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EAPCET Physics · Intermediate second year

Wave Optics

Wave Optics treats light as a wave: Huygens’ principle, interference, diffraction and polarisation.

Subject
Physics
Taught in
Intermediate second year
Physics in the paper
40 of 160 questions

Key formulas and results

  • Fringe width in Young’s experiment: β=λDd\beta = \frac{\lambda D}{d}
  • Bright fringe at path difference nλn\lambda; dark fringe at (2n−1)λ2\frac{(2n - 1)\lambda}{2}
  • Single-slit first minimum: asin⁡θ=λa\sin\theta = \lambda
  • Brewster’s law: tan⁡ip=μ\tan i_{p} = \mu
  • Malus’s law: I=I0cos⁡2θI = I_{0}\cos^{2}\theta
  • Phase difference ϕ=2πλΔx\phi = \frac{2\pi}{\lambda}\Delta x; in Young’s experiment the path difference Δx=dsin⁡θ≈dyD\Delta x = d\sin\theta \approx \frac{dy}{D}
  • Position of the nnth bright fringe y=nλDdy = \frac{n\lambda D}{d}; of the nnth dark fringe y=(2n−1)λD2dy = \frac{(2n - 1)\lambda D}{2d}
  • Angular fringe width: θ=λ/d\theta = \lambda/d
  • Resultant intensity with two equal sources: I=4I0cos⁡2(ϕ/2)I = 4I_{0}\cos^{2}(\phi/2), so it ranges from 0 to 4I04I_{0}
  • Optical path: a distance xx in a medium of index μ\mu equals μx\mu x in air; a slab of thickness tt adds a path (μ−1)t(\mu - 1)t
  • Fringe shift when a slab of thickness tt covers one slit: Δy=(μ−1)tDd\Delta y = \frac{(\mu - 1)tD}{d}, towards the covered slit
  • Single slit: width of the central maximum =2λDa= \frac{2\lambda D}{a}; secondary maxima have intensities about I0/22I_{0}/22 and I0/62I_{0}/62
  • Angular limit of resolution of a circular aperture: sin⁡θ=1.22λa\sin\theta = \frac{1.22\lambda}{a}

What this chapter covers

  • Huygens’ principle
  • Interference and Young’s double slit
  • Diffraction at a single slit
  • Polarisation
  • Resolving power

How it is usually asked

  • Changes in fringe width
  • Intensity at a point
  • Polarisation
  • Diffraction minima

Understanding Wave Optics

Wave optics explains effects that rays cannot. By Huygens’ principle, every point on a wavefront acts as a source of secondary wavelets. When light from two coherent sources overlaps, as in Young’s double-slit experiment, the path difference decides whether the waves reinforce (bright fringe) or cancel (dark fringe). The fringe width λD/d grows with wavelength and screen distance and shrinks with slit separation.

Diffraction is the spreading of light around an obstacle or through a narrow slit. For a single slit, the central bright band is twice as wide as the others, with minima where a sin θ = nλ. Polarisation shows that light is a transverse wave: a polaroid passes only one direction of vibration, and Malus’s law gives the transmitted intensity. At the Brewster angle, reflected light is completely polarised.

EAPCET mostly asks how fringe width changes when λ, D, d or the medium changes, the intensity after one or more polarisers, the positions of diffraction minima, and Brewster’s angle. A quick check for interference: two coherent waves of intensity I₀ each give 4I₀ at a bright fringe and zero at a dark one, and a thin film over one slit shifts the whole pattern without changing the fringe width.

Worked example

EAPCET-style question

In a Young’s double-slit experiment, the slit separation is 0.5 mm, the screen is 1 m away and light of wavelength 500 nm is used. If the whole apparatus is immersed in water of refractive index 4/34/3, the new fringe width is

  1. (1)0.75 mm
  2. (2)1.33 mm
  3. (3)1 mm
  4. (4)0.5 mm

Solution

  1. In air, β=λDd=500×10−9×10.5×10−3=1×10−3 m=1 mm\beta = \frac{\lambda D}{d} = \frac{500 \times 10^{-9} \times 1}{0.5 \times 10^{-3}} = 1 \times 10^{-3}\ \text{m} = 1\ \text{mm}.
  2. In water, the wavelength becomes λ/μ\lambda/\mu, so the fringe width becomes β/μ\beta/\mu.
  3. β′=1 mm÷43=0.75 mm\beta' = 1\ \text{mm} \div \frac{4}{3} = 0.75\ \text{mm}.

Answer: 0.75 mm

Shortcuts and results to remember

  • In Young’s experiment with equal slits, intensity 3I03I_{0}, 2I02I_{0} and I0I_{0} (out of 4I04I_{0}) occur at β/6\beta/6, β/4\beta/4 and β/3\beta/3 from the central maximum.
  • Unpolarised light passing through one polariser always loses half its intensity, whatever the orientation.
  • At Brewster’s angle the reflected and refracted rays are perpendicular: ip+r=90∘i_{p} + r = 90^\circ.
  • Coherent sources must have the same frequency and a constant phase difference.
  • Unit check for fringe problems: 1 A˚=10−7 mm1\ \text{Å} = 10^{-7}\ \text{mm} and 1 nm=10−6 mm1\ \text{nm} = 10^{-6}\ \text{mm}.

More practice questions

Practice question 1

In Young’s experiment (λ=6000 A˚\lambda = 6000\ \text{Å}), a glass plate of μ=1.5\mu = 1.5 is placed over one slit. The minimum thickness that reduces the intensity at the central position by 25% is

  1. (1)0.1 μm
  2. (2)0.2 μm
  3. (3)0.4 μm
  4. (4)0.6 μm

Solution

  1. Intensity becomes 3I03I_{0} out of 4I04I_{0}: 4I0cos⁡2(ϕ/2)=3I04I_{0}\cos^{2}(\phi/2) = 3I_{0}, so ϕ=π/3\phi = \pi/3
  2. Path difference Δx=ϕ2πλ=λ6\Delta x = \frac{\phi}{2\pi}\lambda = \frac{\lambda}{6}
  3. (μ−1)t=λ6(\mu - 1)t = \frac{\lambda}{6}, so t=6000 A˚6×0.5=2000 A˚t = \frac{6000\ \text{Å}}{6 \times 0.5} = 2000\ \text{Å}
  4. t=2×10−7 m=0.2 μmt = 2 \times 10^{-7}\ \text{m} = 0.2\ \mu\text{m}

Answer: 0.2 μm

Practice question 2

Unpolarised light of intensity I0I_{0} passes through two polarisers and emerges with intensity 3I0/83I_{0}/8. The angle between their axes is

  1. (1)30°
  2. (2)45°
  3. (3)60°
  4. (4)90°

Solution

  1. After the first polariser the intensity is I0/2I_{0}/2
  2. Malus’s law: I02cos⁡2θ=3I08\frac{I_{0}}{2}\cos^{2}\theta = \frac{3I_{0}}{8}
  3. cos⁡2θ=34\cos^{2}\theta = \frac{3}{4}, so cos⁡θ=32\cos\theta = \frac{\sqrt{3}}{2}
  4. θ=30∘\theta = 30^\circ

Answer: 30°

Common mistakes to avoid

  • Mixing nm and mm in β = λD/d — convert every length to metres before substituting.
  • Applying Malus’s law to unpolarised light — the first polaroid always passes half the intensity, I₀/2, whatever its orientation.
  • Swapping the conditions for single-slit and double-slit patterns — in a single slit, a sin θ = nλ gives minima, not maxima.
  • Thinking immersing the setup in a liquid increases the fringe width — the wavelength shrinks by μ, so the fringe width decreases.

How to prepare it

Immersing Young’s setup in a medium of index μ reduces the fringe width by a factor of μ.

Once the formulas feel familiar, test them the way the exam does: practise Wave Optics questions chapter-wise, then attempt a full previous year paper to see whether it holds up under a 3-hour timer.

Related Physics chapters

See every chapter in the EAPCET Physics syllabus.

Practise Wave Optics chapter-wise

Chapter-wise quizzes show exactly which topics in this chapter you still get wrong — before a full paper does.

Frequently asked questions

Is Wave Optics part of the EAPCET Physics syllabus?
Yes. Wave Optics is taught in Intermediate second year Physics, and the EAPCET Engineering syllabus is based on the Intermediate first- and second-year syllabus. Check the official notification each year for any changes.
How should I prepare Wave Optics for EAPCET?
Immersing Young’s setup in a medium of index μ reduces the fringe width by a factor of μ.
How many Physics questions are there in EAPCET?
40 of the 160 questions in the Engineering paper are Physics, spread across 30 chapters.
What mistakes do students make in Wave Optics?
Mixing nm and mm in β = λD/d — convert every length to metres before substituting. Applying Malus’s law to unpolarised light — the first polaroid always passes half the intensity, I₀/2, whatever its orientation. Swapping the conditions for single-slit and double-slit patterns — in a single slit, a sin θ = nλ gives minima, not maxima. Thinking immersing the setup in a liquid increases the fringe width — the wavelength shrinks by μ, so the fringe width decreases.