Case Study
Passage with linked questions
Case Set 1
Case AnalysisPassage
Priya is performing Young's double-slit experiment in her school laboratory. She uses a sodium lamp (wavelength 589 nm) as the light source. The two slits are separated by 0.3 mm and the screen is placed 1.5 m away. She observes a clear pattern of alternating bright and dark fringes on the screen. She then replaces the sodium lamp with a red laser (wavelength 650 nm) and notices that the fringe width increases. Her teacher explains that the fringe pattern is a result of the superposition of coherent waves from the two slits and that the positions of maxima and minima depend on the path difference between the waves arriving from the two slits at each point on the screen.
Question 1: Calculate the fringe width observed when the sodium lamp (wavelength 589 nm) is used in the experiment described above.
- Fringe width β = λD/d = (589×10⁻⁹ × 1.5) / (0.3×10⁻³) = 8.835×10⁻⁷ / 3×10⁻⁴ ≈ 2.945×10⁻³ m ≈ 2.95 mm.
- The fringe width is approximately 2.95 mm, meaning consecutive bright (or dark) fringes are separated by this distance on the screen.
Question 2: Priya replaces the sodium lamp with a red laser (650 nm). By what factor does the fringe width change, and what is the new fringe width?
- Fringe width is directly proportional to wavelength: β ∝ λ. The new fringe width β′ = (650/589) × 2.95 mm ≈ 1.104 × 2.95 mm ≈ 3.26 mm.
- The fringe width increases by a factor of 650/589 ≈ 1.10, i.e., by about 10%; the pattern becomes slightly more spread out with the red laser.
Question 3: If Priya covers one slit with a thin glass slab of refractive index 1.5 and thickness t, the central fringe shifts by exactly 5 fringe widths. Find the thickness t of the slab and state the direction of the shift.
- The extra optical path introduced by the glass slab over one slit is (n−1)t = 0.5t. For the central fringe to shift by N = 5 fringe widths, the extra path must equal Nλ: 0.5t = 5 × 589×10⁻⁹ m.
- Solving: t = 5 × 589×10⁻⁹ / 0.5 = 5.89×10⁻⁶ m = 5.89 μm.
- The central fringe shifts toward the slit covered by the glass slab, because the optical path from that slit is increased, so the point of zero path difference moves closer to that slit.
ByteNotes