CHAPTER 10: WAVE OPTICS
SAMPLE QUESTION PAPER (ASSIGNMENT-1)
- There are 33 questions in all. All questions are compulsory.
- Section A: 16 questions (12 MCQs and 4 Assertion-Reasoning) of 1 mark each.
- Section B: 5 short answer questions of 2 marks each.
- Section C: 7 short answer questions of 3 marks each.
- Section D: 2 case study-based questions of 4 marks each.
- Section E: 3 long answer questions of 5 marks each.
SECTION A (16 Marks)
Q.1. Two coherent sources of light have an intensity ratio of $\displaystyle {4 : 1}$. The ratio of maximum to minimum intensity in the interference pattern produced by them is:
Explanation: Since $\displaystyle {I \propto a^2}$, the amplitude ratio is $\displaystyle {a_1/a_2 = \sqrt{I_1/I_2} = \sqrt{4/1} = 2/1}$.
$\displaystyle {{\frac{{I_{\text{max}}}{{I_{\text{min}}}}} = {\left( {\frac{{a_1 + a_2}}{{a_1 - a_2}}} \right)}^2 = {\left( {\frac{{2 + 1}}{{2 - 1}}} \right)}^2 = 9 : 1}$.
Q.2. Assertion (A): In Young's double-slit experiment, the fringe width for dark and bright fringes is the same.
Reason (R): Fringe width is given by $\displaystyle {\beta = \lambda D / d}$, which does not depend on the order of the fringe.
Q.3. The shape of the wavefront of light diverging from a point source is:
Explanation: A isotropic point source emits light uniformly in all 3D directions, forming spherical wave surfaces.
Q.4. In a single-slit diffraction experiment, the first minimum of the diffraction pattern is obtained when the path difference between wavelets from the two edges of the slit is:
Explanation: First minimum occurs when $\displaystyle {a \sin\theta = \lambda}$, dividing the slit into two halves whose corresponding wavelets cancel pairwise with path difference $\displaystyle {\lambda/2}$.
Q.5. The phase difference between two superimposing waves that give rise to a bright spot in Young’s double-slit experiment is ($n$ is an integer):
Explanation: Constructive interference requires a path difference of $\displaystyle {\Delta = n\lambda}$, corresponding to a phase difference $\displaystyle {\phi = (2\pi/\lambda)\Delta = 2n\pi}$.
Q.6. According to Huygens' principle, the amplitude of secondary wavelets emanating from a wavefront is:
Explanation: Huygens-Fresnel inclination factor $\displaystyle {(1 + \cos\theta)/2}$ gives maximum factor $\displaystyle {1}$ at $\displaystyle {\theta = 0^\circ}$ (forward) and zero at $\displaystyle {\theta = 180^\circ}$ (backward).
Q.7. In Young's double-slit experiment, a dark fringe is obtained at a point where path difference between the two interfering waves is ($n = 0, 1, 2, \dots$):
Explanation: Destructive interference occurs when waves arrive $\displaystyle {180^\circ}$ out of phase, requiring path difference equal to an odd multiple of $\displaystyle {\lambda/2}$.
Q.8. Light passes from a rarer medium (speed $\displaystyle {v_1}$) into a denser medium (speed $\displaystyle {v_2}$). Which of the following is correct?
Explanation: Refractive index of denser medium $\displaystyle {n_2 > n_1}$, so wave speed $\displaystyle {v_2 = c/n_2 < v_1 = c/n_1}$.
Q.9. Assertion (A): Two independent sodium lamps cannot act as coherent sources of light.
Reason (R): Light emitted by an ordinary independent source undergoes abrupt, random phase changes in times of the order of $\displaystyle {10^{-10}\text{ s}}$.
Q.10. In a single-slit diffraction experiment, if the width of the slit is doubled, the linear width of the central maximum becomes:
Explanation: Central maximum width $\displaystyle {W = 2\lambda D / a \propto 1/a}$. Doubling slit width $\displaystyle {a}$ halves $\displaystyle {W}$.
Q.11. What fundamental property of light does Young's double-slit experiment demonstrate?
Explanation: Interference fringes formed by superposition of overlapping light beams prove that light propagates as waves.
Q.12. In Young's double-slit experiment, if the distance between screen and slits ($\displaystyle {D}$) is doubled, the fringe width becomes:
Explanation: $\displaystyle {\beta = \lambda D / d \implies \beta \propto D}$. Doubling $\displaystyle {D}$ doubles fringe width $\displaystyle {\beta}$.
Q.13. When monochromatic light travels from air into glass, the physical property that remains completely unchanged is its:
Explanation: Frequency is an intrinsic property of the light source and does not change across medium interfaces.
Q.14. Assertion (A): In interference and diffraction patterns, light energy that disappears from dark regions reappears in bright regions.
Reason (R): Redistribution of light energy obeys the principle of conservation of energy.
Q.15. In a single-slit diffraction experiment, if the slit width is halved, the width of central maximum becomes:
Explanation: Linear width $\displaystyle {W = 2\lambda D / a}$. Halving $\displaystyle {a}$ doubles central maximum width.
Q.16. Diffraction of sound waves is much more noticeable in daily life than diffraction of light waves because:
Explanation: Significant diffraction requires obstacle/aperture size $\displaystyle {a \sim \lambda}$. Sound wavelength ($\sim\text{meters}$) matches doors/walls, whereas visible light wavelength ($\sim 10^{-7}\text{ m}$) is tiny.
SECTION B (10 Marks)
Q.17. State Huygens' principle. How did Huygens explain the absence of the backwave associated with a wavefront?
Huygens' Principle: (1) Every point on a primary wavefront acts as a source of secondary spherical wavelets spreading in all directions at wave speed in that medium. (2) The forward envelope (tangent surface) to these wavelets at a later instant gives the new wavefront.
Absence of Backwave: Huygens postulated an inclination factor $\displaystyle {(1 + \cos\theta)/2}$ for secondary wavelet amplitude. For backward direction ($\displaystyle {\theta = 180^\circ}$), factor $\displaystyle {(1 + \cos 180^\circ)/2 = 0}$, so zero amplitude wave is formed backward.
Q.18. Write the conditions in terms of path difference for obtaining (i) a bright fringe, and (ii) a dark fringe in Young's double-slit experiment.
State two mandatory conditions that must be satisfied for two light sources to act as coherent sources.
(i) Bright Fringe (Constructive): Path difference $\displaystyle {\Delta = n\lambda}$, where $n = 0, 1, 2, \dots$
(ii) Dark Fringe (Destructive): Path difference $\displaystyle {\Delta = (n + 1/2)\lambda}$, where $n = 0, 1, 2, \dots$
OR Part (Coherent Sources Conditions):
1. Sources must emit waves of identical frequency (or wavelength).
2. Sources must maintain a zero or constant phase difference over time.
Q.19. In YDSE, two slits are separated by $\displaystyle {0.5\text{ mm}}$ and screen is placed $\displaystyle {1.0\text{ m}}$ away. If fringe width is $\displaystyle {1.2\text{ mm}}$, calculate wavelength of light used.
Formula: $\displaystyle {\beta = {\frac{{\lambda D}}{{d}}} \implies \lambda = {\frac{{\beta d}}{{D}}} = {\frac{{(1.2 \times 10^{-3}) \times (0.5 \times 10^{-3})}}{{1.0}}} = 0.6 \times 10^{-6}\text{ m} = 600\text{ nm}}$.
Q.20. Write two points of difference between YDSE interference pattern and single-slit diffraction pattern.
1. Fringe Width & Intensity: In YDSE interference, all bright fringes are equally wide and have equal intensity. In diffraction, the central maximum is twice as wide as secondary maxima, and side maxima decrease rapidly in intensity.
2. Origin: Interference is caused by superposition of two distinct waves from two separate slits; diffraction is caused by superposition of secondary wavelets originating from different parts of the same slit.
Q.21. Unpolarised light of intensity $\displaystyle {I_0}$ is incident on polaroid $\displaystyle {P_1}$. A second polaroid $\displaystyle {P_2}$ is placed after $\displaystyle {P_1}$ with pass axis at $\displaystyle {30^\circ}$ to $\displaystyle {P_1}$. Calculate emergent intensity from $\displaystyle {P_2}$.
By Malus' Law across $\displaystyle {P_2}$: $\displaystyle {I_2 = I_1 \cos^2\theta = {\left( {\frac{{I_0}}{{2}}} \right)} \cos^2 30^\circ = {\left( {\frac{{I_0}}{{2}}} \right)} {\left( {\frac{{\sqrt{3}}}{{2}}} \right)}^2 = {\frac{{3 I_0}}{{8}}} = 0.375 I_0}$.
SECTION C (21 Marks)
Q.22. Using Huygens' construction, draw a diagram to show refraction of a plane wavefront at a plane boundary separating a rarer medium ($\displaystyle {v_1}$) from a denser medium ($\displaystyle {v_2}$). Hence verify Snell's law.
In right $\displaystyle {\Delta ABC}$: $\displaystyle {\sin i = BC / AC = v_1 \tau / AC}$.
In right $\displaystyle {\Delta ADC}$: $\displaystyle {\sin r = AD / AC = v_2 \tau / AC}$.
Dividing: $\displaystyle {{\frac{{\sin i}}{{\sin r}}} = {\frac{{v_1}}{{v_2}}} = {\frac{{c/n_1}}{{c/n_2}}} = {\frac{{n_2}}{{n_1}}} \implies n_1 \sin i = n_2 \sin r}$ (Snell's Law).
Q.23. In YDSE, slit separation is $\displaystyle {1.0\text{ mm}}$ and screen is $\displaystyle {1.0\text{ m}}$ away. Light of wavelengths $\displaystyle {600\text{ nm}}$ and $\displaystyle {480\text{ nm}}$ is used.
(a) Calculate distance of 3rd bright fringe for $\displaystyle {600\text{ nm}}$ from central maximum.
(b) Calculate least distance from central maximum where bright fringes of both wavelengths coincide.
(a) $\displaystyle {x_3 = {\frac{{3 \lambda_1 D}}{{d}}} = {\frac{{3 \times (600 \times 10^{-9}) \times 1.0}}{{10^{-3}}}} = 1.8 \times 10^{-3}\text{ m} = 1.8\text{ mm}}$.
(b) Coincidence condition: $\displaystyle {n_1 \lambda_1 = n_2 \lambda_2 \implies n_1 (600) = n_2 (480) \implies {\frac{{n_1}}{{n_2}}} = {\frac{{480}}{{600}}} = {\frac{{4}}{{5}}}}$.
Smallest integers: $\displaystyle {n_1 = 4}$ for $\displaystyle {600\text{ nm}}$ and $\displaystyle {n_2 = 5}$ for $\displaystyle {480\text{ nm}}$.
Least coincidence distance: $\displaystyle {x = {\frac{{4 \lambda_1 D}}{{d}}} = {\frac{{4 \times (600 \times 10^{-9}) \times 1.0}}{{10^{-3}}}} = 2.4 \times 10^{-3}\text{ m} = 2.4\text{ mm}}$.
Q.24. Two coherent sources produce displacements $\displaystyle {y_1 = a \cos\omega t}$ and $\displaystyle {y_2 = a \cos(\omega t + \phi)}$. Derive expression for resultant intensity $\displaystyle {I = 4 I_0 \cos^2(\phi/2)}$. State conditions for maxima and minima.
Resultant amplitude $\displaystyle {A = 2a \cos(\phi/2)}$.
Since intensity $\displaystyle {I \propto A^2}$ and single wave intensity $\displaystyle {I_0 \propto a^2}$:
$\displaystyle {I = 4 I_0 \cos^2(\phi/2)}$.
Maximum Intensity ($\displaystyle {4I_0}$): $\displaystyle {\cos^2(\phi/2) = 1 \implies \phi = 2n\pi}$ ($\displaystyle {\Delta = n\lambda}$).
Minimum Intensity ($\displaystyle {0}$): $\displaystyle {\cos^2(\phi/2) = 0 \implies \phi = (2n+1)\pi}$ ($\displaystyle {\Delta = (n+1/2)\lambda}$).
Q.25. A parallel beam of light of wavelength $\displaystyle {500\text{ nm}}$ is incident normally on a single slit of width $\displaystyle {0.20\text{ mm}}$. Find angular position from central maximum of: (i) first minimum, (ii) first secondary maximum.
(i) First minimum ($\displaystyle {a \sin\theta_1 = \lambda}$): $\displaystyle {\theta_1 \approx \sin\theta_1 = {\frac{{\lambda}}{{a}}} = {\frac{{500 \times 10^{-9}}}{{2 \times 10^{-4}}}} = 2.5 \times 10^{-3}\text{ rad}}$.
(ii) First secondary maximum ($\displaystyle {a \sin\theta'_1 = 1.5\lambda}$): $\displaystyle {\theta'_1 \approx {\frac{{1.5 \lambda}}{{a}}} = 1.5 \times (2.5 \times 10^{-3}) = 3.75 \times 10^{-3}\text{ rad}}$.
Q.26. State three characteristic points of difference between double-slit interference pattern and single-slit diffraction pattern.
1. Fringe Spacing: Interference fringes are equally spaced throughout ($\displaystyle {\beta = \lambda D/d}$). In diffraction, the central maximum is twice as wide as secondary maxima.
2. Intensity Distribution: In interference, all bright fringes have nearly equal intensity. In diffraction, the central peak is dominant and secondary peaks rapidly decay in brightness.
3. Minima Darkness: Interference minima are completely dark ($\displaystyle {I_{\text{min}} = 0}$). In diffraction, minima are not completely dark due to incomplete secondary wavelet cancellation.
Q.27. In YDSE with monochromatic light, fringe width is $\displaystyle {0.4\text{ mm}}$. If the setup is immersed in liquid of refractive index $\displaystyle {1.6}$, calculate new fringe width. Does central maximum position shift?
New fringe width: $\displaystyle {\beta' = {\frac{{\lambda' D}}{{d}}} = {\frac{{\beta}}{{\mu}}} = {\frac{{0.4\text{ mm}}}{{1.6}}} = 0.25\text{ mm}}$.
Position of central maximum does NOT shift because path difference at central bisector remains zero ($\displaystyle {\Delta = 0}$) for both paths in the surrounding medium.
SECTION D: CASE STUDY QUESTIONS (8 Marks)
Q.29. Case Study 1: Young's Double-Slit Experiment Numerical Analysis.
In a YDSE setup, two slits separated by $\displaystyle {2\text{ mm}}$ are illuminated by monochromatic light of wavelength $\displaystyle {600\text{ nm}}$, and fringes are observed on a screen kept $\displaystyle {3\text{ m}}$ away.
(a) Calculate fringe width $\displaystyle {\beta}$.
(b) Find distance of 4th bright fringe from central maximum.
(c) What happens to fringe width if screen is moved closer to slits?
(d) If one slit is completely covered, what pattern is seen on screen?
(a) $\displaystyle {\beta = \lambda D / d = (600 \times 10^{-9} \times 3) / (2 \times 10^{-3}) = 9 \times 10^{-4}\text{ m} = 0.9\text{ mm}}$.
(b) $\displaystyle {x_4 = 4\beta = 4 \times 0.9\text{ mm} = 3.6\text{ mm}}$.
(c) Moving screen closer ($\displaystyle {D}$ decreases) decreases fringe width ($\displaystyle {\beta \propto D}$).
(d) Closing one slit destroys interference; a broad single-slit diffraction pattern appears instead.
Q.30. Case Study 2: Single-Slit Diffraction and Sound vs Light Bending.
Diffraction is the bending of light around obstacles whose size is comparable to wavelength $\displaystyle {\lambda}$. Light of wavelength $\displaystyle {\lambda}$ falls normally on a slit of width $\displaystyle {a}$.
(a) Write condition for position of $n$-th secondary minimum.
(b) How does angular width of central maximum change if slit width $\displaystyle {a}$ is decreased?
(c) Compare intensity of secondary maxima with central maximum.
OR (c) Why can two students converse across a 5m high wall but cannot see each other?
(a) $\displaystyle {a \sin\theta_n = n\lambda}$ ($n = 1, 2, 3, \dots$).
(b) Angular width $\displaystyle {2\theta = 2\lambda/a}$; decreasing $\displaystyle {a}$ increases angular width (spreads pattern).
(c) Secondary maxima are much weaker (1st secondary max is $\sim 4.5\%$ of central peak intensity).
OR: Sound wavelength ($\sim 1\text{ m}$) matches wall size, causing strong diffraction around wall. Light wavelength ($\sim 5 \times 10^{-7}\text{ m}$) is negligible relative to wall, so light travels straight without bending.
SECTION E (15 Marks)
Q.31. (a) State Huygens' principle. Using Huygens' construction, derive the law of reflection ($\displaystyle {i = r}$) for a plane wavefront at a plane reflecting surface.
(b) Refractive index of glass is $\displaystyle {1.5}$. Calculate speed of light in glass. Does violet or red light travel slower in a glass prism?
(a) In time $\displaystyle {\tau}$, disturbance travels $\displaystyle {BC = v\tau}$ from B to C. Secondary wavelet from A spreads into radius $\displaystyle {AD = v\tau}$. Right triangles $\displaystyle {\Delta ABC \cong \Delta ADC}$ (RHS congruence: $\displaystyle {BC = AD = v\tau}$, hypotenuse AC common). Thus $\displaystyle {\angle BAC = \angle DCA \implies i = r}$.
(b) Speed $\displaystyle {v = c/n = (3 \times 10^8) / 1.5 = 2 \times 10^8\text{ m/s}}$. Refractive index for violet light is higher ($\displaystyle {n_{\text{violet}} > n_{\text{red}}}$), so violet light travels slower ($\displaystyle {v = c/n}$).
Q.32. (a) Describe Young's double-slit experiment with a labeled diagram and derive expression for fringe width $\displaystyle {\beta = \lambda D / d}$.
(b) In YDSE, slits are separated by $\displaystyle {0.28\text{ mm}}$ and screen is $\displaystyle {1.4\text{ m}}$ away. Distance between central bright fringe and 4th bright fringe is $\displaystyle {1.2\text{ cm}}$. Calculate wavelength of light.
(a) Derive expression for resultant intensity $\displaystyle {I = 4I_0 \cos^2(\phi/2)}$ for two coherent waves of amplitude $\displaystyle {a}$.
(b) If intensity at path difference $\displaystyle {\lambda}$ is $\displaystyle {K}$, find intensity at path difference $\displaystyle {\lambda/3}$.
(b) Given: $\displaystyle {d = 0.28 \times 10^{-3}\text{ m}, D = 1.4\text{ m}, x_4 = 1.2 \times 10^{-2}\text{ m}}$.
$\displaystyle {x_4 = {\frac{{4 \lambda D}}{{d}}} \implies \lambda = {\frac{{x_4 d}}{{4 D}}} = {\frac{{(1.2 \times 10^{-2}) \times (0.28 \times 10^{-3})}}{{4 \times 1.4}}} = 6.0 \times 10^{-7}\text{ m} = 600\text{ nm}}$.
OR Part (b): At path difference $\displaystyle {\lambda}$, phase difference $\displaystyle {\phi_1 = 2\pi \implies I_1 = 4I_0 = K \implies I_0 = K/4}$.
At path difference $\displaystyle {\lambda/3}$, phase difference $\displaystyle {\phi_2 = (2\pi/\lambda)(\lambda/3) = 2\pi/3}$.
$\displaystyle {I_2 = 4 I_0 \cos^2(\pi/3) = 4 I_0 (1/2)^2 = I_0 = K/4}$.
Q.33. (a) Using Huygens' principle, explain single-slit diffraction and obtain condition for minima position ($\displaystyle {a \sin\theta_n = n\lambda}$). Explain why secondary maxima weaken with order $n$.
(b) A slit of width $\displaystyle {0.30\text{ mm}}$ is illuminated normally by light of wavelength $\displaystyle {600\text{ nm}}$. Find linear width of central maximum on a screen $\displaystyle {2\text{ m}}$ away.
(a) Minima condition: Slit of width $a$ divided into $2n$ equal zones. Opposite zones cancel in pairs with $\displaystyle {\lambda/2}$ phase shift when $\displaystyle {a \sin\theta_n = n\lambda}$. Secondary maxima weaken because slit is divided into odd numbers of zones ($3, 5, 7, \dots$), leaving only $1/3\text{rd}, 1/5\text{th}, 1/7\text{th}$ fraction of slit uncancelled.
(b) Linear width $\displaystyle {W = {\frac{{2 \lambda D}}{{a}}} = {\frac{{2 \times (600 \times 10^{-9}) \times 2}}{{0.30 \times 10^{-3}}}} = 8 \times 10^{-3}\text{ m} = 8\text{ mm}}$.
ASSIGNMENT – 2 (MCQ PRACTICE)
1. If one slit of Young's double-slit apparatus is blocked, what is observed on the screen?
2. In YDSE with white light, if one slit is covered with green filter and the other with red filter:
Explanation: Waves of different frequencies (red and green) cannot maintain a constant phase relationship required for interference.
3. Phenomenon that cannot be explained by classical wave theory of light is:
4. Two coherent waves of intensity $\displaystyle {I}$ each superimpose at a point where path difference is $\displaystyle {\lambda/3}$. Resultant intensity is:
Explanation: Phase shift $\displaystyle {\phi = 2\pi/3 \implies I_R = 4I \cos^2(\pi/3) = 4I (1/4) = I}$.
ASSIGNMENT – 3 (ASSERTION & REASON)
1. Assertion (A): Light added to light can produce darkness.
Reason (R): When two coherent light waves meet in opposite phase, destructive interference produces dark fringes.
2. Assertion (A): Phase difference between any two points lying on the same wavefront is zero.
Reason (R): All points on a wavefront are equidistant from the source and oscillate in identical phase.
ASSIGNMENT – 4 (CASE STUDY QUESTIONS)
Case Study: Coincidence of Fringes with Two Wavelengths
Light containing wavelengths $\displaystyle {600\text{ nm}}$ and $\displaystyle {750\text{ nm}}$ illuminates slits separated by $\displaystyle {0.5\text{ mm}}$ with screen at $\displaystyle {1.5\text{ m}}$.
1. Fringe width for 600 nm: (a) $\displaystyle {1.8\text{ mm}}$ (b) $\displaystyle {2.25\text{ mm}}$
2. Least distance from central maximum where bright fringes coincide: (a) $\displaystyle {9.0\text{ mm}}$ (b) $\displaystyle {4.5\text{ mm}}$
1: (a) $\displaystyle {1.8\text{ mm}}$ ($\displaystyle {\beta_1 = 600 \times 10^{-9} \times 1.5 / 0.5 \times 10^{-3} = 1.8\text{ mm}}$).
2: (a) $\displaystyle {9.0\text{ mm}}$ ($\displaystyle {n_1 (600) = n_2 (750) \implies n_1/n_2 = 5/4 \implies x = 5 \times 1.8\text{ mm} = 9.0\text{ mm}}$).
ASSIGNMENT – 5 (FORMULAE & CONCEPTUAL QUESTIONS)
CORE FORMULAE SUMMARY
Fringe Width: $\displaystyle {\beta = {\frac{{\lambda D}}{{d}}}}$ | Maxima Position: $\displaystyle {x_n = n {\frac{{\lambda D}}{{d}}}}$ | Minima Position: $\displaystyle {x_n = {\left(n + {\frac{{1}}{{2}}}\right)} {\frac{{\lambda D}}{{d}}}}$
Resultant Intensity: $\displaystyle {I = 4 I_0 \cos^2{\left({\frac{{\phi}}{{2}}}\right)}}$ where phase shift $\displaystyle {\phi = {\frac{{2\pi}}{{\lambda}}} \Delta}$
Minima Condition: $\displaystyle {a \sin\theta_n = n\lambda}$ | Central Max Linear Width: $\displaystyle {W = {\frac{{2\lambda D}}{{a}}}}$
CONCEPTUAL SHORT QUESTIONS
1. In YDSE, two slits are illuminated by two independent sodium lamps. Will an interference pattern be observed?
2. How does the fringe width in YDSE change when the entire apparatus is immersed in water ($\displaystyle {\mu = 4/3}$)?
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