Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Wave Optics question

2024 · Shift 2 · Q49
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Advanced
  3. /Physics
  4. /Wave Optics
  5. /2024 · Shift 2 · Q49

Wave Optics question

2024 · Shift 2 · Q49

JEE AdvancedPhysicsWave OpticsNumerical+3 / −1
In a Young's double slit experiment, each of the two slits A and B, as shown in the figure, are oscillating about their fixed center and with a mean separation of 0.8 mm0.8 \mathrm{~mm}0.8 mm. The distance between the slits at time ttt is given by d=(0.8+0.04sin⁡ωt)mmd=(0.8+0.04 \sin \omega t) \mathrm{mm}d=(0.8+0.04sinωt)mm, where ω=0.08rads−1\omega=0.08 \mathrm{rad} \mathrm{s}^{-1}ω=0.08rads−1. The distance of the screen from the slits is 1 m1 \mathrm{~m}1 m and the wavelength of the light used to illuminate the slits is 600060006000Å . The interference pattern on the screen changes with time, while the central bright fringe (zeroth fringe) remains fixed at point OOO. JEE Advanced 2024 Paper 2 Online Physics - Wave Optics Question 4 English ComprehensionThe maximum speed in μm/s\mu \mathrm{m} / \mathrm{s}μm/s at which the 8th 8^{\text {th }}8th  bright fringe will move is ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 24

  1. Position of the nnnth bright fringe in YDSE

For slit separation ddd, screen distance DDD, and wavelength λ\lambdaλ, the position of the nnnth bright fringe is

yn=nλDd.y_n = \frac{n\lambda D}{d}.yn​=dnλD​.

Here, for the 8th8^{\text{th}}8th bright fringe,

y8=8λDd(t).y_8 = \frac{8\lambda D}{d(t)}.y8​=d(t)8λD​.

Since ddd changes with time, y8y_8y8​ also changes with time.


  1. Given time-dependent slit separation

d(t)=(0.8+0.04sin⁡ωt) mmd(t) = (0.8 + 0.04\sin \omega t)\,\text{mm}d(t)=(0.8+0.04sinωt)mm

with

ω=0.08 rad s−1.\omega = 0.08\,\text{rad s}^{-1}.ω=0.08rad s−1.

Convert to SI units:

d(t)=(0.8+0.04sin⁡ωt)×10−3 m.d(t) = \left(0.8 + 0.04\sin \omega t\right)\times 10^{-3}\,\text{m}.d(t)=(0.8+0.04sinωt)×10−3m.

Also,

λ=6000 A˚=6×10−7 m,D=1 m.\lambda = 6000\,\text{\AA} = 6\times 10^{-7}\,\text{m}, \qquad D=1\,\text{m}.λ=6000A˚=6×10−7m,D=1m.


  1. Differentiate to get fringe speed

y8=8λDdy_8 = \frac{8\lambda D}{d}y8​=d8λD​

So,

dy8dt=−8λDd2dddt.\frac{dy_8}{dt} = -\frac{8\lambda D}{d^2}\frac{dd}{dt}.dtdy8​​=−d28λD​dtdd​.

Hence the speed is

v=∣dy8dt∣=8λDd2∣dddt∣.v = \left|\frac{dy_8}{dt}\right| = \frac{8\lambda D}{d^2}\left|\frac{dd}{dt}\right|.v=​dtdy8​​​=d28λD​​dtdd​​.

Now,

d(t)=(0.8+0.04sin⁡ωt)×10−3d(t) = (0.8+0.04\sin\omega t)\times 10^{-3}d(t)=(0.8+0.04sinωt)×10−3

Therefore,

dddt=0.04ωcos⁡ωt×10−3 m/s.\frac{dd}{dt} = 0.04\omega \cos\omega t \times 10^{-3}\,\text{m/s}.dtdd​=0.04ωcosωt×10−3m/s.

Its maximum value is when ∣cos⁡ωt∣=1|\cos\omega t|=1∣cosωt∣=1:

∣dddt∣max⁡=0.04×0.08×10−3=3.2×10−6 m/s.\left|\frac{dd}{dt}\right|_{\max} = 0.04\times 0.08\times 10^{-3} = 3.2\times 10^{-6}\,\text{m/s}.​dtdd​​max​=0.04×0.08×10−3=3.2×10−6m/s.


  1. For maximum fringe speed

Since

v∝∣dd/dt∣d2,v \propto \frac{|dd/dt|}{d^2},v∝d2∣dd/dt∣​,

the maximum occurs when ∣dd/dt∣|dd/dt|∣dd/dt∣ is maximum and ddd is approximately at its mean value. Using the standard small-oscillation treatment here,

d≈0.8×10−3 m.d \approx 0.8\times 10^{-3}\,\text{m}.d≈0.8×10−3m.

Thus,

vmax⁡=8×6×10−7×1(0.8×10−3)2×3.2×10−6.v_{\max} = \frac{8\times 6\times 10^{-7}\times 1}{(0.8\times 10^{-3})^2}\times 3.2\times 10^{-6}.vmax​=(0.8×10−3)28×6×10−7×1​×3.2×10−6.

Now compute:

(0.8×10−3)2=0.64×10−6=6.4×10−7.(0.8\times 10^{-3})^2 = 0.64\times 10^{-6} = 6.4\times 10^{-7}.(0.8×10−3)2=0.64×10−6=6.4×10−7.

So,

8×6×10−76.4×10−7=486.4=7.5.\frac{8\times 6\times 10^{-7}}{6.4\times 10^{-7}} = \frac{48}{6.4} = 7.5.6.4×10−78×6×10−7​=6.448​=7.5.

Hence,

vmax⁡=7.5×3.2×10−6=24×10−6 m/s.v_{\max} = 7.5\times 3.2\times 10^{-6} = 24\times 10^{-6}\,\text{m/s}.vmax​=7.5×3.2×10−6=24×10−6m/s.

Convert to μ\muμm/s:

24×10−6 m/s=24 μm/s.24\times 10^{-6}\,\text{m/s} = 24\,\mu\text{m/s}.24×10−6m/s=24μm/s.


  1. Final answer

The maximum speed of the 8th8^{\text{th}}8th bright fringe is

24\boxed{24}24​

in μm/s\mu\text{m/s}μm/s.


  1. Comparison with stored answer

Stored correct answer = 242424.

This matches the derived answer.

PreviousNext

More from Wave Optics

  • A double slit setup is shown in the figure. One of the slits is in medium 2 of refractive index n2​. The other slit is at the interface of this medium with another medium 1 of refractive index n1​(eqn2​). The line… Includes diagram2022 · Multiple correct
  • A parallel beam of light strikes a piece of transparent glass having cross section as shown in the figure below. Correct shape of the emergent wavefront will be (figures are schematic and not drawn to scale) Includes diagram2020 · MCQ
  • In a Young's double slit experiment, the slit separation d is 0.3 mm and the screen distance D is 1 m. A parallel beam of light of wavelength 600 nm is incident on the slits at angle α as shown in figure. On the screen, the point O… Includes diagram2019 · Multiple correct
  • Two coherent monochromatic point sources S1​ and S2​ of wavelength λ=600nm are placed symmetrically on either side of the center of the circle as shown. The sources are separated by a distance d=1.8mm. This… Includes diagram2017 · Multiple correct
  • While conducting the Young's double slit experiment, a student replaced the two slits with a large opaque plate in the XY-plane containing two small holes that act as two coherent point sources (S1, S2) emitting light of wavelength 600 mm.… Includes diagram2016 · Multiple correct
  • A Young's double slit interference arrangement with slits S1 and S2 is immersed in water (refractive index = 4/3) as shown in the figure. The positions of maxima on the surface of water are given by x2 = p2m2 λ 2 − d2, where λ… Includes diagram2015 · Numerical
  • A light source, width emits two wavelengths λ 1 = 400 nm and λ 2 = 600 nm, is used in a Young's double-slit experiment. If recorded fringe widths for λ 1 and λ 2 are β 1 and β 2 and the number of…2014 · Multiple correct
  • In the Young's double-slit experiment using a monochromatic light of wavelength λ, the path difference (in terms of an integer n) corresponding to any point having half the peak intensity is2013 · MCQ