JEE Advanced 2024 Paper 2 · Q15 · Wave Optics
[PARAGRAPH I] 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 mm. The distance between the slits at time $t$ is given by $d = (0.8 + 0.04\sin\omega t)$ mm, where $\omega = 0.08$ rad s$^{-1}$. The distance of the screen from the slits is 1 m and the wavelength of the light used to illuminate the slits is 6000 Å. The interference pattern on the screen changes with time, while the central bright fringe (zeroth fringe) remains fixed at point O.
The maximum speed in μm/s at which the $8^{\text{th}}$ bright fringe will move is _____.
Reveal answer + step-by-step solution
Correct answer:24
Solution
Position of 8th bright fringe: $y_8 = 8\lambda D/d(t)$, with $d(t) = (0.8 + 0.04\sin\omega t)\times 10^{-3}$ m.
Speed: $\dfrac{dy_8}{dt} = -\dfrac{8\lambda D}{d^2}\cdot \dfrac{dd}{dt} = -\dfrac{8\lambda D}{d^2}\cdot 0.04\times 10^{-3}\,\omega\cos\omega t$.
Magnitude maximum when $|\cos\omega t| = 1$ and $d^2$ is smallest (so use $d = 0.76\times 10^{-3}$ m, but for maximum speed we look at the largest magnitude):
Using $d^2$ at $d = 0.8\times 10^{-3}$ m for the leading-order estimate (the small variation in $d$ from 0.76 to 0.84 mm has minor effect on $1/d^2$ at peak velocity, but $\cos\omega t = 1$ requires $\sin\omega t = 0$ ⇒ $d = 0.8\times 10^{-3}$ m exactly):
$\left|\dfrac{dy_8}{dt}\right|_{\max} = \dfrac{8\lambda D}{d^2}\cdot 0.04\times 10^{-3}\cdot \omega$.
Numerator: $8\cdot 6\times 10^{-7}\cdot 1\cdot 0.04\times 10^{-3}\cdot 0.08 = 8\cdot 6\cdot 0.04\cdot 0.08 \times 10^{-10} = 0.1536\times 10^{-10} = 1.536\times 10^{-11}$.
Denominator: $(0.8\times 10^{-3})^2 = 0.64\times 10^{-6}$.
$v_{\max} = \dfrac{1.536\times 10^{-11}}{0.64\times 10^{-6}} = 2.4\times 10^{-5}$ m/s $= 24$ μm/s.
Units: m, m, m, rad/s in numerator → m/s; m$^2$ in denominator → cancels one…[truncated]
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