Wave Optics flash cards

1. 1.What is a wavefront?

   A wavefront is the locus of all points of a medium that vibrate in the same phase at a given instant. The direction of propagation (a ray) is always perpendicular to the wavefront.

   Hint: Locus of points in same phase.

2. 2.Name the three types of wavefronts based on the source.

   (1) Spherical wavefront — from a point source. (2) Cylindrical wavefront — from a linear/slit source. (3) Plane wavefront — from a source at infinity (or a point source very far away).

   Hint: Point, line, infinity.

3. 3.State Huygens' principle.

   Every point on a given wavefront acts as a source of secondary wavelets, which spread out in all directions with the speed of the wave. The new wavefront at a later time is the forward envelope (tangential surface) of these secondary wavelets.

   Hint: Every point → secondary wavelets → envelope.

4. 4.According to Huygens, why do we take only the forward envelope of secondary wavelets and not the backward one?

   The intensity of the secondary wavelets is maximum in the forward direction and zero in the backward direction (obliquity/inclination factor $\tfrac{1}{2}(1+\cos\theta)$). Hence only the forward-moving wavefront is physically real.

   Hint: Obliquity factor is zero backward.

5. 5.Using Huygens' principle, what shape does a plane wavefront retain as it propagates in a homogeneous medium?

   It remains a plane wavefront. Each point emits spherical wavelets of equal radius, and their common tangent is again a plane parallel to the original wavefront.

   Hint: Plane stays plane.

6. 6.How does Huygens' construction explain the laws of reflection?

   When a plane wavefront strikes a reflecting surface, secondary wavelets from points on the surface construct a reflected wavefront. Geometry of equal wavelet radii gives angle of incidence = angle of reflection, both in the same plane.

   Hint: Equal wavelet radii → i = r.

7. 7.How does Huygens' principle explain refraction (Snell's law)?

   As a wavefront enters a denser medium, its speed drops, so the part in the denser medium travels less distance, bending the wavefront. This gives $\dfrac{\sin i}{\sin r}=\dfrac{v_1}{v_2}=n_{21}$, i.e. Snell's law.

   Hint: Speed change bends wavefront.

8. 8.On refraction into a denser medium, what happens to the wave's speed, wavelength and frequency?

   Frequency stays the same (set by the source). Speed decreases: $v=c/n$. Wavelength decreases: $\lambda_{med}=\lambda_{vac}/n$.

   Hint: Frequency fixed; v and λ drop by n.

9. 9.Why does frequency remain unchanged when light passes from one medium to another?

   Frequency is determined by the source. At the boundary the fields must oscillate continuously; the number of wavefronts arriving per second equals the number leaving per second, so $f$ is conserved.

   Hint: Source-controlled; continuity at boundary.

10. 10.Define the principle of superposition of waves.

    When two or more waves overlap in a region, the resultant displacement at any point is the vector sum of the displacements due to the individual waves at that point.

    Hint: Resultant = sum of displacements.

11. 11.What is interference of light?

    Interference is the modification (redistribution) of intensity of light in a region due to the superposition of two coherent light waves — producing alternate bright (constructive) and dark (destructive) regions.

    Hint: Redistribution of intensity by superposition.

12. 12.Distinguish constructive and destructive interference in terms of phase.

    Constructive: waves arrive in phase (phase difference $=2n\pi$), amplitudes add → maximum intensity. Destructive: waves arrive out of phase (phase difference $=(2n+1)\pi$), amplitudes subtract → minimum intensity.

    Hint: In phase adds; opposite phase cancels.

13. 13.Give the path-difference conditions for constructive and destructive interference.

    Constructive (bright): path difference $\Delta x = n\lambda$, where $n=0,1,2,\dots$ Destructive (dark): $\Delta x = (2n-1)\dfrac{\lambda}{2}$, i.e. an odd multiple of $\lambda/2$.

    Hint: nλ bright; odd λ/2 dark.

14. 14.What is the relation between path difference $\Delta x$ and phase difference $\phi$?

    $\phi = \dfrac{2\pi}{\lambda}\,\Delta x$. A path difference of one full wavelength corresponds to a phase difference of $2\pi$.

    Hint: φ = (2π/λ)·Δx.

15. 15.What are coherent sources?

    Two sources are coherent if they emit light waves of the same frequency (and wavelength) with a constant phase difference that does not change with time.

    Hint: Same frequency, constant phase difference.

16. 16.Why can't two independent light bulbs (or two separate lamps) produce a sustained interference pattern?

    Ordinary sources emit light in random, uncorrelated bursts, so the phase difference between them changes rapidly and randomly. The pattern shifts too fast to be observed — they are incoherent.

    Hint: Random, fluctuating phase difference.

17. 17.How are two coherent sources practically obtained in Young's experiment?

    By splitting light from a single source using two slits ($S_1$ and $S_2$). Since both derive from one wavefront, they maintain a constant phase relationship.

    Hint: Single source split into two.

18. 18.State the two broad methods of producing coherent sources for interference.

    (1) Division of wavefront — e.g. Young's double slit, Fresnel's biprism. (2) Division of amplitude — e.g. thin films, Newton's rings.

    Hint: Divide wavefront vs divide amplitude.

19. 19.For two waves of amplitudes $a_1$ and $a_2$ with phase difference $\phi$, write the resultant intensity.

    $I = I_1 + I_2 + 2\sqrt{I_1 I_2}\,\cos\phi$, where $I_1\propto a_1^2$ and $I_2\propto a_2^2$. The $2\sqrt{I_1I_2}\cos\phi$ term is the interference term.

    Hint: I = I₁+I₂+2√(I₁I₂)cosφ.

20. 20.For two equal-intensity coherent sources each $I_0$, what are $I_{max}$ and $I_{min}$?

    $I_{max}=4I_0$ (when $\cos\phi=+1$) and $I_{min}=0$ (when $\cos\phi=-1$). The resultant varies between 0 and $4I_0$.

    Hint: Equal sources: 4I₀ and 0.

21. 21.Give $I_{max}$ and $I_{min}$ in terms of amplitudes $a_1$ and $a_2$.

    $I_{max}\propto(a_1+a_2)^2$ and $I_{min}\propto(a_1-a_2)^2$. Bright fringes have amplitudes adding, dark fringes have them subtracting.

    Hint: (a₁+a₂)² and (a₁−a₂)².

22. 22.Does interference violate conservation of energy? Explain.

    No. Energy is not destroyed at dark fringes; it is redistributed — the energy missing from dark regions appears in bright regions. Average intensity over the pattern equals $I_1+I_2$.

    Hint: Energy redistributed, not lost.

23. 23.Describe the setup of Young's Double Slit Experiment (YDSE).

    Monochromatic light passes through a single slit $S$, then through two close parallel slits $S_1$ and $S_2$ (spacing $d$). Light from $S_1,S_2$ overlaps on a screen a distance $D$ away, producing alternate bright and dark fringes.

    Hint: Single slit → double slit → screen.

24. 24.In YDSE, write the expression for the path difference at a point on the screen at distance $y$ from the centre.

    $\Delta x = \dfrac{y\,d}{D}$, where $d$ = slit separation and $D$ = slit-to-screen distance (valid for $D \gg d$, small angles).

    Hint: Δx = yd/D.

25. 25.Give the positions of bright fringes (maxima) in YDSE.

    For constructive interference $\dfrac{y_n d}{D}=n\lambda$, so $y_n=\dfrac{n\lambda D}{d}$, with $n=0,1,2,\dots$ ($n=0$ is the central bright fringe).

    Hint: yₙ = nλD/d.

26. 26.Give the positions of dark fringes (minima) in YDSE.

    For destructive interference $\dfrac{y_n d}{D}=(2n-1)\dfrac{\lambda}{2}$, so $y_n=\dfrac{(2n-1)\lambda D}{2d}$, with $n=1,2,3,\dots$

    Hint: yₙ = (2n−1)λD/2d.

27. 27.Define fringe width $\beta$ and give its formula in YDSE.

    Fringe width is the distance between two consecutive bright (or two consecutive dark) fringes: $\beta=\dfrac{\lambda D}{d}$. It is the same for bright and dark fringes.

    Hint: β = λD/d.

28. 28.In YDSE, how does fringe width depend on $\lambda$, $D$ and $d$?

    $\beta=\dfrac{\lambda D}{d}$: fringe width increases with wavelength $\lambda$ and screen distance $D$, and decreases with slit separation $d$.

    Hint: β ∝ λ, ∝ D, ∝ 1/d.

29. 29.What happens to the fringe pattern if the slit separation $d$ is increased?

    Since $\beta=\lambda D/d$, increasing $d$ decreases the fringe width — fringes come closer together (pattern gets more crowded).

    Hint: Larger d → narrower fringes.

30. 30.Why must $d$ (slit separation) be very small in YDSE?

    For fringes to be resolvable, $\beta=\lambda D/d$ must be large enough to see. Since $\lambda$ is tiny (~$10^{-7}$ m), $d$ must be small (fraction of a mm) so that $\beta$ is measurable.

    Hint: Small d gives observable β.

Open the interactive deck for the other 69 cards, with self-grading so the ones you keep missing come back.