.
AnswerFalse — Correction: Interference requires two coherent sources, but diffraction occurs from a single slit. Diffraction is the bending of light at a single aperture, not the superposition from two sources.
2. Polarisation can be observed in both longitudinal and transverse waves.
AnswerFalse — Correction: Only transverse waves can be polarised. Longitudinal waves cannot be polarised because their vibrations are along the direction of propagation.
3. Increasing the wavelength of light in YDSE increases the fringe width.
AnswerTrue — Since beta = lambda D / d, fringe width is directly proportional to wavelength.
4. The central maximum in single-slit diffraction is the widest and brightest part of the pattern.
AnswerTrue — The central maximum is twice as wide as the secondary maxima and has the highest intensity.
5. Brewster's angle depends on the wavelength of incident light.
AnswerTrue — Since tan ip = n and refractive index n varies with wavelength (dispersion), the Brewster angle also depends on wavelength.
6. In YDSE, the intensity of all bright fringes is the same.
AnswerTrue — In ideal YDSE with equal amplitude slits, all bright fringes have the same intensity I_max = 4I0.
7. Huygens' principle can explain the wave theory of light but cannot explain the photoelectric effect.
AnswerTrue — Huygens' principle is a wave theory and cannot explain phenomena like the photoelectric effect, which require quantum (particle) theory.
IV. Match the Following
| Column A | Column B |
|---|
| 1. Interference | (a) Bending of light around obstacles |
| 2. Diffraction | (b) Restriction of vibrations to one plane |
| 3. Polarisation | (c) Superposition of coherent waves |
| 4. Brewster's law | (d) tan ip = n |
| 5. Malus's law | (e) I = I0 cos² theta |
| 6. Huygens' principle | (f) Secondary wavelets |
Answer1-c, 2-a, 3-b, 4-d, 5-e, 6-f
V. Answer Briefly (2-mark)
1. What is interference of light? State the conditions for sustained interference.
Answer
Interference of light is the phenomenon where two coherent light waves superpose to form alternate bright and dark fringes. The conditions for sustained interference are: (i) the two sources must be coherent (same frequency and constant phase difference), (ii) the amplitudes should be nearly equal for good contrast, and (iii) the sources should be narrow and close to each other.
2. Distinguish between interference and diffraction.
Answer
Interference results from the superposition of waves from two coherent sources, producing equally spaced bright and dark fringes of equal intensity. Diffraction results from light bending around a single aperture or obstacle, producing a pattern with a broad central maximum and narrower secondary maxima. In interference, all bright fringes have the same intensity, while in diffraction, the central maximum is much brighter than the secondary maxima.
3. State and explain Brewster's law.
Answer
Brewster's law states that when unpolarised light is incident on a transparent medium at the Brewster angle ip, the reflected light is completely polarised. The Brewster angle is given by tan ip = n, where n is the refractive index of the medium. At this angle, the reflected and refracted rays are perpendicular to each other, and the reflected ray contains only the component of light vibrating perpendicular to the plane of incidence.
4. What is Huygens' principle?
Answer
Huygens' principle states that every point on a wavefront acts as a source of secondary wavelets that travel forward with the speed of light in that medium. The new wavefront at any later time is the forward envelope (tangent surface) of all these secondary wavelets. This principle explains reflection, refraction, and diffraction of waves and is fundamental to wave optics.
5. Define resolving power of an optical instrument.
Answer
Resolving power is the ability of an optical instrument to distinguish between two closely spaced objects as separate entities. For a microscope, it is the minimum distance between two objects that can be distinguished. For a telescope, it is the minimum angular separation between two distant objects. Resolving power is directly proportional to the aperture of the instrument and inversely proportional to the wavelength of light used.
VI. Answer in Detail (5-mark)
1. Describe Young's double-slit experiment and derive the expression for fringe width.
Answer
Young's double-slit experiment demonstrates the wave nature of light by producing an interference pattern. A monochromatic light source illuminates two narrow parallel slits S1 and S2 separated by a distance d. The slits act as coherent sources. Light from these slits falls on a screen placed at a distance D from the slits (where D is much greater than d). Alternate bright and dark fringes are observed on the screen.
For a point P on the screen at a distance y from the centre O, the path difference between the two waves is delta = d sin theta ≈ d y / D (for small angles). For the nth bright fringe, the path difference must be an integral multiple of lambda: d y_n / D = n lambda, giving y_n = n lambda D / d. The position of the (n+1)th bright fringe is y_(n+1) = (n+1) lambda D / d. The fringe width beta = y_(n+1) - y_n = lambda D / d. For the nth dark fringe, the path difference is (2n + 1) lambda / 2, and the position is y_n = (2n + 1) lambda D / (2d). The fringe width is the same for both bright and dark fringes. The key observations are: (i) fringe width increases with wavelength and screen distance, (ii) fringe width decreases with increasing slit separation, and (iii) the fringes are equally spaced.
2. Explain diffraction of light at a single slit and derive the condition for minima.
Answer
Diffraction is the bending of light around the edges of an obstacle or aperture. In single-slit diffraction, a narrow slit of width a is illuminated by monochromatic light. According to Huygens' principle, every point on the slit acts as a source of secondary wavelets. These wavelets interfere to produce a diffraction pattern on a screen at a distance D.
For a point P on the screen making an angle theta with the central axis, consider two points on the slit separated by a/2. The path difference between waves from the top and bottom of the slit is a sin theta. When this path difference equals lambda, 2 lambda, 3 lambda, etc., waves from different parts of the slit cancel each other out. The condition for minima is a sin theta = n lambda, where n = 1, 2, 3, ... The central maximum (n = 0) is the brightest and widest. The first secondary minimum occurs at sin theta = lambda / a. The angular width of the central maximum is 2 theta = 2 lambda / a. The intensity of secondary maxima decreases rapidly — the first secondary maximum has only about 4.5% of the intensity of the central maximum. If the slit width a is much larger than lambda, diffraction is negligible and geometrical optics applies. If a is comparable to lambda, significant diffraction occurs.
3. What is polarisation? Explain polarisation by reflection and state Brewster's law.
Answer
Polarisation is the process of restricting the vibrations of a transverse wave to a single plane. Unpolarised light has vibrations in all directions perpendicular to the direction of propagation. After polarisation, the vibrations occur only in one plane. Polarisation can be achieved by reflection, refraction, double refraction, and selective absorption.
When unpolarised light is incident on a transparent medium like glass, the reflected light is partially polarised. At a particular angle of incidence called the Brewster angle (ip), the reflected light is completely polarised with vibrations perpendicular to the plane of incidence. Brewster's law states that tan ip = n, where n is the refractive index of the reflecting medium. At the Brewster angle, the reflected and refracted rays are perpendicular to each other (ip + ir = 90°). Brewster's law can be derived: since ip + ir = 90°, we have ir = 90° - ip. By Snell's law, n = sin ip / sin ir = sin ip / sin(90° - ip) = sin ip / cos ip = tan ip.
Applications of polarisation include: polaroid sunglasses that reduce glare by blocking horizontally polarised light reflected from surfaces, liquid crystal displays (LCDs) that use polarised light to control pixel brightness, and stress analysis in engineering using photoelasticity.
4. Explain the construction of a polarimeter and how Malus's law is verified experimentally.
Answer
A polarimeter consists of a monochromatic light source, a polariser (which produces plane-polarised light), and an analyser (a second polariser that can be rotated). When unpolarised light passes through the polariser, it becomes plane-polarised with intensity I0. This polarised light then falls on the analyser, which is rotated through an angle theta about the axis of propagation. According to Malus's law, the intensity of light emerging from the analyser is I = I0 cos² theta.
When theta = 0° (polariser and analyser have parallel axes), I = I0 (maximum transmission). When theta = 90° (axes are perpendicular — crossed polarisers), I = 0 (no light passes through). When theta = 45°, I = I0/2 (half the intensity passes). Malus's law can be verified by placing a light detector after the analyser and measuring the intensity at different angles. A plot of I versus cos² theta gives a straight line through the origin, confirming the law. This experiment demonstrates that light is a transverse wave because only transverse waves can be polarised and show the cos² theta dependence predicted by Malus's law.
Worked Examples
Example 1 — Young's double-slit fringe width
In a double-slit experiment, λ = 600 nm, slit separation d = 1 mm, screen distance D = 1 m. Find the fringe width.
β = λD / d = (600×10⁻⁹ × 1) / (1×10⁻³)
= 6×10⁻⁴ m = 0.6 mm
Example 2 — Wavelength from fringe width
If fringes are 0.5 mm wide with d = 1.2 mm and D = 1 m, find λ.
λ = βd / D = (0.5×10⁻³ × 1.2×10⁻³) / 1
= 6×10⁻⁷ m = 600 nm
Example 3 — Position of the nth bright fringe
Find the distance of the 3rd bright fringe from the centre (β = 0.6 mm).
yₙ = n × β = 3 × 0.6 = 1.8 mm
Common Mistakes to Avoid
1. ❌ Not converting nanometre to metre → ✅ Always convert wavelength: 1 nm = 10⁻⁹ m. Write lambda = 600 nm = 600 × 10⁻⁹ m before substituting into the formula.
2. ❌ Confusing slit separation d with screen distance D → ✅ In beta = lambda D / d, D is the large distance (in metres, typically 1 m) and d is the small slit separation (in mm, typically 0.5-2 mm). Always label which is which.
3. ❌ Mixing bright and dark fringe conditions → ✅ Bright fringe: path difference = n lambda. Dark fringe: path difference = (2n + 1) lambda / 2. Write the condition clearly before solving.
4. ❌ Saying sound can be polarised → ✅ Only transverse waves can be polarised. Sound is a longitudinal wave and cannot be polarised. Light can be polarised because it is a transverse wave.
5. ❌ Forgetting that the central maximum in single-slit diffraction is twice as wide as secondary maxima → ✅ The central maximum extends from n = -1 to n = +1, giving an angular width of 2 lambda / a. Each secondary maximum has width lambda / a.
6. ❌ Writing Brewster's law as sin ip = n → ✅ Brewster's law is tan ip = n, not sin ip = n. Use tan⁻¹(n) to calculate the Brewster angle.
Exam Checklist
- Wavelength is converted from nm to m before substituting into any formula.
- Fringe width formula
beta = lambda D / d is memorised and applied with correct units. - Bright fringe condition: path difference = n lambda (n = 0, 1, 2, ...).
- Dark fringe condition: path difference = (2n + 1) lambda / 2.
- Single-slit diffraction minima condition: a sin theta = n lambda.
- Central maximum in diffraction is twice as wide as secondary maxima.
- Polarisation proves the transverse nature of light.
- Brewster's law:
tan ip = n; at Brewster angle, reflected light is completely polarised. - Malus's law:
I = I0 cos² theta. - Huygens' principle: every point on a wavefront is a source of secondary wavelets.
- Difference between interference (two sources) and diffraction (single aperture) is clearly stated.
- Resolving power improves with larger aperture and shorter wavelength.
- White light produces a white central fringe and coloured higher-order fringes in YDSE.
