How to Use This Problem Set

Your full workout for Wave Optics, grouped by theme: wave model & media, Huygens geometry, interference intensities, Young's fringes, single-slit diffraction, and polarisation.

Keep these handy:

  • v=c/nv = c/n; λmed=λ/n\lambda_{med} = \lambda/n; ν\nu unchanged; sin⁡isin⁡r=v1v2\frac{\sin i}{\sin r} = \frac{v_1}{v_2}
  • Coherent: Δ=nλ\Delta = n\lambda bright (I=4I0I = 4I_0), (n+12)λ(n+\tfrac12)\lambda dark; general I=4I0cos⁡2(ϕ/2)I = 4I_0\cos^2(\phi/2); incoherent I=2I0I = 2I_0
  • Unequal: Imax/Imin=(a1+a2a1−a2)2I_{max}/I_{min} = \left(\frac{a_1+a_2}{a_1-a_2}\right)^2
  • Young: bright x=nλD/dx = n\lambda D/d; β=λD/d\beta = \lambda D/d; angular θ=λ/d\theta = \lambda/d
  • Single slit: minima asin⁡θ=nλa\sin\theta = n\lambda; central width 2λD/a2\lambda D/a (twice the others)
  • Polaroids: half rule I0/2I_0/2; Malus I=I0cos⁡2θI = I_0\cos^2\theta; three-polaroid I08sin⁡22θ\frac{I_0}{8}\sin^2 2\theta

State the condition, then substitute.

Solved Examples - Wave Model & Media

Example 1. Speed of 500 nm light in glass (n = 1.5)?

Solution: v=c/n=2×108v = c/n = 2\times10^8 m/s.

Example 2. Its wavelength in the glass?

Solution: λ/n=500/1.5=333\lambda/n = 500/1.5 = 333 nm; frequency unchanged at 6×10146\times10^{14} Hz.

Example 3. A plane wave refracts at i = 45 degrees, r = 30 degrees. Find v1/v2v_1/v_2.

Solution: sin⁡45/sin⁡30=0.707/0.5=1.414\sin45/\sin30 = 0.707/0.5 = 1.414.

Example 4. Frequency of 400 nm violet light?

Solution: ν=3×108/4×10−7=7.5×1014\nu = 3\times10^8/4\times10^{-7} = 7.5\times10^{14} Hz.

Example 5. Number of 600 nm (air) wavelengths in 1.2 mm of glass (n = 1.5)?

Solution: λg=400\lambda_g = 400 nm; 1.2×10−3/4×10−7=30001.2\times10^{-3}/4\times10^{-7} = 3000.

Example 6. A 2 ns plane-wave advance in glass (n = 1.5)?

Solution: vτ=(2×108)(2×10−9)=0.4v\tau = (2\times10^8)(2\times10^{-9}) = 0.4 m.

Solved Examples - Interference Intensities

Example 7. Two coherent equal sources, path difference 2λ2\lambda: intensity?

Solution: Integer multiple: I=4I0I = 4I_0 (bright).

Example 8. Path difference 2.5λ2.5\lambda?

Solution: Half-integer: I=0I = 0 (dark).

Example 9. Path difference λ/6\lambda/6?

Solution: ϕ=π/3\phi = \pi/3; I=4I0cos⁡2(π/6)=4I0×0.75=3I0I = 4I_0\cos^2(\pi/6) = 4I_0\times0.75 = 3I_0.

Example 10. Amplitude ratio 3:1: find Imax/IminI_{max}/I_{min}.

Solution: (a1+a2)2(a1−a2)2=(3+1)2(3−1)2=164=4\frac{(a_1+a_2)^2}{(a_1-a_2)^2} = \frac{(3+1)^2}{(3-1)^2} = \frac{16}{4} = 4.

Example 11. Two incoherent equal sources (I0I_0 each): resultant?

Solution: 2I02I_0 — intensities add, no fringes.

Solved Examples - Young's Double Slit

Example 12. d = 0.8 mm, D = 1.2 m, λ\lambda = 480 nm. Fringe width?

Solution: β=λD/d=(4.8×10−7)(1.2)/(8×10−4)=7.2×10−4\beta = \lambda D/d = (4.8\times10^{-7})(1.2)/(8\times10^{-4}) = 7.2\times10^{-4} m = 0.72 mm.

Example 13. β=0.6\beta = 0.6 mm with d = 1 mm, D = 1 m. Find λ\lambda.

Solution: λ=βd/D=(6×10−4)(10−3)/1=6×10−7\lambda = \beta d/D = (6\times10^{-4})(10^{-3})/1 = 6\times10^{-7} m = 600 nm.

Example 14. That setup dipped in water (n = 1.33): new β\beta?

Solution: 0.6/1.33=0.450.6/1.33 = 0.45 mm.

Example 15. Position of the 4th bright fringe (β=0.72\beta = 0.72 mm)?

Solution: 4β=2.884\beta = 2.88 mm.

Example 16. 5th of 480 nm coincides with which order of 600 nm?

Solution: 5×480=4×6005\times480 = 4\times600: the 4th order.

Example 17. Angular fringe width for d = 0.25 mm, λ\lambda = 500 nm?

Solution: θ=λ/d=5×10−7/2.5×10−4=2×10−3\theta = \lambda/d = 5\times10^{-7}/2.5\times10^{-4} = 2\times10^{-3} rad.

Example 18. Fringes in a 1 cm region with β=0.5\beta = 0.5 mm?

Solution: 10/0.5=2010/0.5 = 20 fringes.

Example 19. Doubling D and halving d: factor change in β\beta?

Solution: β∝D/d\beta \propto D/d: 2×2=4×2 \times 2 = 4\times wider.

Solved Examples - Single-Slit Diffraction

Example 20. a = 0.1 mm, λ\lambda = 500 nm: first minimum angle?

Solution: sin⁡θ=λ/a=5×10−3\sin\theta = \lambda/a = 5\times10^{-3} rad.

Example 21. Central maximum width on a screen 2 m away?

Solution: 2λD/a=2(5×10−7)(2)/10−4=0.022\lambda D/a = 2(5\times10^{-7})(2)/10^{-4} = 0.02 m = 2 cm.

Example 22. Halving a: new central width?

Solution: Doubles to 4 cm.

Example 23. a = 0.2 mm, λ\lambda = 600 nm: 3rd minimum angle?

Solution: sin⁡θ=3λ/a=9×10−3\sin\theta = 3\lambda/a = 9\times10^{-3} rad.

Example 24. Compare a single slit (a) and double slit (d) with a = d: the single-slit first minimum vs the double-slit first bright fringe both at:

Solution: the same angle λ/a\lambda/a — but one is dark, the other bright. Identical equation, opposite meaning.

Solved Examples - Polarisation

Example 25. Unpolarised 120 W/m2^2 through one polaroid?

Solution: 120/2=60120/2 = 60 W/m2^2, polarised.

Example 26. Polarised 60 W/m2^2 at 45 degrees?

Solution: 60cos⁡245=60×0.5=3060\cos^2 45 = 60\times0.5 = 30 W/m2^2.

Example 27. Polarised 60 W/m2^2 at 60 degrees?

Solution: 60cos⁡260=60×0.25=1560\cos^2 60 = 60\times0.25 = 15 W/m2^2.

Example 28. Unpolarised 100 W/m2^2 through two polaroids at 30 degrees?

Solution: 50cos⁡230=50×0.75=37.550\cos^2 30 = 50\times0.75 = 37.5 W/m2^2.

Example 29. Three polaroids, unpolarised I0I_0, middle at 45 degrees, last crossed with first: output?

Solution: I08sin⁡290=I0/8\frac{I_0}{8}\sin^2 90 = I_0/8 — the maximum.

Example 30. Through two polaroids, intensity falls to one-quarter of the polarised maximum. Angle?

Solution: cos⁡2θ=1/4\cos^2\theta = 1/4: θ=60\theta = 60 degrees.

Example 31. Crossed polaroids: transmitted fraction of unpolarised light?

Solution: Zero.

Example 32. Two coherent sources I0I_0 and 4I04I_0: Imax/IminI_{max}/I_{min}?

Solution: a1:a2=1:2a_1:a_2 = 1:2; (3)2/(1)2=9(3)^2/(1)^2 = 9.

Example 33. Light 0.6 micrometre vs a 0.3 mm slit: ratio?

Solution: λ/a=6×10−7/3×10−4=2×10−3\lambda/a = 6\times10^{-7}/3\times10^{-4} = 2\times10^{-3} — small, so diffraction is slight.

Example 34. Two polaroids give maximum transmission when their axes are:

Solution: Parallel (θ=0\theta = 0): I=I0cos⁡20=I0I = I_0\cos^2 0 = I_0.